$$$$

27  $★$ Examples

In this optional chapter we integrate familiar functions directly from the definition. I’ve written this merely to illustrate its possible (inspired by a challenge posed by past students to me), not because its useful: this entire chapter is rendered entirely superflouous following our proof of the fundamental theorem of calculus!

Recall our definition of axiomatically integrable says that $f$ is integrable on $[a,b]$ only if *every possible definition of $\int$ satisfying the axioms agrees on the value of ${\int }_{[a,b]}f$. This is quite a slippery concept to work with, so we developed the concepts of upper and lower sums to help us out.

27.1 Powers

Here

Proposition 27.1 (Integrating $f(x)=x$) Let $[a,b]$ be any closed interval in $\mathbb{R}$. Then $f(x)=x$ is integrable on $[a,b]$ and $${\int }_{[a,b]}x=\frac{b^2-a^2}{2}$$

Proof. Start with $[0,b]$, then look at $0<a<b$ using interval subdivision. To show $x$ is integrable, it is enough to find a sequence $P_n$ of shrinking partitions where $limL(f,P_n)=limU(f,P_n)$. Then $U(f)=L(f)$ necessarily.

For each $n$, let $P_n$ be the evenly spaced partition with $n$ subintervals, of width ${\mathrm{\Delta }}_n=(b-a)/n$. Since $f(x)=x$ is monotone increasing, we know on each subinterval $[t_{i-1},t_i]$ that $$m_i=t_{i-1}=(i-1){\mathrm{\Delta }}_n{M}_i=t_i=i{\mathrm{\Delta }}_n$$

Thus, the upper and lower sums for these partitions are

$$\begin{array}{rl}L(x,P_n)& =\sum _{1\le i\le n}{m}_i{\mathrm{\Delta }}_n=(i-1){\mathrm{\Delta }}_n{\mathrm{\Delta }}_n\\ & ={\mathrm{\Delta }}_n^2(0+1+2+\cdots +(n-1))\end{array}$$

$$\begin{array}{rl}U(x,P_n)& =\sum _{1\le i\le n}{M}_i{\mathrm{\Delta }}_n=i{\mathrm{\Delta }}_n{\mathrm{\Delta }}_n\\ & ={\mathrm{\Delta }}_n^2(1+2+\cdots +n)\end{array}$$

These are nearly identical formulae: the upper sum is just one term longer than the lower sum and so their difference is

$$U(x,P_n)-L(x,P_n)=n{\mathrm{\Delta }}_n^2=n\frac{b^2}{n^2}=\frac{b^2}{n}$$

As $n\to \mathrm{\infty }$ this converges to zero: thus, if either the upper or lower sum converges, then both do, and both converge to the same value by the limit theorems. For example, if we prove $U(f,P_n)$ converges then

$$\begin{array}{rl}limL(x,P_n)& =lim(U(x,P_n)-U(f,P_n)+L(x,P_n))\\ & =limU(x,P_n)-lim(U(x,P_n)-L(x,P_n))\\ & =limU(xs,P_n)+0s\end{array}$$

So, we focus on just proving that $U(x,P_n)$ converges and finding its value.

Exercise 27.1 Use the sum of the first $n$ integers we have previously derived to prove that $limU(x,P_n)=\frac{b^2}{2}lim(1+1/n)=\frac{b^2}{2}$.

Thus $x$ is integrable on $[0,b]$ and

$${\int }_{[0,b]}x=\frac{b^2}{2}$$

Knowing this, we complete the case for a general positive interval $[a,b]$ with $0<a<b$ by subdivision:

Exercise 27.2 Show that $${\int }_{[a,b]}x=\frac{b^2-a^2}{2}$$

*Hint: do $0<a<b$ first, then deal with the case where one or both may be negative, with another subdivision.

Proposition 27.2 (Integrating $f(x)=x^2$) Let $[a,b]$ be any closed interval in $\mathbb{R}$. Then $f(x)=x^2$ is integrable on $[a,b]$ and $${\int }_{[a,b]}{x}^{2}dx=\frac{b^3-a^3}{3}$$

Exercise 27.3 Following the same technique as above, show that $x^2$ is integrable on $[a,b]$:

• First, restrict yourself to intervals of the form $[0,b]$ for $b>0$.
• Use the monotonicity of $x^2$ on these intervals to explicitly write out upper and lower sums.
• Use the following identity on sums of squares from elementary number theory to compute their value $$\sum _{1\le k\le N}{k}^2=\frac{N(N+1)(2N+1)}{6}$$
• Explain how to generalize this to intervals of the form $[a,0]$ for $a<0$, and finally to general intervals $[a,b]$ for any $a<b\in \mathbb{R}$ using subdivision.

27.2 Exponentials

Here’s a quite long calculation showing that it’s possible to integrate exponential functions directly from first principles. The length of this calculation alone is a good selling point for the fundamental theorem of calculus! There are several facts about exponentials we will need from our previous investigations; listed here for ease of reference.

