1 Review

A few useful identities involving summation notation

$\sum_{i=1}^n i = 1+ 2 + 3+ \dots + n = \underbrace{(n+1)+(n+1) + \dots + (n+1)}_{n/2 \text{ times}} = \frac{n(n+1)}{2}$
$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$
$\sum_{i=1}^ni^3 = \frac{n^2(n+1)^2}{4}$

We call $i$ the index of summation.
Let $f(x) = 2+3x^2$ , then $\sum_{i=2}^4 f(i) = \sum_{i=2}^4 (2+3i^2) = 2+ 3(2)^2 + 2+ 3(3)^2 + 2+ 3(4)^2 = 93$
Geometric series: $\sum_{i=0}^{n-1} ar^i = a + ar + ar^2 + \dots + ar^{n-1} = \frac{a(1-r^n)}{1-r}$
Reindexing: $\sum_{i=1}^5 \frac{(-1)^{i+1}}{i^2} = \sum_{i=0}^4 \frac{(-1)^{i+2}}{(i+1)^2}$
Theorem: $\sum_{i = 1}^n \left(cf(x_i) + dg(x_i)\right) = c\sum_{i=1}^n f(x_i) + d\sum_{i=1}^n g(x_i)$ for all constants $c,d$ .

1.1 Estimating the area of regions bounded by a curve $y= f(x)$ and the $x$ -axis

Let $f(x)= x^2$ on the interval $[1,10]$ . For the area under the curve, we subdivide the interval into $n$ small intervals of size $\Delta x$ and make rectangles of heights $f(x_i)$ . Then an approximate area under the curve is given by the sum of the area of these approximating rectangles.

That is, $\Delta x= (10-1)/n = 9/n$ and the sample points are $x_i = 1 + i\Delta x = 1 + 9i/n$ , so $\begin{align*} \text{Approximate area }&= \sum_{i=1}^n A_i = \sum_{i=1}^n f(x_i)\cdot\Delta x \\ &= \sum_{i=1}^n (1+9i/n)^2 \cdot 9/n\\ &=\frac{9}{n}\sum_{i=1}^n \left( 1+ \frac{18i}{n} + \frac{81i^2}{n^2}\right)\\ &= \frac{9}{n}\left(n + \frac{18}{n}\sum_{i=1}^n i + \frac{81}{n^2}\sum_{i=1}^ni^2\right)\\ &=\frac{9}{n}\left(n + \frac{18}{n}\frac{n(n+1)}{2} + \frac{81}{n^2}\frac{n(n+1)(2n+1)}{6}\right)\\ &=\frac{9}{n}\left(n + 9(n+1) + \frac{27(n+1)(2n+1)}{2n}\right)\\ &= 9 + 81 + \frac{81}{n} + \frac{243(2n^2 + 3n+ 1)}{2n^2}\\ &= 90 + \frac{81}{n} + 243 + \frac{243(3)}{2n} + \frac{243}{2n^2} \end{align*}$ The actual area is given by allowing the number of rectangles to become infinite (hence, arbitrarily small), thus letting $n$ tend to infinity. That is, the actual area is found as $\lim_{n\to \infty} \left(90 + \frac{81}{n} + 243 + \frac{243(3)}{2n} + \frac{243}{2n^2}\right) = 90+243 = 333.$ Note: the choice of sample points were the right hand sides of the rectangles, which gave us an over-estimate. One could also have chosen the left-hand sides of the rectangles and obtained an under-estimate and both would have the same limit.

2 The Definite Integral

Let $f(x)$ be a continuous function on $[a,b]$ and let $\Delta x = \frac{b-a}{n}$ for a known integer $n>0$ . Let $x_i^*$ be a sample point in each $\Delta x$ interval , then the expression $\sum_{i=1}^n f(x_i^*)\Delta x$ is called a Riemann sumover $[a,b]$ . The number obtained by $\lim_{n\to \infty} \sum_{i=1}^n f(x_i^*)\Delta x$ is called a definite integral of $f(x)$ over $[a, b]$ . The definite integral is then denoted by $\begin{equation} \label{definiteintegral}\boxed{\int_a^bf(x)dx = \lim_{n\to \infty} \sum_{i=1}^n f(x_i^*)\Delta x} \end{equation}$ (1) where

$f(x)$ is the integrand
$\int$ is the integral sign
$a,b$ are the limits of integration

and this process is called integration. If $f(x)$ is a function on $[a,b]$ such that $\int_a^bf(x)dx$ exists, we say that $f(x)$ is integrable on $[a, b]$ .

Theorem 1. If $f(x)$ is continuous on $[a, b]$ , then it is integrable on $[a, b]$ (i.e. $\int_a^b f(x)dx$ exists).

Definition 1. If $f(x)$ is continuous and non-negative on $[a, b]$ , then we define the area of the region bounded by $f(x)$ and the $x$ -axis over $[a, b]$ as being the positive number obtained by $\int_a^bf(x)dx$ . In the case that $f(x)$ is negative, we take the absolute value and if $f(x)\geq 0$ on $[a,c]$ then $f(x)\leq 0$ on $[c,b]$ where $a<c<b$ , then the area is $\int_a^cf(x)dx + \left(-\int_c^b f(x)dx\right)$ .

Example 1. Using Riemann sums to find a formula for $\int_a^b x dx$ (where $0<a<b$ ).

$\begin{align*} \int_a^b xdx &= \lim_{n\to \infty}\sum_{i=1}^n \left(a + \frac{(b-a)i}{n}\right)\cdot \frac{b-a}{n}\\ &= \lim_{n\to \infty}\sum_{i=1}^n \left( a\Delta x + i \Delta x^2 \right)\\ &= \lim_{n\to \infty}\left(a\Delta x \sum_{i=1}^n 1 + \Delta x^2 \sum_{i=1}^n i\right)\\ &= \lim_{n\to \infty}\left(a\Delta x n + \Delta x^2 \frac{n(n+1)}{2}\right)\\ &= \lim_{n\to \infty}\left(a(b-a) + \frac{(b-a)^2}{n^2}\cdot\frac{n^2 + n}{2}\right)\\ &= a(b-a) + \lim_{n\to \infty}\left( \frac{(b-a)^2}{2}+ \frac{(b-a)^2}{2n}\right)\\ &= a(b-a) + \frac{(b-a)^2}{2}\\ &= \frac{b^2 - a^2}{2} \end{align*}$

2.1 Properties of the definite integral

We provide the following list of properties without proof. Note that, geometrically, these results should be intuitive.

$\int_a^bf(x)dx = - \int_b^af(x)dx$
$\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx$
$\int_a^b (f(x)\pm g(x)) dx = \int_a^b f(x)dx \pm \int_a^b g(x) dx$
$\int_a^b \lambda f(x)dx = \lambda \int_a^b f(x)dx$
if $f(x)\geq 0$ on $[a,b]$ , then $\int_a^b f(x)dx\geq 0$
if $f(x)$ is odd (i.e. $f(-x) = - f(x)$ ), then $\int_{-a}^a f(x)dx = 0$
if $f(x)$ is even (i.e. $f(-x) = f(x)$ ), then $\int_{-a}^a f(x)dx = 2\int_0^af(x)dx$
If $f(x)\geq g(x)$ on $[a,b]$ then $\int_a^bf(x)dx \geq \int_a^b g(x) dx,$
If $m\leq f(x) \leq M$ for all $x$ on $[a,b]$ , then $m(b-a)\leq \int_a^bf(x)dx\leq M(b-a)$
$\int_a^af(x) dx = 0$

Example 2. Show that $\int_2^3\frac{1}{x^2 + 1} dx$ is in the interval $[1/10, 1/5]$ .

Note, for $x\in [2,3]$ that $x^2 +1 \in [5,10]$ or equivalently that $5\leq x^2 + 1\leq 10$ for all $x\in [2, 3]$ . Hence, also, $m:= 1/10\leq \frac{1}{x^2+1} \leq 1/5=:M$ and our properties guarantee, that $\frac{1}{10} = \frac{3-2}{10} \leq \int_2^3 \frac{1}{x^2+1}dx \leq \frac{1}{5}.$

Example 3. Evaluate; $\int_{-1}^3 2|x|dx = 2\left[\int_{-1}^0 -x dx + \int_0^3 xdx\right] = 2(\tfrac{1}{2} +\tfrac{9}{2})$

2.2 Average Value of a Function

Suppose $f(x)$ is integrable on $[a,b]$ with $a<b$ . The average value of $f$ on $[a,b]$ is defined as $\begin{equation} \label{averagevalue}\boxed{A(f)_{[a,b]} = \frac{1}{b-a}\int_a^b f(x)dx} \end{equation}$ (2)

Example 4. Find the average value of $f(x)=100-3x^2$ on $[1, 5]$ .

$A(f)_{[1,5]} = \frac{1}{5-1} \int_1^5 (100-3x^2)dx = \frac{1}{4}(276) = 69$

Theorem 2 (Average-value Theorem). Suppose $f$ is continuous on $[a,b]$ , with $a<b$ . Then there exists an $x^*\in [a,b]$ such that $f(x^*) = A(f)_{[a,b]}$ . That is, a continuous function actually achieves its averages value somewhere within the interval $[a,b]$ .

Proof. Since $f(x)$ is continuous there exist a minimum and maximum at some $c,d\in [a,b]$ respectively . Then, certainly $f(c)\leq f(x)\leq f(d)$ for all $x\in [a,b]$ . Now, from property 9, we have $f(c)(b-a)\leq \int_a^bf(x)dx\leq f(d)(b-a)$ or equivalently (dividing by $(b-a)$ ) $f(c)\leq \frac{1}{b-a}\int_a^bf(x)dx\leq f(d).$ Now, by the intermediate value theorem, there exists $x^*\in[c,d]$ such that $f(x^*) = \frac{1}{b-a}\int_a^bf(x)dx$ . ◻

2.3 Fundamental Theorem of Calculus - Part I

Theorem 3 (FTC I). Suppose $f(x)$ is continuous on $[a, b]$ . Let $F(x)$ be a function on $[a, b]$ defined as $F(x) = \int_a^x f(t) dt$ . Then $F'(x) = f(x)$ .

Proof. Let $x\in[a,b]$ and proceed computing the derivative by first principles. $\begin{align*} F'(x) &= \lim_{h\to 0} \frac{F(x+h) - F(x)}{h}\\ &= \lim_{h\to 0} \frac{1}{h}\left[ \int_a^{x+h}f(t) dt - \int_a^x f(t) dt\right]\\ &= \lim_{h\to 0} \frac{1}{h}\left[ \int_a^{x}f(t) dt + \int_x^{x+h}f(t)dt - \int_a^x f(t) dt\right]\\ &= \lim_{h\to 0} \frac{1}{h} \int_x^{x+h}f(t) dt \end{align*}$ By the average value theorem there is an $x^*\in [x,x+h]$ so that $f(x^*)= \frac{1}{h} \int_x^{x+h}f(t) dt$ so then $\begin{align*} F'(x) = \lim_{h\to 0} \frac{1}{h} \int_x^{x+h}f(t) dt = \lim_{h\to 0} f(x^*) = \lim_{x^*\to x} f(x^*) = f\left(\lim_{x^*\to x} \right) = f(x) \end{align*}$ as required. ◻

Example 5.

$\frac{d}{dt}\int_0^t\sqrt{x^2 + 1} dt = \sqrt{t^2+ 1}$
$\frac{d}{dt}\int_t^{-3} \sin^2(x)dx = - \frac{d}{dt}\int_{-3}^t \sin^2(x)dx = -\sin^2(t)$
$\begin{align*} \frac{d}{dt}\int_0^{t^2} \cos(x^2)dx &= \frac{d}{dt}\int_0^{u(t)}\cos(x^2) dx \\ &= \frac{d}{du}\int_0^u\cos(x^2) dx \cdot\frac{du}{dt}\\ &= \cos(u^2) \cdot 2t\\ & =2t \cos(t^4) \end{align*}$
$\frac{d}{dt}\int_{-2t}^t \frac{1}{1+ x^2} dx = \frac{d}{dt}\left(-\int_0^{-2t}\frac{1}{1+x^2}dx + \int_0^t \frac{1}{1+x^2}dx\right) = - \frac{-2}{1+4t^2} + \frac{1}{1+t^2}$

Lemma 1. If $f$ and $g$ are two functions whose derivative is $F$ , then $f(x) = g(x) + c$

Proof. Given that $f'(x) = F(x) = g'(x)$ , then $\frac{d}{dx}(f(x) - g(x)) = F(x) - F(x) = 0\Rightarrow f(x) - g(x) = c$ since the derivative of a constant is 0. ◻

2.4 Fundamental Theorem of Calculus II

Definition 2. A function $F$ is called an antiderivative of $f$ on an interval $I$ if $F'(x) = f(x)$ for all $x\in I$ .

Theorem 4 (FTC II). Suppose $f(x)$ is continuous on $[a, b]$ and let $F(x)$ be any anti-derivative of $f(x)$ (i.e. $F'(x) = f(x)$ ). Then $\begin{equation} \label{FTC2}\boxed{ \int_a^bf(x)dx = F(b) - F(a)} \end{equation}$ (3)

Proof. Given that $f(x)$ is continuous on $[a,b]$ and $F'(x) = f(x)$ , let $G(x)=\int_a^xf(t) dt$ so that $G'(x) = f(x)$ too. By the previous lemma, then, there is some constant $c$ such that $G(x) = F(x) + c$ . Now, $\begin{align*} \int_a^bf(t)dt = G(b) = G(b)-0 = G(b) - G(a) = (F(b) + c) - (F(a) + c) = F(b) - F(a) \end{align*}$ ◻

This theorem says that knowing any anti-derivative $f(x)$ allows us to compute any definite integral.