• Exponentials are always nonzero
• Exponentials are strictly increasing, or strictly decreasing
• Exponentials are differentiable everywhere

Proposition 27.3 (Integrating Exponentials) Let $E$ be an exponential function, and $[a,b]$ an interval. Then $E$ is integrable on $[a,b]$ and $${\int }_{[a,b]}E=\frac{E(b)-E(a)}{E^{\prime }(0)}$$

Proof. We will show the argument for $E$ an increasing exponential (its base $E(1)>1$): an identical argument applies to decreasing exponentials (only switching $U$ and $L$ in the computations below).

To show $E(x)$ is integrable, it is again enough to find a sequence $P_n$ of shrinking partitions where $limL(f,P_n)=limU(f,P_n)$. Indeed - for each $n$, let $P_n$ denote the evenly spaced partition of $[a,b]$ with widths ${\mathrm{\Delta }}_n=(b-a)/n$ $$P_n=\{a,a+{\mathrm{\Delta }}_n,a+2{\mathrm{\Delta }}_n,\cdots ,a+n{\mathrm{\Delta }}_n=b\}$$

We will begin by computing the lower sum. Because $E$ is continuous, it achieves a maximum and minimum value on each interval $P_i=[t_i,t_{i+1}]$. And, since $E$ is monotone increasing, this value occurs at the leftmost endpoint. Thus,

$$\begin{array}{rl}L(E,P_n)& =\sum _{0\le i<n}\underset{P_i}{inf}\{E(x)\}|P_i|\\ & =\sum _{0\le i<n}E(t_i){\mathrm{\Delta }}_n\\ & =\sum _{0\le i<n}E(a+i{\mathrm{\Delta }}_n){\mathrm{\Delta }}_n\end{array}$$

Using the law of exponents for $E$ we can simplify this expression somewhat:

$$\begin{array}{rl}E(a+i{\mathrm{\Delta }}_n)& =E(a)E(i{\mathrm{\Delta }}_n)\\ & =E(a)E({\mathrm{\Delta }}_n+{\mathrm{\Delta }}_n+\cdots +{\mathrm{\Delta }}_n)\\ & =E(a)E({\mathrm{\Delta }}_n)E({\mathrm{\Delta }}_n)\cdots E({\mathrm{\Delta }}_n)\\ & =E(a)E({\mathrm{\Delta }}_n{)}^i\end{array}$$

Plugging this back in and factoring out the constants, we see that the summation is actually a partial sum of a geometric series:

$$\begin{array}{rl}\sum _{0\le i<n}E(a+i{\mathrm{\Delta }}_n){\mathrm{\Delta }}_n& =\sum _{0\le i<n}E(a)E({\mathrm{\Delta }}_n{)}^i{\mathrm{\Delta }}_n\\ & =E(a){\mathrm{\Delta }}_n\sum _{0\le i<n}E({\mathrm{\Delta }}_n{)}^i\end{array}$$

Having previously derived the formula for the partial sums of a geometric series, we can write this in closed form:

$$\sum _{0\le i<n}E({\mathrm{\Delta }}_n{)}^i=\frac{1-E({\mathrm{\Delta }}_n{)}^n}{1-E({\mathrm{\Delta }}_n)}$$

But, we can simplify even further! Using again the laws of exponents we see that $E({\mathrm{\Delta }}_n{)}^n$ is the same as $E(n{\mathrm{\Delta }}_n)$, and $n{\mathrm{\Delta }}_n$ is nothing other than the width of our entire interval, so $b-a$. Thus the numerator becomes $1-E(b-a)$, and putting it all together yields a simple expression for $L(E,P_n)$:

$$L(E,P_n)=E(a){\mathrm{\Delta }}_n\frac{1-E(b-a)}{1-E({\mathrm{\Delta }}_n)}$$

Some algebraic re-arrangement is beneficial: first, note that by the laws of exponents we have

$$\begin{array}{rl}E(a)(1-E(b-a))& =E(a)-E(b-a)E(a)\\ & =E(a)-E(b)\end{array}$$

Thus for every $n$ we have

$$L(E,P_n)=(E(a)-E(b))\frac{{\mathrm{\Delta }}_n}{1-E({\mathrm{\Delta }}_n)}$$

We are interested in the limit as $n\to \mathrm{\infty }$: by the limit laws we can pull the constant $E(a)-E(b)$ out front, and only concern ourselves with the fraction involving ${\mathrm{\Delta }}_n$. There’s one final trick: look at the negative reciprocal of this fraction:

$$\frac{-1}{\frac{{\mathrm{\Delta }}_n}{1-E({\mathrm{\Delta }}_n)}}=\frac{E({\mathrm{\Delta }}_n)-1}{{\mathrm{\Delta }}_n}$$