Example 6. Now we are able to compute a few more definite integrals like $\int_2^3 e^xdx = e^x|_2^3 = e^3 - e^2$ and $\int_{-7}^{500} \sin xdx = - \cos(500) - (-\cos(-7))$

Notation: If $F(x)$ is an antiderivative of $f(x)$ denote by $\int f(x)dx$ the family of antiderivatives $F(x)+C$ and call this the indefinite integral of $f(x)$ .

2.5 A few antiderivatives

$\int kdx = kx+C$
$\int x^ndx = \frac{x^{n+1}}{n+1} + C$
$\int e^x dx = e^x + C$
$\int \frac{1}{x}dx = \ln|x| + C$
$\int \sin x dx = -\cos x + C$
$\int\cos xdx = \sin x + C$
$\int\sec^2 dx = \tan x + C$
$\int\sec x \tan x dx = \sec x + C$

Example 7. Find the reigon bounded by $\cos x$ and the $x$ -axis over the interval $[0, \pi]$ .

Note: $\int_0^\pi \cos xdx$ won’t work since the function is negative on the second half of this interval. Must use, $A = A_1 + A_2 = \int_0^{\pi/2} \cos xdx + \left(-\int_{\pi/2}^\pi\cos x dx \right) = \sin x|_0^{\pi/2} - \sin x|_{\pi/2}^\pi = (1-0) - (0-1) = 2$

Example 8. $\int_{-1}^1\frac{1}{x^2}$ does not exist since $\frac{1}{x^2}$ is not continuous at $0$ and (3) does not apply.

Definition 3. Let $y = f(x)$ be continuous, we define the differential of $y$ , denoted by $dy$ , as being $dy = f'(x) \Delta x$ where $\Delta x = x_1 - x$ and $x_1 >x$ .

Observation: if $y=x$ , then $dy = dx = \frac{dx}{dx}\cdot \Delta x = \Delta x$ so $\Delta x = dx$ and so, if $y = f(x)$ then $dy = f'(x) dx$ .

3 Techniques of integration

3.1 Integration by Substitution

Theorem 5 (The substitution rule). Let $u = g(x)$ be differentiable and having as its image, the interval $I$ . Suppose $f(x)$ is continuous on the interval $I$ . Then, $\begin{equation} \label{substitution}\boxed{ \int f(g(x)) \cdot g'(x) dx = \int f(u) du} \end{equation}$ (4)

Proof. Given that $u = g(x)$ , we are required to show that the general antiderivative of $f(g(x)) \cdot g'(x)$ is the same as the general antiderivative (AD) of $f(u)$ .

Suppose $F(u)$ is an AD of $f(u)$ , then $\int f(u) du = F(u) + C = F(g(x)) + C$ .

Now we claim that $\int f(g(x)) g'(x) = F(g(x) ) + C$ . Indeed, it remains to show that $\frac{d}{dx} (F(g(x)) + C = f(g(x))g'(x)$ which is true by the chain rule so the theorem holds. ◻

Remark 1. What the substitution rule is saying is that $\int \underbrace{f(g(x))}_{f(u)}\underbrace{g'(x)dx}_{du}$ and thats it!

Example 9. letting $u = x^3+9$ so that $du = 3x^2 dx$ which is equivalent to $x^2dx = \frac{1}{3}du$ , we find $\begin{align*} \int_0^3 x^2\sqrt{x^3+9}dx &= \int_{x=0}^{x=3}\frac{1}{3}\sqrt{u}du\\ & = \frac{1}{3}\left(\frac{2}{3}u^{2/3}\right)|_{x=0}^{x=3}\\ &=\frac{2}{9}(x^3+9)^{2/3}|_0^3 = 42 \end{align*}$

Example 10. Here will will use $u = \sin x$ . $\int\sin^3 x\cos xdx = \int u^3 = \frac{u^4}{4} + C = \frac{\sin^4 x}{4} + C$

3.2 Integration by Parts

Consider two functions $u(x), v(x)$ . The product rule says that $\frac{d}{dx}(u(x)\cdot v(x)) = u'(x) v(x) + u(x) v'(x).$ Integrating both sides of this equation now says that $u(x)v(x) = \int \left[u(x)v(x)\right]' = \int u'(x)v(x)dx + \int u(x) v'(x)dx = \int v(x) du + \int u(x) dv.$ Rearranging, that is $\int u(x) dv = u(x) v(x) - \int v(x) du$ or in more compact notation $\begin{equation} \label{IBP}\boxed{\int u dv = uv -\int vdu} \end{equation}$ (5)

Recall that $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}\Rightarrow \int\frac{dx}{\sqrt{1-x^2}} = \arcsin x + C$ $\frac{d}{dx}\arccos x = \frac{-1}{\sqrt{1-x^2}}\Rightarrow- \int\frac{dx}{\sqrt{1-x^2}} = \arccos x + C$ $\frac{d}{dx}\arctan x = \frac{1}{1+x^2}\Rightarrow \int\frac{dx}{1+x^2} = \arctan x + C$

Example 11. Now for a long list of examples with solutions!

$\int\arcsin xdx = x\arcsin x + \sqrt{1-x^2} + C.$ Solution: With $u = \arcsin x ,dv = dx$ , so that $du = \frac{1}{\sqrt{1-x^2}}, v = x$ (using integration by parts - IBP) we find $\begin{align*} \int \arcsin x dx = x\arcsin x - \int \frac{x}{\sqrt{1-x^2}} dx. \end{align*}$ Now, the substitution $y = 1-x^2$ so that $dy = -2xdx$ or $xdx = -\frac{dy}{2}$ reveals the above as $x\arcsin x- \frac{1}{2}\int y^{-1/2}dy = x\arcsin x + \frac{1}{2} 2\sqrt{y} + C = x\arcsin x + \sqrt{1-x^2} + C.$
$\int x^2 e^{-x}dx = -e^{-x}(x^2 + 2x + 2) + C$ Solution: This will require two iterations of IBP. First we will have $u = x^2$ and $dv = e^{-x}dx$ so that $du = 2xdx$ and $v = -e^{-x}$ . The first iteration becomes $\int x^2 e^{-x}dx = -x^2 e^{-x} - \int (-2x)e^{-x}dx = -x^2 e^{-x} +2 \int xe^{-x}dx.$ For the second term we use $u = x, dv = e^{-x}dx$ so that $du = dx, v = -e^{-x}$ and the above becomes $-x^2 e^{-x} +2\left[-xe^{-x} + \int e^{-x}dx \right] = -x^2 e^{-x}-2xe^{-x} - 2e^{-x}+C$ as required.
$\int e^{2x}\sin(3x)dx = \frac{13}{2}e^{2x}\left(2\sin(3x) - \frac{3}{2}\cos(3x)\right) + C$ Solution: This will also require two iterations of IBP, but has a slightly different solution.

Let $u= \sin(3x), dv = e^{2x}dx$ so $du = 3\cos(3x)dx, v = \frac{1}{2}e^{2x}$ ,

$\int e^{2x}\sin(3x)dx = \tfrac{1}{2}\sin(3x)e^{2x} - \tfrac{3}{2}\int e^{2x}\cos(3x)dx$ Now for the second term, use $u = \cos(3x) , dv = e^{2x}dx$ so that $du = -3\sin(3x) dx, v = \frac{1}{2}e^{2x}$ and then the integral from the second term above becomes $\int e^{2x}\cos(3x)dx = \tfrac{1}{2}\cos(3x)e^{2x} +\tfrac{3}{2}\int e^{2x}\sin(3x)dx.$ Notice that the original integral $y = \int e^{2x}\sin(3x)dx$ has reappeared. However, this doesn’t mean we have gone in a circle since other terms have come out. Putting these together reads $\begin{align*} y &= \tfrac{1}{2}\sin(3x)e^{2x} - \tfrac{3}{2}\int e^{2x}\cos(3x)dx \\ &= \tfrac{1}{2}\sin(3x)e^{2x} - \tfrac{3}{2}\left[\tfrac{1}{2}\cos(3x)e^{2x} +\tfrac{3}{2}\int e^{2x}\sin(3x)dx\right] \\ &= \tfrac{e^{2x}}{2}\left(\sin(3x) - \tfrac{3}{2}\cos(3x)\right) - \tfrac{9}{4}y \end{align*}$ or in other words, $(1+\tfrac{9}{4}) y = \frac{e^{2x}}{2}\left(\sin(3x) - \tfrac{3}{2}\cos(3x)\right)\Rightarrow\boxed{ y = \tfrac{2}{13}e^{2x} (\sin(3x)- \tfrac{3}{2}\cos(3x))}$
$\int \sec^3xdx = \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln|\sec x + \tan x|+ C$ Solution: Here again we will find a similar solution to the previous problem. Let $I$ denote the value of interest, $\int \sec^3 x dx$ , (for later when we need to rearrange). For IBP, use $u = \sec x, dv = sec^2 x dx$ so that $du = \sec x\tan x dx, v = \tan x$ . Now $\begin{align*} I &= \sec x\tan x - \int\sec x\tan^2 xdx \\ &=\sec x\tan x - \int\sec x(\sec^2 x - 1) dx\\ & = \sec x\tan x - \int\sec^3 x xdx + \int\sec xdx = \sec x\tan x - I + \ln |\sec x\tan x| + C \end{align*}$ So, rearranging, $2I = \sec x\tan x + \ln |\sec x\tan x| + C\Rightarrow \boxed{I =\frac{1}{2}\left( \sec x\tan x + \ln |\sec x\tan x| \right)+ C}$

3.3 Trigonometric Integrals

Recall and remember the following trigonometric identities: $\sin(A\pm B ) = \sin A \cos B \pm \cos A \sin B$ $\cos (A\pm B) = \cos A \cos B \mp \sin A \sin B$ $\cos ^2 x = \frac{1}{2}\left(1+ \cos 2x\right), \quad\sin^2 x = \frac{1}{2} \left(1 - \cos 2x\right),\text{ and } \sin^2 x + \cos^2 x = 1$

Now, we will $\boxed{\text{Evaluate: } \int\sin^m x\cos ^n xdx}$ by considering three different cases.

Solution:

when $n$ is odd;

Write $n = 2k+1$ , then by the substitution $u = \sin x$ $\begin{align*} \int \sin^m x\cos^{2k+1}x dx & = \int \sin^m x (\cos^2 x)^k \cos xdx \\ &= \int \sin^m x (1- \sin^2 x)^k \cos xdx\\ &= \int u^m(1-u^2)^k du \end{align*}$ and this is easy to evaluate (begin a polynomial).
when $m$ is odd;

Write $m = 2k+1$ and the substitution $u = \cos x$ $\begin{align*} \int (\sin^2 x)^k \cos^n x \sin xdx& =- \int (1-u^2)^ku^n du \end{align*}$
when both $m$ and $n$ are even;

The identity $\cos^2 x= \tfrac{1}{2}( 1+ \cos(2x)) \text{ or } \sin^2 x = \tfrac{1}{2}( 1- \cos(2x))$ followed by $\sin x \sin y = \tfrac{1}{2} (\sin(x+ y) + \sin(x-y)) \text{ or } \cos x\cos y = \tfrac{1}{2}(\cos (x+y) + \cos(x - y))$ will bring one of the even powers down to either case (i) or case (ii).

and similarly,

$\boxed{\text{Evaluate: }\int\tan^m x \sec^n x dx}$

If $n$ is even, say $n=2k$ , then the substitution $u = \tan x$ and $du = \sec^2 xdx$ gives $\begin{align*} \int \tan^mx (\sec^2x)^kdx&= \int\tan^m \sec^{2k-2}x \sec^2 xdx \\ &= \int\tan^m (\sec^2x)^{k-1} \sec^2 xdx \\ &= \int\tan^m (1+\tan^2 x)^{k-1} \sec^2 xdx \\ &= \int u^m(1+u^2)^{k-1}du \end{align*}$
If $m$ is odd, then we will substitute $u = \sec x$ with $du = \sec x \tan xdx$ like so $\begin{align*} \int \tan^{2k+1}x \sec ^n xdx&= \int (\tan^2x)^k \sec^{n-1}x \sec x\tan x dx\\ &= \int (\sec^2 x - 1)^k \sec^{n-1}x \sec x\tan x dx\\ &= \int(u^2 -1 )^k u^{n-1}du \end{align*}$

In any other case, we won’t find a closed expression for the integral $\int \tan^mx \sec^n x dx$ .

Example 12. Now for a list of examples with solutions!

$\begin{align*} \int\sin^3x\cos^2xdx &=\int (1- \cos^2x)\cos^2x \sin xdx, \qquad u = \cos x, du = - \sin xdx\\ &= - \int (1-u^2)u^2du\\ &=- \int(u^2 - u^4)du\\ &= \frac{u^5}{5} - \frac{u^3}{3} + C\\ &= \frac{\cos^5x}{5} - \frac{\cos^3 x}{3} + C \end{align*}$
$\begin{align*} \int \cos^5 xdx &= \int(1-\sin^2 x)^2 \cos x dx, \qquad u = \sin x, du = \cos xdx\\ &= \int(1-u^2)^2du\\ &=\int (1 - 2u^2 + u^4)du\\ &- u - \frac{2u^3}{3} + \frac{u^5}{5} + C\\ &= \sin x - \frac{2\sin^3 x}{3} + \frac{\sin^5 x}{5} + C \end{align*}$
$\begin{align*} \int \sin^2 x \cos^2 x dx & = \int(\sin x \cos x )^2 dx \\ &= \int \left(\tfrac{1}{2}\sin(2x)\right)^2dx\\ &= \tfrac{1}{4}\int \sin^2 (2x) dx\\ &= \tfrac{1}{4}\int \tfrac{1}{2}(1- \cos (4x)) dx\\ &= \tfrac{1}{8}\int (1-\cos (4x)) dx\\ &= \tfrac{1}{8}x + \tfrac{1}{8}\frac{\sin(4x)} {4} + C \end{align*}$
$\begin{align*} \int\tan^3 x\sec xdx & =\int(\sec^2 x - 1) \sec^2x \sec x\tan x dx, \qquad u = \sec x, du = \sec x\tan x dx\\ &= \int(u^2- 1)u^2du\\ &= \frac{u^5 }{5}- \frac{u^3}{3} + C \end{align*}$
$\begin{align*} \int \tan^4x\sec^4xdx &= \int\tan^4 x(\tan^2x + 1) \sec^2xdx, \qquad u = \tan x, du = \sec^2 x\\ &= \int u^4(u^2 + 1)du\\ &= \frac{u^7}{7} + \frac{u^5}{5} + C \end{align*}$

3.4 Trigonometric Substitution

This method will apply to integrands containing $a^2 + x^2, a^2- x^2$ or $x^2 - a^2$ .

So, if

Integrand has	substitute	apply identity
$a^2 - x^2$	$x = a\sin x, -\pi/2 \leq x \leq \pi/2$	$1- \sin^2 \theta = \cos ^2 x\theta$
		or $\cos^2 \theta = 1/2(1 + \cos 2\theta)$
$x^2- a^2$	$x = a\sec x, -\pi/2 \leq x \leq \pi/2$	$\sec^2 \theta -1 = \tan^2 x\theta$
	or $\pi \leq x \leq 3\pi/2$
$a^2 + x^2$	$x = a\tan x, -\pi/2 \leq x \leq \pi/2$	$1+ \tan^2 \theta = \sec^2 x\theta$

Example 13. These ideas will be best demonstrated through a list of examples

Letting $x = \sin \theta$ so that on the interval $-\pi/2 \leq \theta \leq \pi/2$ we have $\theta = \arcsin x$ and $dx = \cos \theta d\theta$ , then $\begin{align*} \int \frac{x^3}{\sqrt{1-x^2}} dx & = \int \frac{\sin^3\theta}{\sqrt{1-\sin^2 \theta}} \cos\theta d\theta\\ &= \int \frac{\sin^3 \theta } {|\cos \theta|} \cos \theta d\theta\\ & = \int \sin^3 \theta d\theta\\ &= \int (1-\cos^2\theta )\sin\theta d\theta, \qquad \text{let } u = \cos\theta, du = -\sin \theta d\theta\\ &= -\int(1-u^2)du \\ &= -u +\frac{u^3}{3}+C\\ &-\cos \theta + \frac{\cos^3\theta}{3}+C\\ &= -\sqrt{1-x^2} + \frac{\sqrt{1-x^2}^3}{3}+C \end{align*}$
The trigonometric substitution here is $x = \tfrac{3}{2}\tan \theta$ on $-\pi/2\leq \theta \leq \pi/2$ so that $\theta = \arctan (2x/3)$ and $dx = \tfrac{3}{2}\sec^2 \theta d\theta$ , then $\begin{align*} \int \frac{1}{(4x^2+9)^2}dx&= \int\frac{1}{(4(\tfrac{3}{2}\tan\theta)^2 + 9)^2} (\tfrac{3}{2}\sec^2 \theta)d\theta\\ &= \tfrac{1}{81}\cdot\tfrac{3}{2}\int \frac{\sec^2\theta}{\sec^4\theta} d\theta\\ &= \tfrac{1}{54}\int \frac{1}{\sec^2 \theta}d\theta\\ &= \tfrac{1}{54}\int \cos^2\theta d\theta\\ &= \tfrac{1}{54}\int\tfrac{1}{2}(1+ \cos(2\theta)) d\theta\\ &= \tfrac{1}{108}\int(1+ \cos(2\theta)) d\theta\\ &= \tfrac{1}{108}(\theta + \tfrac{1}{2}\sin(2\theta)) + C\\ &=\tfrac{1}{108}(\arctan x + \tfrac{1}{2}2\sin\theta\cos \theta) + C\\ &=\tfrac{1}{108}(\arctan x +\sin\theta\cos \theta) + C\\ &=\tfrac{1}{108}(\arctan x +\tfrac{2x}{4x^2 + 9}\cdot\tfrac{3}{4x^2 + 9}) + C \end{align*}$ where the last line is found from the fact that $\tan\theta = \frac{2x}{3}$ so the hypotenuse is necessarily $4x^2 + 9$ hence $\cos\theta = \tfrac{3}{4x^2 + 9}$ and $\sin\theta = \tfrac{2x}{4x^2 + 9}$ .
Here we use $x = \sin \theta$ on $\pi/2\leq \theta \leq \pi/2$ so that $dx = \cos \theta d\theta$ . Note that this is definite integral so we have to change the bounds of integration. That is, when $x = 1/2, \theta= \pi/6$ and when $x = \sqrt{3}/2, \theta= \pi/3$ . $\begin{align*} \int_{1/2}^{\sqrt{3}/2}\frac{x^3}{\sqrt{1-x^2}}dx & = \int_{\pi/6}^{\pi/3} \frac{\sin^3\theta\cos\theta}{\sqrt{1-\sin^2 \theta}}d\theta\\ &= \int_{\pi/6}^{\pi/3}\sin^3\theta d\theta, \text{ let } u = \cos \theta\\ &= \int_{\pi/6}^{\pi/3}(1-\cos^2\theta)\sin\theta d\theta, \text{ let } u = \cos \theta\\ &=- \int_{u = \sqrt{3}/2}^{u = 1/2}(1-u^2)du\\ &= \left[-u + \frac{u^3}{3}\right]_{\sqrt{3}/2}^{1/2}\\ &= \left(\tfrac{1}{24} - \tfrac{1}{2} \right) - \left(\tfrac{3\sqrt{3}}{24} - \tfrac{\sqrt{3}}{2}\right)\\ &= \tfrac{9 \sqrt{3} - 11}{24} \end{align*}$
Here we will use $x+1 = 2\sin \theta$ for $|\theta|\leq \pi/2$ so $\theta= \arcsin\left(\tfrac{x+1}{2}\right)$ and $dx = 2\cos \theta d\theta$ . Now $\begin{align*} \int\frac{x}{\sqrt{3 - 2x - x^2}}dx& = \int \frac{x}{4- (x+1)^2}dx\\ &= \int \frac{(2\sin \theta-1)2\cos\theta}{\sqrt{4(1-\sin^2\theta)}}d\theta\\ &= \int \frac{2\sin \theta- 1) \cos\theta}{\cos\theta}d\theta\\ &= 2\int\sin \theta d\theta- \int d\theta\\ &= - 2\cos\theta- \theta+ C\\ &=- 2\sqrt{4 - (x+1)^2}- \arcsin\left(\tfrac{x+1}{2}\right)+ C \end{align*}$

3.5 Integration with Partial Fractions

This method is used to solve a rational integrand (i.e. of the form $\frac{p(x) }{q(x)}$ for polynomials $p$ and $q$ ).

Case 1: $\deg p \geq \deg q$ , use long division.

Case 2: $\deg p < \deg q$ , factor the denominator and split up with partial fractions.

Example 14. As usual, this is best demonstrated by example.

$\begin{align*} \int\frac{x^3+ x^2+ x -1}{x^2 + 2x + 2}dx & = \int (x-1)dx + \int\frac{x+1}{x^2 + 2x + 2}dx, \qquad \text{(long division)}\\ &= \frac{x^2}{2} - x+ \frac{\ln|x^2 + 2x+2|}{2} + C\\ \end{align*}$
$\begin{align*} \int \frac{4x^2 - 3x - 4}{x^3+ x^2 -2x}dx &= \int \frac{4x^2 - 3x - 4}{x(x-1)(x-2)}dx \end{align*}$ and here we write $\begin{align*} \frac{4x^2 - 3x - 4}{x(x-1)(x-2)} &= \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+2}\\ &= \frac{A(x-1)(x+2) + Bx(x+2) + C x(x-1)}{x(x-1) (x+2)}\\ &= \frac{(A+B+C)x^2 + (A + 2B - C)x - 2A}{x(x-1) (x+2)}. \end{align*}$ This is the linear system $A+ B + C = 4,$ $A+ 2B - C = -3$ and $- 2A = -4$ which has solution $A = 2, B= -1, C = 3$ . So, $\begin{align*} \int \frac{4x^2 - 3x - 4}{x^3+ x^2 -2x}dx &= \int \frac{4x^2 - 3x - 4}{x(x-1)(x-2)}dx \\ &=\int\frac{2}{x}dx - \int\frac{1}{x-1}dx + \int\frac{3}{x+2}dx\\ &= 2\ln|x| - ln|x-1| + 3 ln|x+2|_C \end{align*}$
Using that $\frac{x^3 + 4x - 1}{x(x-1)^3} = \frac{A}{x} + \frac{B}{x-1}+ \frac{C}{(x-1)^2} + \frac{D}{(x-1)^3}$ has solution $A= 1, B = 0 , C = 3, D = -4$ , we find that $\begin{align*} \int\frac{x^3 + 4x - 1}{x(x-1)^3}dx &= \int\frac{1}{x}dx + 3\int \frac{1}{(x-1)^2}dx -4\int \frac{1}{(x-1)^3}dx\\ &= \ln|x| - 3(x-1)^{-1} +2(x-1)^{-2}+ C \end{align*}$ where we have used the substitution $u = x-1, du = dx$ to integrate the second two terms.
Here, the partial fractions look slightly different because of the irreducible quadratic factor in the denominator $\frac{5x^3 - 3x^2 + 2x -1}{x^2(x^2+1)} = \frac{A}{x} + \frac{ B}{x^2} + \frac{Cx+D}{x^2 +1}.$ Notice, the linear $Cx+D$ above the quadratic denominator. This is solved the same way as above and we find that $A = 2, B = -1, C = 3, D = -2$ , so $\begin{align*} \int\frac{5x^3 - 3x^2 + 2x -1}{x^2(x^2+1)}dx& = 2\int \frac{1}{x}dx - \int\frac{1}{x^2} dx + \int \frac{3x-2}{x^2 + 1} dx\\ & = 2 \ln |x| + \frac{1}{x} + 3\int \frac{x}{x^2 + 1} - 2\int\frac{1}{x^2 + 1}dx\\ & = 2 \ln |x| + \frac{1}{x} + \tfrac{3}{2}\ln|x^2+1| - 2 \arctan x + C \end{align*}$

3.6 Non-Elementary Functions

The elementary functions are polynomials, trigonometric, exponential, rational or any combination of these using $+, -, \times, \div, \exp$ or composition.

Consider, the function $F:\mathbb{N}\to \mathbb{N}$ defined by the rule that $F(n) = \text{the } n^{th} \text{ Fibonacci number}$ . Is $F$ an elementary function? The answer here is YES! Now, is every function elementary? NO.

For example, $f(t):= e^{t^2}$ is continuous on $\mathbb{R}$ , so integrable. Then define $F(x) := \int_0 ^x e^{t^2}dt$ which has derivative $F'(x) = f(x)$ (by FTC I). It can be shown that $F(x)$ is not elementary, meaning you can’t find a formula for $F(x)$ .

4 Applications of integration

4.1 Areas between curves

Consider a region bounded above by $y = f(x)$ and below by $y = g(x)$ .

$\begin{align*} A &= A_1- A_2\\ & = \int_a^bf(x)dx - \int_a^b g(x) dx\\ & = \int_a^b(f(x) - g(x))dx \end{align*}$
$\begin{align*} A &= A_1+A_2\\ & = \int_a^bf(x)dx + \left(-\int_a^b g(x) dx\right)\\ & = \int_a^b(f(x) - g(x))dx \end{align*}$
$\begin{align*} A &= A_1- A_2\\ & = -\int_a^bg(x) dx -\left(- \int_a^b f(x) dx\right)\\ & = \int_a^b(f(x) - g(x))dx \end{align*}$

Notice that in all three cases thus far, we have the area between $f$ and $g$ on $[a, b]$ when $f(x) \geq g(x)$ is given by the formula $\int_a^b (f(x)- g(x) ) dx.$

Now, suppose that $f(x)\geq g(x)$ on $[a, c]$ and then $g(x) \geq f(x)$ on $[c, b]$ . Then

$\begin{align*} A &= A_1+ A_2\\ & = \int_a^c(f(x)-g(x)) dx + \int_c^b( g(x) - f(x)) dx\\ & = \int_a^b(f(x) - g(x))dx \end{align*}$

Example 15. Find the area of the region bounded by...

$Y = x$ and $y = 6-x^2$ ;

Solution: Since there is no given interval on which to find the area, we must compute where these two curves meet. That is, solve $x = y = 6-x^2$ which is equivalent to $0= (x+3)(x-2)$ so that these curves meet at $x = -3$ and $x = 2$ . On this interval we find that $6-x^2 \geq x$ so the area between them is computed by $\begin{align*} \int_{-3}^2(6- x^2 - x) dx = \left[6x - \frac{x^3}{3} - \frac{x^2}{2}\right]|_{-3}^2 = \tfrac{125}{6} \end{align*}$
$y = \tfrac{x}{2}$ and $y^2 = 8 - x$ . These curves meet when $\frac{x^2}{4} = 8 - x$ which has solutions $x = -8$ and $x = 4$

Solution 1: Here we integrate along the $x$ -axis as follows; $A = \int_{-8}^4 \left(\frac{x}{2} - (-\sqrt{8 - x})\right)dx + 2\int_4^8 \sqrt{8-x}dx$

Solution 2: If we tilt our heads, this is actually easier done by integrating along the $y$ -axis. The intersection points are $(- 8, -4)$ and $(4, 2)$ and the area is given by $\int_{-4}^2 (8-y^2 - 2y)dy = (8y - \tfrac{y^3}{3} - y^2)|_{-4}^2$

4.2 Arc Length of a Parameterized Curve

Let $\mathcal{C}$ be the smooth curve obtained obtained by graphing $y = f(x)$ over the interval $[a, b)$ . For the arc length of $\mathcal{C}$ we apply the following simple recipe.

Subdivide the interval $(a, b)$ into $n$ equal subintervals of length $\Delta x = \frac{b-a}{x}$
Let $a = x_0, \dots, b = x_n$
Let $p_i$ be the point $(x_i, f(x_i))$ on $\mathcal{C}$ and $d(p_{i-1}, p_i)$ be the Euclidean distance between the points $p_{i-1}$ and $p_i$ . That is, $d(p_{i-1}, p_i) = \sqrt{(x_i - x_{i-1})^2 + (f(x_i)- f(x_{i-1}))^2} = (x_i - x_{i-1}) \sqrt{1+ \left(\frac{f(x_i) - f(x_{i-1})}{x_i - x_{i-1}}\right)^2} = \Delta x\sqrt{1 + f'(x_i^*)^2}$ where the last equality holds by the Mean Value Theorem.

Hence, now the approximate length of $\mathcal{C}$ is given by $\sum_{i=1}^n \sqrt{1 + f'(x_i^*)^2}\Delta x$ another Riemann sum! Finally, the arc length of $\mathcal{C}$ is given by $\begin{equation} \label{arclength} \boxed{\mathcal{L}ength(\mathcal{C}) = \lim_{n\to\infty}\sum_{i=1}^n \sqrt{1 + f'(x_i^*)^2}\Delta x = \int_a^b \sqrt{1 + f'(x)^2} dx } \end{equation}$ (6)

Example 16. Find the length of the curve $\mathcal{C}$ defined by $y = x^{3/2}$ on the interval $[0, 5]$ .

Solution: $\begin{align*} \mathcal{L}ength(\mathcal{C}) &= \int_0^5\sqrt{1+y\prime(x)^2}dx\\ &= \int_0^5\sqrt{1+\tfrac{9}{4}x}dx\\ &= \frac{1}{2}\int_0^5\sqrt{4+9x}dx = \frac{335}{27}\\ \end{align*}$

5 Calculating Volumes

5.1 Cross sectional method

The cross sectional method is often the most practical method used to compute volumes of solids of revolution (a solid obtained by revolving a region bounded by $y = f(x)$ over $[a,b]$ about the $x$ - or $y$ -axis).

Principle behind the cross section method

subdivide $[a,b]$ into $n$ equal sub intervals, each of length $\Delta x = \frac{b-a}{n}$
Let $A(x_i)$ denote the area of the cross section of the solid at $x_i$
Then $A(x_1), A(x_2), \dots$ is a sequence of areas of cross sections, each a function of $x_i$ .
If we then keep $A(x)$ constant over the interval $(x_i, x_{i+1})$ we obtain an approximation $V_n(x_i) = A(x_i)\Delta x$
We now have a sequence of volumes $V_n(x_1) , V_n(x_2), \dots$
The $n^{th}$ approximation of the volume of this solid is then $V_n = \sum_{i=1}^n V_n(x_i) = \sum_{i=1}^nA(x_i)\Delta x$ which is a Riemann sum.
So, theoretically, the actual volume of the solid is $V = \int_a^bA(x_i) dx = \lim_{n\to \infty} \sum_{i=1}^n A(x_i)\Delta x$

Example 17. Using the cross sectional method to calculate the volume of a square based pyramid, with base $b\times b$ and height h.

Solution: The side length at $x$ can be found using similar triangles as $s(x) = \frac{xb}{h}$ , so then the area at $x$ is then $A(x) = s(x)^2 = \frac{x^2 b^2}{h^2}$ . Now, the volume can be computed by $V = \int_0^h A(x) dx =\frac{b^2}{h^2} \int_0^hx^2 dx =\frac{b^2}{h^2}\left[ \frac{x^3}{3}\right]_0^h = \frac{b^2h}{3}$

Note that this method would also work using the $y$ -axis and integrating with respect to $y$ .

The above example was not a solid of revolution. However, there is a nice formula for the cross sectional area (and hence the volume) of a solid of revolution.

Notice, that when $y = f(x)$ is revolved about, say, the $x$ -axis, the cross sectional area is a circle, of radius $f(x)$ centred at the $x$ axis. So then $A(x) = \pi f(x)^2$ (the area of a circle of radius $f(x)$ ). Hence, $\begin{equation} \label{revolution} \boxed{V = \int_a^b \pi f(x)^2 dx} \end{equation}$ (7)

If this was about the $y$ -axis then $V = \int_c^d \pi g(y) dy$ .

Example 18. All of the following are solids of revolution.

Show that the volume of a sphere of radius $r$ is $\frac{4}{3}\pi r^3$ .

Solution: Here we use the function $y = \sqrt{r^2 - x^2}$ which is the top half of the circle to be revolved. So then the volume is achieved by integrating $A(x) = \pi y^2$ along the interval $[-r, r]$ . That is, $V = \int_{-r}^r \pi (\sqrt{r^2 - x^2})^2dx =\pi \int_{-r}^r (r^2 - x^2)dx = \pi\left[r^2x - \frac{x^3}{3}\right]_{-r}^r = \frac{4}{3}\pi r^2$
Find the volume of a cone with height $h$ and radius $r$ .

Solution: This will be done along the $y$ -axis, since standing cones make more sense than sideways cones. The relationship between $x$ and $y$ here is that $y = \frac{h}{r}x$ (which can be thought of as the slope of a cone whose tip is at the origin). So the radius at $y$ is $x = \frac{r}{h} y$ and the volume is then given by $V =\pi \int_0^h\left( \frac{r}{h}y\right)^2dy = \pi \frac{r^2}{h^2} \int_0^h y^2dy = \pi \frac{r^2}{h^2}\cdot \frac{h^3}{3} = \frac{1}{3}\pi r^2 h$

Now, we are ready for something slightly more involved.

Example 19. Find the volume of the solid formed by revolving the region bounded by $y^2 = x$ and $y = x^3$ about the (i) $x$ -axis, (ii) $y$ -axis, and (iii) the $x = -1$ axis.

Notice beforehand that the intersection occurs at $(0,0)$ and $(1,1)$ . So,

Revolving about the $x$ -axis, $V = \int_0^1 \pi(\sqrt{x}^2 - (x^3)^2)dx = \pi\left[\frac{x^2}{2}- \frac{x^7}{7}\right]_0^1 = \frac{5}{14} \pi$
About the $y$ -axis, $V = \int_0^1\pi\left((\sqrt[3]{y})^2 - (y^2)^2\right)dy = \pi\left[\frac{3}{5}y^{5/3}- \frac{y^5}{5}\right]_0^1 = \frac{2}{5}\pi$
About $x = -1$ axis, we need to readjust the integrand. $\begin{align*} V &= \int_0^1\pi\left((\sqrt[3]{y}+1)^2 - (y^2+1)^2 \right)dy \\ &= \pi\int_0^1 \left(y^{2/3}+ 2y^{1/3} + 1 - y^4 - 2y^2 -1\right)dy \\ &= \pi\left[\frac{2}{5}y^{5/2} + \frac{6}{4}y^{4/3} - \frac{y^5}{5} - \frac{2}{3}y^3\right]_0^1\\ = \frac{37}{30}\pi \end{align*}$

5.2 Cylindrical Shell Method

The volume of a shell is equal to the outer volume minus the inner volume. Mathematically if $r<R$ represent the inner and outer radius of the cylinder respectively, then $\begin{align*} V = \text{outer } - \text{ inner}\\ &= \pi R^2h - \pi r^2h\\ &= \pi h (R^2 - r^2) \\ &= 2\pi h\frac{R+r}{2}(R-r)\\ &= 2\pi h R^* t \end{align*}$ where $R^* = \frac{R + r}{2}$ is the average radius and $t = R-r$ is the thickness of the shell.

Steps for this method:

Suppose we have a region bounded by $y = f(x)$ an suppose it is revolved about the $y$ -axis,
Subdivide the interval $[a,b]$ into $n$ equal subintervals $\Delta x = \frac{b-a}{n}$ ,
Let $V(x_i)$ represent the volume of the cylindrical shell with outer radius $x_i$ , inner radius $x_{i-1}$ and height $f(x_i)$ ,
then we have a sequence $V(x_1), V(x_2), \dots$ of volumes of shells which fill our solid so the approximate volume is $\sum_{i=1}^n V(x_i) = \sum_{i=1}^n 2\pi f(x_i) x_i^* \Delta x, \text{ where } x_i^* = \frac{x_i + x_{i-1}}{2}$ which is approximately $\sum_{i=1}^n 2\pi x_i f(x_i) \Delta x$ a Riemann sum! Then the volume is given by $\begin{equation} \label{shells} \boxed{V = \lim_{n\to \infty} \sum_{i=1}^n 2\pi x_i f(x_i) \Delta x = 2\pi \int_a^b xf(x) dx} \end{equation}$ (8)

Note: This is most often used when there is a hole punched in the solid.

Example 20. Find the volume of the solid that remains after a hole of radius $a$ has been bored through the centre of a sphere of radius $b(>a)$ .

Solution: $\begin{align*} V &= 2\int_a^b2\pi x \sqrt{b^2 - a^2}dx , \quad \text{ let } x = b\sin\theta, dx = b\cos \theta\\ &= 4\pi \int_{x= a}^{x = b}b^2 \sin \theta\sqrt{b^2(1-\cos^2 \theta)} b\cos\theta d\theta\\ &= 4\pi b^3 \int_{x= a}^{x = b}\sin \theta\cos^2\theta d\theta, u = \cos\theta\\ &= 4\pi b^3 \int_{x=a}^{x=b} u^2 du \\ & = 4\pi b^3 \left[\frac{u^3}{3}\right]_{x = a}^{x = b} \end{align*}$ Now, with $u = \cos\theta= \frac{\sqrt{b^2 - x^2}}{b}$ we have $4\pi b^3 \left[\frac{u^3}{3}\right]_{x = a}^{x = b} = 4\pi b^3\left[\frac{(b^2-x^2)^{3/2}}{b^3}\right]_{x = a}^{x = b} = \frac{4}{3}\pi(b^2- a^2)^{3/2}$

Example 21. Using the method of cylindrical shells, find the volume obtained by revolving the region bounded by $y = x^2$ and $y^2 = x$ about the $y$ -axis.

Solution: $V = \int_0^12\pi x \left(\sqrt{x} - x^3\right)dx = 2\pi \int_0^1(x^{3/2} - x^4)dx = 2 \pi\left[\frac{2}{5}x^{5/2} - \frac{x^5}{5}\right]_0^1 = \frac{2}{5}\pi$

In general we have obtained the following two formulas $\boxed{V = \int_a^b \pi f(x) dx \text{ and } V = \int_a^b 2\pi x f(x)dx}$

6 Numerical Approximation

This is used to approximate definite integrals when they cannot be integrated.

There are 3 common methods:

Midpoint Rule: Given $\int_a^b f(x) dx = \lim_{n\to \infty } \sum_{i=1}^n f(x_i)\Delta x$ , recall that we had a choice of sample points $x_i$ in the interval $(x_{i-1}, x_{i})$ and we have just been using the right hand side of the interval. Now, consider $R_n = \sum_{i=1}^n f(x_i)\Delta x\quad - \quad\text{The right-hand approximation}$ $L_n = \sum_{i=1}^n f(x_{i-1})\Delta x\quad - \quad\text{The left-hand approximation}$ and finally $\boxed{M_n = \sum_{i=1}^n f(m_i)\Delta x \quad - \quad\text{The Midpoint approximation}}$ where $m_i = \frac{x_{i-1} + x_i}{2}$
Trapezoidal Rule: Here we use $\boxed{T_n = \frac{L_n + R_n}{2} = \frac{1}{2}\left[\sum_{i=1}^n f(x_{i-1})\Delta x + \sum_{i=1}^n f(x_{i})\Delta x\right] = \frac{\Delta x}{2}\left[f(x_0) + 2f(x_1) +\dots + 2f(x_{n-1}) + f(x_n) \right]}$
Simpson’s Rule: Here we use $2n$ sample points $x_1, x_2, \dots, x_n, \dots, x_{2n}$ , but our intervals remain of size $\Delta x = \frac{b-a}{n}$

$\boxed{S_{2n} := \frac{1}{3} \left[2M_n + T_{n}\right]}$ Now, $M_n = 2\Delta x\left[\sum_{i = 1} ^{n} f(x_{2i-1})\right]$ because midpoints here are the odd indexed sample points and $T_{2n} = 2\frac{\Delta x}{2}\left(f(x_0) + 2f(x_2) + \dots 2f(x_{2n-2}) + f(x_{2n})\right)$ So,

$\begin{align*} S_{2n} &= \frac{\Delta x}{3}\left[4\sum_{i = 1} ^{n} f(x_{2i-1}) +(f(x_0) + 2f(x_2) + \dots 2f(x_{2n-2}) + f(x_{2n}))\right]\\ &= \frac{\Delta x}{3}\left(f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \dots + 4f(x_{2n-1}) + f(x_{2n})\right) \end{align*}$

Example 22. Approximate the definite integral $\int_0^3 x^2 dx$ with $n = 6$ (i.e. so that $\Delta x= \frac{3 - 0 }{6 } = .5$ using; (i) midpoint rule, (ii) trapezoid rule, (iii) Simpson’s rule.

Solution:

For this, we need the following table of values:

$n$ $m_i$ $f(m_i)$

1 .25 0.0625

2 .75 0.5625

3 1.25 1.5625

4 1.75 3.0625

5 2.25 5.0625

6 2.75 7.5625

So, $\begin{align*} M_n &= \sum_{i=1}^6 f(m_i)\Delta x = \frac{1}{2}(19.875)= 8.9375 \end{align*}$
$\begin{align*} T_n &= \frac{\Delta x}{2}(f(0) + 2f(0.5)+ 2f(1) + + \dots + 2f(5.5) + f(6)) \\ &= \frac{1}{4}( 0^2 + 2(0.5)^2 + \dots + 3^2)\\ & = 9.125 \end{align*}$
$\begin{align*} S_n = \frac{\Delta x } {3} (0^2 + 4(0.5)^2 + 2(1)^2 + 4(1.5)^2 + 2(2)^2 + 4(2.5)^2 + 3^2) = \frac{1}{6}(54) = 9 \end{align*}$

$n$	$m_i$	$f(m_i)$
1	.25	0.0625
2	.75	0.5625
3	1.25	1.5625
4	1.75	3.0625
5	2.25	5.0625
6	2.75	7.5625

6.1 Error bounds on approximations

Here we will just provide the error bounds without examining the required analysis.

For $M_n$ ; the error bound is $\boxed{|E(M_n)| \leq \frac{k(b-a)^3}{24n^2}}$ where $k = \max_{x\in (a,b)}\{f''(x)\}$ .

For $T_n$ : Error is $\boxed{|E(T_n)| \leq \frac{k(b-a)^3}{12 n^2},\qquad k = \max_{x\in (a,b)}\{f''(x)\}}$

For $S_n$ ; Error is $\boxed{|E(S_n)| \leq \frac{k (b-a)^5}{180n^4}, \qquad k = \max_{x\in (a,b)}\{f^{(4)}(x)\}.}$

Example 23. Estimate the error bound when using $T_{10}$ or $S_{10}$ approximation for $\int_1^2 \frac{1}{x}dx$ .

Solution: Note that $\frac{d^2}{dx^2}\left(\frac{1}{x}\right) = \frac{2}{x^3}$ and $\frac{d^4}{dx^4}\left(\frac{1}{x}\right) = \frac{d^2}{dx^2}\left(\frac{2}{x^3}\right) = \frac{d}{dx}\left(\frac{-6}{x^4}\right) =\frac{24}{x^5}$ so that the constant $k$ for $T_{10}$ is $k_T = \max_{[1,2]} \left(\frac{2}{x^3}\right) = 2$ and for $S_{10}$ is $k_S = \max_{[1,2]} \left(\frac{24}{x^5}\right) = 24.$ Now, $|E(T_{10})| \leq \frac{k_T(2-1)^3}{12(10)^2} = \frac{1}{600}$ and $|E(S_{10})| \leq \frac{k_S(2-1)^5}{180(10)^4} = \frac{24}{1800000}.$

7 Improper Integrals

The following integrals $\int_a^\infty f(x)dx = \lim_{b\to \infty} \int_a^bf(x)dx \text{ OR } \int_{-\infty}^af(x)dx = \lim_{b\to -\infty}f(x)dx$ are called improper. Similarly, $\int_{-\infty}^\infty f(x)dx = \int_{-\infty}^af(x)dx + \int_a^\infty f(x)dx$ is improper (provided, both limits exist).

Warning: Do not do $\int_{-\infty}^\infty f(x)dx = \lim_{t\to \infty}\int_{-t}^t f(x)dx$ , since one may not converge.

Example 24. Evaluate the improper integral $\int_{-\infty}^\infty \frac{1}{1+ x^2}dx$ .

Solution: $\begin{align*} \int_{-\infty}^\infty \frac{1}{1+x^2}dx & = \lim_{b\to -\infty}\int_b^0 \frac{1}{1+x^2}dx + \lim_{t\to \infty}\int_0^t\frac{1}{1+x^2}dx\\ &= \lim_{b\to -\infty}\left[\arctan x\right]_b^0 + \lim_{t\to \infty}\left[\arctan x\right]_0^t\\ &= \left(0 - \frac{-\pi}{2}\right) + \left(\frac{\pi}{2} - 0\right)\\ &= \pi \end{align*}$

7.1 Improper integrals of another type

Suppose the function $f$ is defined only on the interval $(a, b]$ with no specified value at $a$ . Then we can still define the definite integral ‘improperly’ as $\int_a^bf(x)dx = \lim_{t\to a^-} \int_t^bf(x)dx.$ The same can be done when $f(x)$ is defined on $[a, b)$ , or even if, say $f(x)$ is only defined on $[a, c)\cup (c, b]$ (with a hole in the middle of the interval), then $\int_a^b f(x)dx = \lim_{t\to c^-} \int_a^tf(x)dx + \lim_{t\to c^+}\int_t^bf(x)dx$ Note that the above integral will only converge if both limits exist.

Example 25.

Proposition 1. [Comparison test for improper integrals] Suppose that $f(x)\geq g(x)$ on $[a, \infty)$ . If $\int_a^\infty f(x)dx$ converges, then $\int_a^\infty g(x) dx$ also converges.

Furthermore, if $\int_a^\infty g(x) dx$ diverges, then so does $\int_a^\infty f(x)dx$ .

Example 26. Does $\int_0^\infty e^{-x^2}dx$ converge?

Soloution: We have $x^2\geq x$ on $[1, \infty]$ so $e^{x^2}\geq e^x$ on $[1,\infty)$ or $\frac{1}{e^{x^2}}\leq \frac{1}{e^x}$ . So $\begin{align*} \int_0^\infty e^{-x^2} dx &= \int_0^1e^{-x^2}dx + \int_1^\infty e^{-x^2}dx\\ &\leq \int_0^1e^{-x^2}dx + \int_1^\infty e^{-x}dx\\ &= \int_0^1e^{-x^2}dx + \lim_{t\to \infty}\int_1^te^{-x}dx\\ &= \int_0^1e^{-x^2}dx + \lim_{t\to \infty}\left[-e^{-t} - (-e^{-1})\right]\\ & = \int_0^1e^{-x^2}dx + e <\infty \end{align*}$ so our integral converges by comparison test.

8 Differential Equations

A differential equation (DE) is an equation which contains a derivative expression. The order of a DE is the highest derivative which appears in the equation.

Example 27. $x y''(x) = \sin xe^{y(x)}$ is order 2.

Another, simpler, DE is $y'(x) = \sin x + e^x$ . This is easy to solve (integrate both sides), $y(x) = - \cos x + e^x + C$ is a solution.

Lets see a more exciting example.

Example 28. Let $P(t)$ be a population function (with $t$ for time). After studying the population over time, and discovering a proportionality amongst the $P(t)$ and its derivative. That is to say $P'(t) = k P(t)$ , for some constant $k$ . The goal now, is to obtain a formula for $P(t)$ . Rearranging slightly is $\begin{align*} \frac{P'(t)}{P(t)} = k & \Rightarrow \int \frac{P'(t)}{P(t)} dt = \int k dt\\ &\Leftrightarrow \ln |P(t)| = kt+C\\ & e^{\ln P(t)} = e^{kt+C}\\ & P(t) = e^{kt}e^C\\ &P(t) = C'e^{kt} \end{align*}$ for some positive constant $C$ that is determined by specified initial conditions (such as $P(0) = 100$ ).

Example 29. Verify that $y = x^2\ln x$ is a solution to $x^2y'' - 3xy' + 4y = 0$ in the region $x>0$ .

Solution: With $y = x^2\ln x$ , we compute $y' = 2x\ln x + x$ and $y'' = 2\ln x + 2+ 1$ . Substituting into the DE, we find $\begin{align*} x^2y'' - 3xy' + 4y &= x^2( 2\ln x + 3) - 3x(2x\ln x + x) + 4x^2\ln x\\ &= 2x^2 \ln x + 3x^3 - 6x^2 \ln x - 3x^2 + 4x^2\ln x\\ &= 0 \end{align*}$

A particular solution to a DE is one without arbitrary constants. That is, any particular function satisfying the DE. The solution $y = x^2\ln x$ from the previous is example is a particular solution.

Example 30. Suppose we want a solution to $y' + 2y = e^{-x}$ with $y(0) = 3$ [This is called an initial value problem].

Check that $y = e^{-x} + Ce^{-2x}$ is a general solution. Indeed, $y' = - e^{-x}- 2Ce^{-2x}$ and $- e^{-x}- 2Ce^{-2x} + 2(e^{-x} + C e^{-2x} = 0.$ Now since $y(0) = 3$ , we can solve for the constant $C$ as $3 = e^{0} + Ce^{0} = 1 + C \Rightarrow C = 2$ so that $y = e^{-x} + 2e^{-2x}$ is a particular solution to the initial value problem (IVP).

8.1 Separable Differential Equations

If a DE can be expressed in the form $y' = g(x) f(y)$ , we say it is separable. Equivalently $\frac{1}{f(y)} \frac{dy}{dx} = g(x) \Leftrightarrow \frac{1}{f(y)} dy = g(x) dx.$

A separable DE is solved by integrating both sides of this rearrangement. $\int \frac{1}{f(y)} dy = \int g(x) dx,$ but this requires some justification: If we are given a DE of the form $N(y)dy = M(x)dx$ or equivalently $N(y) \frac{dy }{dx} - M(x) = 0$ . Now let $H_1(y)$ be an anti-derivative of $N(y)$ and $H_2(x)$ an anti-derivative of $M(x)$ . This equation is then equivalent to the following: $\begin{align*} &H'_1(y) \frac{dy}{dx} - H'_2(x) = 0\\ &\frac{d}{dx}(H_1(y)) - H'_2(x) = 0\\ & = \frac{d}{dx}\left(H_1(y) - H_2(x) )\right)= 0 \end{align*}$ Now, this implies that $H_1(y) - H_2(x) = K$ (is constant) and we can find $y(x)$ from $H_2(y)$ .

Example 31. Is $\frac{dy}{dx} = \frac{3x^2 + 4x + 2}{2(y-1)}$ separable? If so, solve it.

Solution: Yes, it is solvable and it can be re-expressed as $2(y-1)dy = (3x^2 - 4x + 2)dx \Rightarrow 2\int(y-1)dy = \int(3x^2 - 4x + 2)dx.$ This integrates to $y^2 - 2y = x^3 + 2x^2 + 2x + C.$ Now, although this is separable an we managed to integrate it, we find that the solution cannot be explicitly expressed as $y = \cdots$ . In attempt to extract an explicit solution, we treat this as a quadratic equation in $y$ . $y^2 - 2y - (x^3 + 2x^2 + 2x + C) = 0$ which has solutions, $y = \frac{2\pm \sqrt{4 - 4(-1)(x^3 + 2x^2 + 2x + C)}}{2} = 1\pm\sqrt{x^3 + 2x^2 + 2x +1+ C}.$ If we had been given the IVP of $y(0) = -1$ , then this would imply that the solution be $y = 1-\sqrt{x^3 + 2x^2 + 2x +1+ C}$ .

8.2 Linear Differential Equations

A linear differential equation is most generally expressed as $\begin{equation} \label{LDE} y' + p(x) y = q(x) \end{equation}$ (9)

Theorem 6. The solution to the linear differential equation (9) is $y = \frac{1}{I(x)} \int I(x) q(x) dx$ where $I(x) = e^{\int p(x) dx}$ .

This function $I(x)$ is called the integrating factor and is a function satisfying that $I(x) \left(y ' + p(x) y\right) = \left(I(x) y\right)'.$

Proof. First, verifying that $I(x) = e^{\int p(x)dx}$ satisfies the above claimed equation. Indeed, expanding the right by the product rule, we have $I(x) \left(y ' + p(x) y\right) = \left(I(x) y\right)' = I'(x) y + I(x)y' \Leftrightarrow I(x) p(x) y = I'(x) y$ so that, provided $I(x)\neq 0$ , $p(x) = \frac{I'(x)}{I(x)}$ . Integrating this says that $\int p(x) dx = \int\frac{I'(x) }{I(x)} dx \Leftrightarrow \ln|I(x)| = \int p(x) dx \Leftrightarrow I(x) = e^{\int p(x) dx}.$

Now, that this holds, the left hand side of our DE is expressed as $(I(x)y)'$ and now reads, $(I(x)y) = q(x)$ . Hence, integrating this, $I(x)y = \int q(x) dx$ so $y = \frac{1}{I(x)} \int I(x) q(x) dx$ as claimed. ◻

Note that this method of solving a DE requires it be of first order (i.e. linear).

Example 32.

Solve the IVP $y' - 2xy = 1$ with $y(0) =1$ .

Solution: The integrating factor here is $I(x) = e^{-\int 2 x dx} = e^{-x^2}$ , so $y = \frac{1}{e^{-x^2}} \int e^{-x^2}dx = e^{x^2}\left(\int_{0}^x e^{-t^2}dt + C\right) = e^{x^2}\int_{a}^x e^{-t^2}dt +C e^{x^2}$ and with $y(0)= 1$ we find $C$ as $1 = e^0 \int_0^0 e^{-t^2}dt + e^{0}C \Rightarrow C =1$
Solve the IVP $(y-x\sin x) dx + x dy = 0$ with $y(\pi) =3$ .

Solution: This is equivalent to $y = x\sin x + x y' = 0$ or, better, $y'+ \frac{1}{x}y = \sin x$ . For the integrating factor $I(x) = e^{\int \frac{1}{x} dx} = e^{\ln |x|} = |x|$ . Now, $\begin{align*} y &= \frac{1}{|x|} \int|x| \sin x dx, \\ &= \frac{1}{x}\left[-x \cos x + \int \cos xdx\right], \qquad \text{IBP with } u = x, dv = \sin x dx\\ &= -\cos x + \frac{\sin x}{x} + \frac{C}{x} \end{align*}$

9 Sequences and Series

Recall, the limit of a sequence: A sequence $\{x_n\}$ converges to $L$ if for all $\epsilon>0$ , there exists $k\in \mathbb{N}$ such that $|x_n - L| <\epsilon$ for all $n>k$ .

Example 33. $a_n = (-1)^n \frac{\cos n}{n^2}$ and $\left|(-1)^n\frac{\cos n}{n^2}\right|\leq \frac{1}{n^2}\to 0$ and by squeeze theorem, $a_n\to 0$ .

An increasing (decreasing) sequence is one such that $a_{n+1}\geq(\leq) a_n$ for all $n\geq 1$ . A monotonic sequence is one which is either strictly increasing or decreasing.

Theorem 7 (Monotone sequence theorem). A bounded monotonic sequence always has a limit (i.e. will always converge).

9.1 Series

Let $a_n$ be a sequence, then $\sum_{n=1}^\infty = a_1 + a_2 + a_3+ \dots$ is an infinite series. The $n^{th}$ partial sum of a series is $S_n = \sum_{k=1}^n a_k$ . To evaluate an infinite series we use the limit of the partial sums: $\sum_{n=1}^\infty a_n = \lim_{n\to \infty} S_n = \lim_{n\to \infty} \sum_{k=1}^n a_k.$ If the limit exists, series converges. If limit does not exist, series diverges.

Warning: When discussing series, say $\sum_{n=1}^\infty c_n$ , there are two sequences in play. We have the sequence itself $c_n$ and the partial sums $S_n$ , so remember $\sum_{n=1}^\infty = \lim_{n\to \infty} S_n.$

Example 34. $c_i = i\Rightarrow \sum{n=1}^\infty n = \lim_{n\to \infty} \sum_{k=1}^n k = \lim_{n\to \infty}\frac{n(n+1)}{2} = \infty, \text{ (diverges)}$

Consider the geometric series $\sum_{n=1}^\infty a r^{n-1} = a + ar + ar^2 + \dots + ar^n + \dots.$ The partial sums are given by $S_n = a + ar + ar^2 + \dots + ar^{n-1}$ and multiplying by $r$ says, $r S_n = ar + ar^2 + \dots + ar^n.$ Subtracting these, we have $S_n - rS_n = a - ar^n \Leftrightarrow S_n(1 - r)= a(1-r^n)$ so that $S_n = \frac{1(1-r^n)}{1-r}.$ Now, when does a geometric series converge? We must have that $\lim_{n\to \infty} \frac{a(1-r^n)}{1-r}$ exists, which means, in order for a geometric series to converge we must have $|r|<1$ .

Furthermore, if $|r|<1$ , then the limit can be computed exactly as $\lim_{n\to \infty} \frac{a(1-r^n)}{1-r} = \frac{a}{1-r}$ . That is, the sum a geometric series is $\sum_{n=1}^\infty ar^{n-1} = \begin{cases} \frac{a}{1-r}, |r|<1\\ \infty, |r|\geq 1\end{cases}$

Example 35. Express the repeating decimal $1.2\overline{31}$ as an infinite series and then as a fraction.

Solution: We have $1.2\overline{31} = 1.2 + \frac{31}{10^3} + \frac{31}{10^3}\cdot \frac{1}{10^2} +\frac{31}{10^3}\cdot \frac{1}{(10^2)^2 } + \dots$

Letting $a= \frac{31}{10^3}$ and $r = \frac{1}{10^2}<1$ , so that $1.2\overline{31} = 1.2 + \sum_{n=1}^\infty ar^{n-1} = \frac{12}{10} + \frac{31}{10^3} \cdot \frac{1}{1 - \frac{1}{10^2}} = \frac{1219} {990}.$

Theorem 8. Suppose $\sum_{n=1}^\infty c_n = L$ and $\sum_{n=1}^\infty b_n = M$ and let $k\in \mathbb{R}$ . Then, $\sum_{n=1}^\infty kc_n = k L$ and $\sum_{n=1}^\infty (c_n + b_n) = L + M.$ Note that this theorem requires that both converge.

Theorem 9. If the series $\sum_{n=1}^\infty c_n$ converges, then the sequence $c_n$ must converge to 0.

Proof. Given that $\sum_{n=1}^\infty c_n = L$ , consider the sequence $S_n$ of partial sums. Observe, that for each $n$ , $c_n = s_n - S_{n-1}$ and now, $\lim_{n\to \infty} c_n = \lim_{n\to \infty} S_n - S_{n-1} = L - L = 0.$ ◻

9.2 Divergence test

Suppose that $\sum_{n=1}^\infty c_n$ is a series. If $\lim_{n\to \infty} c_n \neq 0$ then the series diverges. This is just the converse of the previous result. Otherwise, divergence test provides no information.

Example 36. Discuss the convergence/divergence of the series

$\sum_{n=1}^\infty\left(\frac{3}{n} + \frac{5}{4^n}\right)$ .

Solution: $\begin{align*} \sum_{n=1}^\infty\left(\frac{3}{n} + \frac{5}{4^n}\right) & = \sum_{n=1}^\infty \frac{3}{n}+ \sum_{n=1}^\infty \frac{5}{4^n}\\ & = 3\underbrace{\sum_{n=1}^\infty \frac{1}{n}}_{diverges}+ \underbrace{\sum_{n=1}^\infty \frac{5}{4}\left(\frac{1}{4}\right)^{n-1}}_{converges (geometric\,r=1/4)}\\ \end{align*}$ Therefore, the series diverges.
$\sum_{n=1}^\infty \frac{1}{(2n+1)(2n-1)}$

Solution: By partial fractions, we find

$\begin{align*} \sum_{n=1}^\infty \frac{1}{(2n+1)(2n-1)} & = \sum_{n=1}^\infty\left( \frac{-1}{2(2n+1)} + \frac{1}{2(2n-1)}\right)\\ &= \lim_{n\to \infty} - \frac{1}{6} + \frac{1}{2} - \frac{1}{10} + \frac{1}{6} - \frac{1}{14} + \frac{1}{10} + \dots + \frac{1}{2}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) \\ & = \lim_{n\to \infty}\frac{1}{2} - \frac{1}{2(2n-1)} = \frac{1}{2} \end{align*}$ This is a telescoping series.

Theorem 10 (Integral test for convergence of a series). Let $\sum_{n=1}^\infty c_n$ be a series and suppose $f(x)$ is a function on $[1, \infty)$ such that

$f(x) > 0$ on $[1,\infty)$ .
$f(x)$ is continuous on $[1, \infty)$
$f(x)$ is non-increasing on $[1, \infty)$ (i.e. $f'(x) \leq 0$ )
$f(n) = c_n$ for all $n$ .

Then, $\sum_{n=1}^\infty c_n \text{ converges } \Leftrightarrow \int_1^\infty f(x)dx\text{ converges}.$

Proof. Suppose that $\sum_{n=1}^\infty c_n$ converges and show that $\int_1^\infty f(x)dx$ converges. Certainly $S_n = c_1+ c_2 + \dots + c_n\geq \int_1^n f(x)dx$ so $S = \lim_{n\to \infty} S_n \geq \int_1^\infty f(x)dx$ which implies convergence.

Suppose now that $\int_1^\infty f(x)dx$ converges and show the converse. Observe that $c_2 + c_3+ \dots + c_n \leq \int_1^\infty f(x)dx$ so $S_n = c_1+c_2 + \dots c_n \leq c_1 + \int_1^\infty f(x)dx$ and hence $\lim_{n\to \infty} S_n \leq c_1 + \int_1^\infty f(x)dx< \infty$ so converges. ◻

9.3 Estimating the value of a series

Suppose that $\sum_{n=1}^\infty c_n$ is a series proved to be convergent by the integral test (say $\sum_{n=1}^\infty c_n = S$ ).

We would like to approximate $S$ by the partial sums $S_n$ and determine what the error will be. Let $R_n = S- S_n$ be the remainder (Note that, we are assuming that our $c_n$ ’s are positive so that the partial sums are an increasing sequence). So, also $R_n = S - S_n = (c_1+ c_2 + \dots + c_n + \dots )- (c_1 + c_2 + \dots + c_n) = c_{n+1} + c_{n+2} + \dots$ and we claim that $\int_{n+1}^\infty f(x)dx \leq R_n \leq \int_n^\infty f(x)dx$ for $f(n) = c_n$ .

First, it is easy to see that $R_n = c_{n+1} + c_{n+2} + \dots \leq \int_n^\infty f(x)dx$ . Now, also $R_n\geq \int_{n+1}^\infty f(x)dx$ since these rectangles are an over approximation of the integral.

Now, $\begin{align*} \int_{n+1}^\infty f(x)dx \leq &R_n \leq \int_n^\infty f(x)dx\\ S_n + \int_{n+1}^\infty f(x)dx \leq &S_n +R_n \leq S_n + \int_n^\infty f(x)dx\\ S_n + \int_{n+1}^\infty f(x)dx \leq &S \leq S_n + \int_n^\infty f(x)dx\\ \end{align*}$ So we know that $S$ lies in the interval $\left[S_n + \int_{n+1}^\infty f(x)dx, S_n + \int_n^\infty f(x)dx\right]$ and the error cannot be more than the length of the interval. That is, $\boxed{|\text{Error}| \leq ( S_n + \int_n^\infty f(x)dx) - (S_n + \int_{n+1}^\infty f(x)dx) =\int_n^\infty f(x)dx - \int_{n+1}^\infty f(x)dx = \int_n^{n+1}f(x)dx}$

Example 37. Use $S_{10}$ to help calculate $\sum_{n=1}^\infty \frac{1}{n^4}$ .

Solution: With a calculator we find $S_{10} = 1.082036583$ . Notice that $\frac{1}{x^4}$ is continuous, decreasing on $[1, \infty)$ and matches the values of the sequence. So, a good approximation is the midpoint of the interval $\left[S_{10} + \int_{11}^\infty \frac{1}{x^4}dx, S_{10} + \int_{10}^\infty \frac{1}{x^4}dx\right]$ . That is, $S \approx \frac{2S_{10} + \int_{11}^\infty \frac{1}{x^4}dx+ \int_{10}^\infty \frac{1}{x^4}dx}{2} \approx 1.082328469$

9.4 p-series

Any series of the form $\sum_{n=1}^\infty \frac{1}{n^p}$ where $p$ is a fixed positive number.

Theorem 11. If $\sum_{n=1}^\infty \frac{1}{n^p}$ is a $p$ -series, then it converges if and only if $p>1$ .

Proof. We first consider the special case that $p = 1$ . This series $\sum_{n=1}^\infty \frac{1}{n}$ is called the Harmonic series. By the integral test $\int_1^\infty \frac{1}{x} dx\lim_{t\to \infty} \int_1^t \frac{1}{x}dx = \lim_{t\to \infty} \ln x|_1^t = \infty$ . This diverges and hence, the harmonic series diverges.

Now, consider the case $p\neq 1$ : $\int_1^\infty \frac{1}{x^p}dx = \lim{t\to \infty} \left[\frac{x^{1-p}}{1-p}\right]_1^t = \frac{1}{1-p}\lim_{t\to \infty} \frac{1}{t^{p-1}} -1$ which diverges if $p-1<0 \, (p<1)$ and converges if $p-1>0\, (p>1)$ . ◻

9.5 Comparison test

Theorem 12 (Comparison test). Let $\sum_{n=1}^\infty a_n, \sum_{n=1}^\infty b_n$ be two series with $0\leq a_n \leq b_n$ . Then

$\sum_{n=1}^\infty a_n$ converges if $\sum_{n=1}^\infty b_n$ converges
$\sum_{n=1}^\infty b_n$ diverges if $\sum_{n=1}^\infty a_n$ diverges.

Proof. Let $A_n$ and $B_n$ represent the $n^{th}$ partial sums of $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ respectively. Suppose that $B_n\to L$ . Then since $0\leq a_n\leq b_n$ we have $0\leq A_n\leq B_n$ for all $n>0$ . Now, $A_n$ and $B_n$ are both monotone increasing sequences and with $B_n\to L$ we have $A_n$ is bounded above by $L$ . By monotone sequence theorem, $A_n$ converges.

For the second part, $A_n$ is monotone increasing to infinity. Thus, also $B_n\geq A_n$ is also increasing to infinity. ◻

9.6 Limit comparison test

Theorem 13 (Limit comparison test). Suppose there are two series $\sum_{n=1}^\infty a_n, \sum_{n=1}^\infty b_n$ with both $a_n, b_n>0$ for all $n$ . Then

If $\lim_{n\to \infty} \frac{a_n}{b_n} = L\neq 0,\text{ or } \infty$ then either both converge or both diverge.
If $\lim_{n\to \infty} \frac{a_n}{b_n} = 0$ and $\sum b_n$ converges, then so does $\sum a_n$ .
If $\lim_{n\to \infty} \frac{a_n}{b_n} = \infty$ and $\sum b_n$ diverges, then so does $\sum a_n$ .

Corollary 1. If $\sum_{n = 1}^\infty |a_n|$ converges then so does $\sum_{n=1}^\infty a_n$ .

Proof. Suppose that $\sum_{n = 1}^\infty |a_n|$ converges to $M$ . We have the inequality $0\leq a_n +|a_n| \leq 2|a_n|.$ By the comparison test, then $\sum_{n=1}^\infty (a_n + |a_n|)$ converges (since $\sum 2|a_n| = 2\sum |a_n|= 2M$ ). Assuming that $\sum(a_n + |a_n|) = L$ then $\sum a_n = L-M$ as required. ◻

Definition 4. If $\sum a_n$ is a series such that $\sum |a_n|$ converges, we say that $\sum a_n$ is absolutely convergent.

If $\sum a_n$ converges, but $\sum |a_n|$ does not, then $\sum a_n$ is said to be conditionally convergent.

Any series of the form $\sum_n (-1)^n c_n$ where $c_n$ ’s are positive, then the series is called an alternating series.

Theorem 14 (Alternating series test). Let $\sum_{1}^\infty (-1)^{n-1}c_n$ be an alternating series. If $c_{n+1}\leq c_n$ for all $n>0$ and $\lim_{n\to \infty} c_n = 0$ then the series converges.

Proof. Supposing that $\lim_{n\to \infty} c_n = 0$ and $0\leq c_{n+1}\leq c_n$ , let $S_n$ denote the partial sums of the series $\sum_n (-1)^n c_n$ . Note that $S_{2n}$ is monotone increasing and bounded above by $c_1$ . So it remains to show that the limit of $S_{2n}$ is the same as the sum of the series. Note that $S_{2n+1} = S_{2n} + c_{2n+1}\to S + 0 = S$ so thus, the series converges. ◻

9.7 Alternating series estimation

Theorem 15 (Alternating series estimation theorem). If $S = \sum_{n=1}^\infty(-1)^{n-1}b_n$ is the sum of an alternating series which satisfies $0 \leq b_{n+1}\leq b_n \text{ and } \lim_{n\to \infty} b_n = 0$ then, $R_n \leq |S- S-n| \leq b_{n+1}.$

Proof. We know from above that $S$ lies between two consecutive partial sums $S_n$ and $S_{n+1}$ . It follows that $|S-S_n| \leq |S_{n+1} - S_n| = b_{n+1}$ . ◻

Note; the rule that the error is smaller than the first neglected term is valid only for alternating series.

9.8 Ratio test

Let $\rho = \lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n}\right|$ for a series $\sum_n a_n$ . Then

if $\rho<1$ , the series is absolutely convergent and hence convergent

if $\rho >1$ or $= \infty$ then the series is divergent

if $\rho = 1$ or does not exist then the test is inconclusive.

9.9 Root test

Let $\sum_n a_n$ be a series. Then

if $\lim_{n\to \infty} \sqrt[n]{|a_n|} <1$ then $\sum a_n$ is absolutely convergent,

if $\lim_{n\to \infty} \sqrt[n]{|a_n|} >1$ or $= \infty$ the series is divergent

and otherwise, if $\lim_{n\to \infty} \sqrt[n]{|a_n|} = 1$ the test is inconcolsive

10 Series testing strategies

The following is a recipe for testing convergence of series.

Check if $\sum a_n$ is geometric, harmonic, p-series or telescoping
Is the $\lim_{n\to \infty} a_n = 0$ ? If not, then the series diverges (by divergence test). Otherwise, convergence is still unknown
Does $|a_k|$ resemble $r^k$ or $\frac{1}{k^p}$ for large $k$ ? If so, try the comparison test, or limit comparison test with geometric series or a $p$ -series
If $a_k = f(k)$ for some positive, decreasing continuous function $f$ , try the integral test
If there are powers of $n$ or $n!$ the ratio test is probably a good idea.
- This test may result in the series diverges with $\rho>1$ or $= \infty$ so there would be no need to proceed with alternating series.
- This may show that $\sum_n a_n$ converges absolutely (when $\rho<1$ ) hence converges
- if $\rho =1$ or doesn’t exist, this test has failed and says nothing
Try the Root test, especially if you see powers of $n$ !
- If $\rho>1$ or $= \infty$ series diverges
- if $\rho <1$ it converges absolutely and hence converges
- if $\rho = 1$ , or does not exist the try something else because we still don’t know
if both ratio and root test have already failed, or it is clear that $\sum|a_n|$ is divergent, try (if series is alternating) the alternating series test incase it is conditionally convergent.
Error bounds:
- if series converges by the integral test, use the integral formula for approximate value of series.
- if the series converges by the alternating series test, use the alternating series estimation theorem to approximate the value.

11 Power series

A power series centered at $a$ is a series of the form $\sum_{n=0}^\infty c_n (x-a)^n$ with $x$ a variable. This series differs in values because of the variable $x$ . This means that certain values cause series to converge and some won’t. This should give the impression that a power series is a function whose domain is a collection of values $x$ for which the series converges. When studying these, we will be interested in finding an elementary function $f$ such that $f(x) = \sum _{n=0}^\infty c_n(x-a)^n$ as well as determining its domain.

Observations: We already know that $\sum_{n=0}^\infty r^n = 1 + r + r^2 + \dots$ converges to $\frac{1}{1-r}$ for $|r|<1$ while it diverges everywhere else. So, we could say that $\sum_{n\geq 0} x^n$ , centered at zero converges to the function $\frac{1}{1-x}$ on the interval $(-1, 1)$ and diverges elsewhere.

The function $T_n(x):= \sum_{k=0}^n c_n(x-a)^k$ is a polynomial. Recall, from Taylor polynomials that if $T_1(x) = 1, T_2(x) = 1 + x, \dots, T_k(x) = 1 + x+ x^2 + \dots + x^k$ that $\lim_{n\to \infty} T_n(x) = \frac{1}{1-x}$ . Thus, a power series can be viewed as a function whose curve is a limit of a sequence of polynomials.

Also notice that a power series $\sum_nc_n(x-a)^n$ always converges for $x = a$ .

Here, we will mainly discuss two things;

For what values of $x$ does a power series converge? (this is called the radius and interval of convergence
To what function, if any, does it converge to?

Theorem 16. The power series $\sum_{k=0}^\infty c_k (x-a)^k$ converges either for $x = a$ , for all $x$ or there is a positive number $R$ such that the series converges for all $|x-a|<R$ and diverges for all $|x-a|>R$ .

Note: When a power series is known to converge for all $|x-a|<R$ , one must manually determine convergence at the endpoints of this open interval.

Proof. Given the power series $\sum_{k=0}^\infty c_k (x-a)^k$ , suppose we define $\rho = \lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right|$ . Applying the ratio test to $\lim_{n\to \infty} \left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\right| = \rho |x-a|$ . Then this converges for $\rho|x-a|<1$ or, equivalently, $|x-a|<1/\rho$ . So to say that $R = 1/\rho$ . ◻

Remark 2. The above theorem states convergence on intervals centered at $a$ . This is called the interval of convergence. Series may or may not converge at the endpoints of this interval. The $R = 1/\rho = \lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right|$ . If the power series converges to a function, we say $f(x)=\sum_{k=0}^\infty c_k (x-a)^k$ for all $|x-a|<R$

Example 38. Discuss the convergence of the following power series

$\sum_{n=1}^\infty \frac{(-1)^n}{n4^n}(x-2)^n$ Solution: Here $R =1/\rho = \lim_{n\to \infty} \left|\frac{(-1)^n}{n4^n}\frac{(n+1)4^{n+1}}{(-1)^{n+1}}\right| = \lim_{n\to \infty}\left| 4 + 4/n\right| = 4$ so $|x-2|<4$ or $-2<x<6$ . At the endpoints; when $x = -2$ this is $\sum_{n=1}^\infty \frac{(-1)^n(-4)^n}{n4^n} = \sum_{n=1}^\infty \frac{1}{n}$ when $x = 6$ this becomes $\sum_{n=1}^\infty \frac{(-1)^n} {n}$ which is known to converge! Thus, the interval of convergence is $(-2, 6]$ .
$\sum_{n=1}^\infty \frac{2^n}{n!}x^n$ Solution: Here we find $\rho = \lim_{n\to \infty} \left|\frac{2^{n+1}}{(n+1)!}\frac{n!}{2^n}\right| = \lim_{n\to \infty} \frac{2}{n+1} = 0$ hence $1/\rho = \infty$ which implies radius of convergence is infinite and series converges for all values of $x$ .

11.1 Functions as power series

Using that $\frac{1}{1-x} = \sum_{k\geq 0} x^k$ as a basis, we can find other functions expressed as power series!

For example,

$\frac{1}{3+x} = \frac{1}{3} \frac{1}{1 - (-\frac{x}{3})} = \frac{1}{3}\sum_{n=0}^\infty \left(\frac{-x}{3}\right)^n = \sum_{n=0}^\infty \frac{(-1)^n x^n }{3^{n+1}}$ and the interval of convergence is $\left|\frac{x}{3}\right|<1$ or, $|x|<3$ .
$\frac{1}{1+x^2} = \frac{1}{1-(-x^2)} = \sum_{n=0}^\infty (-1)^n x^2n$ with interval of convergence $x^2<1$ which is really just $|x|<1$ .
$\frac{x^5}{3 + x^2} = \frac{x^5}{3} \frac{1}{1- (-\frac{x^2}{3})} = \frac{x^5}{3}\sum_{n=0}^\infty \frac{(-1)^n x^2n}{3^n} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+5}}{3^{n+1}}$ with radius of convergence $\left|\frac{x^2}{3}\right|<1$ or $\left|x\right|<\sqrt{3}$

Now that we can deal with everything of this form, we will need a new style.

Theorem 17. Suppose $\sum_{n=0}^\infty c_n (x-a)^n$ has radius of convergence $R>0$ and suppose that $f(x) = \sum_{n=0}^\infty c_n (x-a)^n$ . Then $f(x)$ is differentiable of $|x-a|<R$ and $f\prime(x) = \sum_{n=1}^\infty nc_n(x-a)^{n-1}$ $\int_a^x f(x)dx = \int_a^x \sum_{n=0}^\infty c_n (x-a)^n dx = \sum_{n=0}^\infty \int_a^x c_n (x-a)^n dx = \sum_{n=0}^\infty \frac{c_nx^{n+1}}{n+1}$ and both of these resulting power series have $R$ for their radius of convergence.

Example 39. Show that $\frac{1}{(1-x)^2} = \sum_{n=0}^\infty (n+1)x^n$ with $R= 1$ .

Solution: $\begin{align*} \frac{d}{dx} (1-x)^{-1} & = \frac{1}{(1-x)^2}\\& = \frac{d}{dx} \sum_{n=0}^\infty x^n\\ & = \sum_{n=1}^\infty nx^{n-1}\\ & = \sum_{n=0}^\infty (n+1)x^n \end{align*}$

$\begin{align*} \ln (1+x) & = \int_a^x \frac{1}{1+t}dt \\ &= \int_a^x \frac{1}{1-(-t)} dt \\ &= \int_0^x \sum_{n=0}^\infty(-1)^n t^n dt \\ &= \sum_{n=0}^\infty(-1)^n \int_0^x t^n dt \\ &= \sum_{n=0}^\infty \frac{(-1)^n x^{n+1}}{n+1}\\ &= \sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n} \end{align*}$ and radius of convergence $R = 1$ .

$\begin{align*} \arctan x &= \int_0^x \frac{1}{1+x^2}dx \\ &= \int_0^x \frac{1}{1-(-x^2)}dx \\ &= \int_0^x \sum_{n=0}^\infty (-1)^n x^{2n}dx \\ &= \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1} \end{align*}$ for $|x| <1$ .

11.2 Taylor and Mclaurin series

Suppose $f(x)$ has power series representation $f(x)= \sum_{n=0}^\infty c_n(x-a)^n$ . What do the $c_n$ ’s look like?

Theorem 18. Let $f(x)= \sum_{n=0}^\infty c_n(x-a)^n$ on $|x-a|<R$ , then $\boxed{c_n = \frac{f^{(n)}(a)}{n!}}$ That is to say $f(x)= \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n.$

Example 40. We know that $\arctan x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}$ which implies that $\frac{(-1)^n}{2n+1} = \frac{\arctan^{(n)}(0)}{n!}$ so the $n^{th}$ derivative of $\arctan$ evaluated at zero is $\frac{(-1)^n n!}{2n+1}$ .

Definition 5 (Taylor series). Let $f(x)$ be a function which is differentiable infinitely many times, then the series $T_{f, a}(x)= \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$ is called the Taylor series generated by $f$ centred at $a$ .

Any Taylor series centred at 0, is called a McLaurin series.

Example 41.

$\sum_{n=0}^\infty x^n$ is the Mclaurin series for $\frac{1}{1-x}$ .
Find the McLaurin series generated by $f(x)= e^x$ and determine the interval of convergence.

Solution: We need only determine $c_n = \frac{f^{(n)}(0)}{n!} = \frac{e^x}{n!}|_{x=0} = \frac{1}{n!}$ . So $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}.$

Notice that there exists functions which generate taylor and McLaurin series that do not converge to the original function. For example, the piecewise defined function $f(x)= \begin{cases} e^{-1/x^2}, x\neq 0 \\ 0, x = 0 \end{cases}$ has a McLaurin series of 0, but this function is not the zero function.

The $n^{th}$ degree taylor polynomial for $f$ , centred at $a$ is defined as the truncated version of its Taylor series $T_n(x) = \sum_{k=0}^n \frac{f^{(n)}(a)}{n!}(x-a)^n$ . These are the partial sums of the Taylor series and again we can define the $n^{th}$ remainder $R_n(x) = T_{f, a}(x) - T_n(x)$ .

We have that $R_n(x) \to 0 \Leftrightarrow T_n(x) \to f(x)$

To show that $R_n(x)\to 0$ , we will use a tool called Taylor’s inequality.

Theorem 19. Let $\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$ be a Taylor series with radius of convergence $R$ . Then $|R_n(x) |$ can be bounded above in the following way:

If $|f^{(n+1)}(x)|\leq M_n$ on $|x - a|<R$ , then $|R_n(x)| \leq \frac{M_n|x-a|^{n+1}}{(n+1)!}, \text{ on } |x-a|\leq d<R.$ If we can show that $\lim_{n\to \infty}\frac{M_n|x-a|^{n+1}}{(n+1)!} =0$ for all $|x-a|\leq d$ , then $f(x)= \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$

Example 42. Recall, Mclaurin series of $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$ has infinite radius of convergence
(since $\lim_{n\to \infty} \frac{x^n}{n!} = 0$ since $\sum_{n=0}^\infty \frac{x^n}{n!}$ converges by divergence test).

Show that $\sum_{n=0}^\infty \frac{x^n}{n!}$ converges to $e^x$ .

Solution: Consider any interval $[-k,k]$ . Since $e^x$ is increasing let $M_n = e^k$ . Then Taylors inequality guarantees that $|R_n(x)|\leq \frac{e^k|x|^{n+1}}{(n+1)!}$ . Now, $\lim_{n\to \infty} |R_n(x)| \leq e^k \lim_{n\to \infty}\frac{|x|^{n+1}}{(n+1)!} = 0$ as required.

Example 43. Find the McLaurin series generated by $\sin x$ .

Solution: since the derivatives of sine are $\cos, -\sin, -\cos, \sin, \dots$ and repeat every four, we find the values of these derivates to be $0, 1, 0, -1, \dots$ and repeat every four. Hence, then the series is given by $0 + 1\cdot x + 0 \frac{x^2}{2!} - 1 \frac{x^3}{3!} + 0 \frac{x^4}{4!} + \dots = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots = \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1} }{(2k+1)!}$ and $R = \lim_{n\to \infty} \left|\frac{(2n+3)!}{(2n+1)!}\right| = \lim_{n\to \infty} (2n+3)(2n+2) = \infty$

Does the series generated by $\sin x$ converge to $\sin x$ ?

We have $\sum_{n=0}^\infty\frac{\sin^{(n)}(0)}{n!}(x-0)^n = \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$ and $|\sin^{( n)}(x) \leq 1 = M_n$ for all $x\in [-k,k]$ . Thus, $|R_n(x) | \leq \frac{|x|^{n+1}}{(n+1)!} \to 0$ So the Taylor series generated by $\sin x$ converges to $\sin x$ for all $x$ .

Note that if you wanted to approximate $\sin x$ near $x= 50$ , choose $a = 50$ .

Example 44. Estimate $\sin(0.1)$ with error $<10^{-8}$ .

Solution: Recall $\sin x = \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$ and we know that $|R_n(x)| \leq \frac{|x|^{n+1}}{(n+1)!}$ . That is, $\begin{align*} |\sin x - T_n(x) | \leq& \frac{|x|^{n+1}}{(n+1)!}<10^{-8}\\ &\frac{|0.1|^{n+1}}{(n+1)!}<10^{-8}\\ \end{align*}$ Try $n=6$ , and find that $R_6(0.1) \leq \frac{(0.1)^{7}}{(6+1)!} = 1.9\times 10^{-11}<10^{-8}$ . Thus, $T_6(0.1) = T_5(0.1) = (0.1) - \frac{(0.1)^3}{3!} + \frac{(0.1)^5}{5!} = 0.099833$ .

Example 45. Find the McLaurin series for $\cos x$ .

Solution: $\cos x = \frac{d}{dx}\sin x = \frac{d}{dx} \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$

Theorem 20. Let $\sum_n a_n x^n, \sum_n b_nx^n$ be two McLaurin series with radii of convergence $R_a$ and $R_b$ respectively. Then $\sum_n c_n x^n = \left(\sum_n a_n x^n\right)\cdot \left(\sum_n b_n x^n\right)$ where $c_n = a_0b_n + a_1b_{n-1} +\dots +a_{n-1}b_1 + a_n b_0 = \sum_{i =0}^n a_ib_{n-i}$ has interval of convergence $R = \min\{R_a, R_b\}$ .

Example 46. Find the Mclaurin series for $\frac{1}{x-1}\ln(1-x)$ .

Solution: $\frac{1}{x-1} = \frac{-1}{1-x} = (-1)(1+x + x^2 + \dots )$ $\ln(1-x) = - x - \frac{x^2}{2} - \frac{x^3}{3}- \frac{x^4}{4} - \dots$ which implies $\begin{align*} \frac{1}{x-1}\ln(1-x) &= (1+x + x^2 + \dots )(x + \frac{x^2}{2} + \frac{x^3}{3}+\frac{x^4}{4} + \dots )\\ &= 1(0) + \left(1(0) + (1)(1)\right)x + \left(1(0) + 1(\frac{1}{2}) + 1(1)\right)x^2 + \left(1 + 1/2 + 1/3\right)x^3 + \dots \\ &= \sum_{n=0}^\infty \left(1 + 1/2 + 1/3 + \dots +1/n \right)x^n \end{align*}$ and $R = 1$ .

Everyone should know the following list of Taylor expansions

$\begin{align*} e^x &= \sum_{n\geq 0}\frac{x^n}{n!}\\ \cos x &= \sum_{n\geq 0}\frac{(-1)^nx^{2n}}{(2n)!}\\ \sin x &= \sum_{n\geq 0}\frac{(-1)^nx^{2n+1}}{(2n+1)!}\\ \arctan x &= \sum_{n\geq 0}\frac{(-1)^nx^{2n+1}}{2n+1}\\ \frac{1}{1-x} &= \sum_{n\geq 0 } x^n\\ \ln (1-x) & = - \sum_{n\geq 0 } \frac{x^n}{n}\\ \end{align*}$

12 Applications of Taylor and McLaurin series

Find the sum of the series $1 - \frac{\pi^2}{2!} + \frac{\pi^4}{4!} - \frac{\pi^6}{6!} + \dots = \sum_{k\geq 0} \frac{(-1)^k\pi^{2k}}{(2k)!} = \cos (\pi) = -1$ .
Approximate the integral $\int_0^1 \frac{1- \cos x}{x^2} dx$ with error less than $0.00001$ .

Solution: $\begin{align*} \int_0^1 \frac{1- \cos x}{x^2} dx&= \int_0^1 \frac{1}{x^2} \left(1 - (1 - x^2/2! + x^4/4!- x^6/6! + \dots\right)dx\\ &= \int_0^1\frac{1}{x^2}\left(x^2/2! - x^4/4!+ x^6/6! + \dots\right)dx\\ &= \int_0^1\left(1/2! - x^2/4!+ x^4/6! + \dots\right)dx\\ &= \left[\frac{x}{2} - \frac{x^3}{3\cdot 4!} + \frac{x^5}{5\cdot 6!} + \dots \right]_0^1\\ &= \frac{1}{2!} - \frac{1}{3\cdot 4!}+\frac{1}{5\cdot 6!}-\frac{1}{7\cdot 8!} + \dots \end{align*}$ This is an alternating series, so the error bound for $S_n$ is $c_{n+1}$ . Letting $n= 4$ , then $c_4 = \frac{1}{7\cdot 8! } = 0.0000036<0.00001$ . Therefor $S_3$ is appropriate and $S_3 = \frac{1}{2!} - \frac{1}{3\cdot 4!}+\frac{1}{5\cdot 6!} \approx 0.486388$
Approximate the integral $\int_0^1 \frac{1- \cos x}{x^2} dx$ with error less than $10^{-7}$ .

Solution:

$\begin{align*} \left|\sin x - \left(x - \frac{x^3}{3!} + \frac{x^5}{5!} + \right)\right| \leq \left|R_n(x)\right| \end{align*}$ implies $\left|\frac{\sin x}{x} - \left(\frac{1}{x}\right)\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} + \right)\right| \leq \left|\frac{R_n(x)}{x}\right|$ so $\begin{align*} \left|\int_0^{1/2}\frac{\sin x}{x}dx - \int_0^{1/2}\left(\frac{1}{x}\right)\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} + \right)dx\right| &\leq \int_0^{1/2} \left|\frac{R_n(x)}{x}\right|dx \\ & \leq \int_0^{1/2}\frac{1}{x} \frac{x^{n+1}}{(n+1)!} dx \\ &\leq \int_0^{1/2}\frac{x^n}{(n+1)!}dx\\ &= \left[\frac{x^{n+1}}{(n+1)(n+1)!}\right]_0^{1/2}\\ &= \frac{1}{2^{n+1}(n+1)(n+1)!} \end{align*}$

When $n=6$ , this is $\frac{1}{2^7(7)7!}$ .

12.1 Binomial series

Recall, if $n>0$ that $(a+b)^n = \sum_{k=0}^n {n\choose k} a^kb^{n-k}$ where ${n\choose k} = \frac{n! }{k!(n-k)!}$ .

Definition 6. For $\alpha$ and real number and $k$ a positive integer, define ${\alpha\choose k}$ as follows: ${\alpha\choose 0 } = 1, \quad {\alpha\choose k} = \frac{\alpha(\alpha-1)(\alpha-2)\dots (\alpha- k+1)}{k!}$

Theorem 21 (Binomial Series Theorem). Let $\alpha\in \mathbb{R}$ . Then the McLaurin series for $(1+x)^\alpha= \sum_{k=0}^\infty {\alpha\choose k} x^k$ with interval of convergence $(-1, 1)$ .

Example 47. Find a McLaurin series for $\frac{1}{\sqrt{9-x}}$ .

Solution: $\begin{align*} \frac{1}{\sqrt{9-x}} & = \frac{1}{3}\left(1+ (-x/9)\right)^{-1/2}, \qquad \text{let } u = x/9\\ &= \frac{1}{3}\left(1 + \frac{-1/2}{1!}(-u) + \frac{(-1/2)(-1/2-1)}{2!}(-u)^2 + \frac{(-1/2)(-1/2-1)(-1/2-2)}{3!}(-u)^3 +\dots \right)\\ &= \frac{1}{3}\left(1 + \frac{1}{2}u + \frac{1\cdot 3}{2^22!}u^2 + \frac{1\cdot 3\cdot 5}{2^33!}u^3 +\dots + \frac{1\cdot 3 \cdot 5\cdots (2n-1)}{2^n n!}u^n+\dots \right)\\ &= \frac{1}{3} \sum_{n=0}^\infty \frac{1\cdot 3\cdot 5\cdots (2n-1)}{2^n n!}u^n \end{align*}$

So, now the McLaurin series for $f(x)=\frac{1}{3} \sum_{n\geq 0} \frac{1\cdot 3\cdot 5\cdots (2n-1)}{2^n n!9^n}x^n$ we can compute $f^{(12)}(0)$ since we know the series! Indeed,

$\frac{f^{(12)}(0)}{12!} = \frac{1\cdot 3\cdot 5\cdots (2n-1)}{3\cdot 2^{12}12! 9^n }\Rightarrow f^{(12)}(0) = \frac{1\cdot 3\cdot 5\cdots (2n-1)}{3\cdot 2^{12}\cdot 9^n }$

12.2 One last application

The pendulum of length $L$ ;

$T = 4\sqrt{\frac{L}{g}}\int_0^{\pi/2}\frac{1}{\sqrt{1- k^2\sin^2\phi}}d\phi$ We would like to approximate $T$ given $\phi_0$

Express $\frac{1}{\sqrt{1- k^2\sin^2\phi}} = \sum_{n=0}^\infty \frac{1\cdot 3\cdot 5\cdots (2n-1)}{2^n n!}(k^2\sin^2\phi)^n =\sum_{n=0}^\infty \frac{1\cdot 3\cdot 5\cdots (2n-1)}{2^n n!}k^{2n}\sin^{2n}\phi$

Then $4\sqrt{\frac{L}{g}}\int_0^{\pi/2}\frac{1}{\sqrt{1- k^2\sin^2\phi}}d\phi = 4\sqrt{\frac{L}{g}}\int_0^{\pi/2}\sum_{n=0}^\infty \frac{1\cdot 3\cdot 5\cdots (2n-1)}{2^n n!}k^{2n}\sin^{2n}\phi d\phi$

Calculus II

Original typeset notes