Because we know $E(0)=1$ for all exponentials, this latter term is none other than the difference quotient defining the derivative for $E$! Since we have proven $E$ to be differentiable, we know that evaluating this along any sequence converging to zero yields the derivative at zero. And as ${\mathrm{\Delta }}_n\to 0$ this implies

$$lim\frac{E({\mathrm{\Delta }}_n)-E(0)}{{\mathrm{\Delta }}_n}=E^{\prime }(0)$$

Thus, our original limit ${\mathrm{\Delta }}_n/(1-E({\mathrm{\Delta }}_n))$ is the negative reciprocal of this, and

$$\begin{array}{rl}limL(E,P_n)& =lim(E(a)-E(b))\frac{{\mathrm{\Delta }}_n}{1-E({\mathrm{\Delta }}_n)}\\ & =(E(a)-E(b))lim\frac{{\mathrm{\Delta }}_n}{1-E({\mathrm{\Delta }}_n)}\\ & =(E(a)-E(b))\frac{-1}{E^{\prime }(0)}\\ & =\frac{E(b)-E(a)}{E^{\prime }(0)}\end{array}$$

Phew! That was a lot of work! Now we have to tackle the upper sum. But luckily this will not be nearly as bad: we can reuse most of what we’ve done! Since $E$ is monotone increasing, we know that the maximum on any interval occurs at the rightmost endpoint, so

$$\begin{array}{rl}U(E,P_n)& =\sum _{0\le i<n}\underset{P_i}{sup}\{E(x)\}|P_i|\\ & =\sum _{0\le i<n}E(t_{i+1}){\mathrm{\Delta }}_n\\ & =\sum _{0\le i<n}E(a+(i+1){\mathrm{\Delta }}_n){\mathrm{\Delta }}_n\end{array}$$

Comparing this with our previous expression for $L(E,P_n)$, we see (unsurprisingly) its identical except for a shift of $i\mapsto i+1$. The law of exponents turns this additive shift into a multiplicative one:

$$\begin{array}{rl}U(E,P_n)& =\sum _{0\le i<n}E(a+(i+1){\mathrm{\Delta }}_n){\mathrm{\Delta }}_n\\ & =\sum _{0\le i<n}E({\mathrm{\Delta }}_n)E(a+i{\mathrm{\Delta }}_n){\mathrm{\Delta }}_n\\ & =E({\mathrm{\Delta }}_n)\sum _{0\le i<n}E(a+i{\mathrm{\Delta }}_n){\mathrm{\Delta }}_n\\ & =E({\mathrm{\Delta }}_n)L(E,P_n)\end{array}$$

Thus, $U(E,P_n)=E({\mathrm{\Delta }}_n)L(E,P_n)$ for every $n$. Since $E$ is continuous, $$limE({\mathrm{\Delta }}_n)=E(lim{\mathrm{\Delta }}_n)=E(0)=1$$

And, as $L(E,P_n)$ converges (as we proved above) we can apply the limit theorem for products to get

$$\begin{array}{rl}limU(E,P_n)& =lim(E({\mathrm{\Delta }}_n)L(E,P_n))\\ & =(limE({\mathrm{\Delta }}_n))(limL(E,P_n))\\ & =limL(E,P_n)\\ & =\frac{E(b)-E(a)}{E^{\prime }(0)}\end{array}$$

Thus, the limits of our sequence of upper and lower bounds are equal! And, by the argument at the beginning of this proof, that squeezes $L(E)$ and $U(E)$ to be equal as well. Thus, $E$ is integrable on $[a,b]$ and its value is what we have squeezed:

$${\int }_{[a,b]}E=\frac{E(b)-E(a)}{E^{\prime }(0)}$$

Corollary 27.1 (Integrating the Natural Exponential) On any interval $[a,b]$ the natural exponential is integrable, and $${\int }_{[a,b]}\exp =\exp (b)-\exp (a)$$

27.3 Trigonometric Functions

Theorem 27.1 For $x\in [0,\pi /2]$, the sine function is integrable and $${\int }_{[0,x]}\sin =1-\cos (x)$$

Proof. On the interval $[0,x]$, we use the sequence of evenly spaced shrinking partitions $P_n$ of width $\mathrm{\Delta }=x/n$, and prove integrability by showing $limL(\sin ,P_n)=limU(\sin ,P_n)$. Because $\sin$ is monotonically increasing on $[0,\pi /2]$ on any subinterval $I=[a,b]$ that $m=\sin a$ and $M=\sin b$. Thus $$L(\sin ,P_n)=\sum _{i=1}^n\sin ((i-1)\mathrm{\Delta })\mathrm{\Delta }$$ $$U(\sin ,P_n)=\sum _{i=1}^n\sin \left(i\mathrm{\Delta }\right)\mathrm{\Delta }$$

Using $\sin (0)=0$ we see the sums agree except for the final term of $U$, meaning

$$U(\sin ,P_n)-L(\sin ,P_n)=\sin (n\mathrm{\Delta })\mathrm{\Delta }=\sin (x)\frac{x}{n}$$

As $x$ is a fixed constant this tends to zero as $n\to \mathrm{\infty }$, so $\sin$ is integrable on $[0,x]$ and we can compute its value as the limit of either the upper or lower sum.
We use the identity for $\sum _{1\le k\le n}\sin kx$ proven in ?exr-summing-angles: $$U(\sin ,P_n)=\sum _{1\le i\le n}\sin (i\mathrm{\Delta })\mathrm{\Delta }=\frac{\sin \left(\frac{n}{2}\mathrm{\Delta }\right)\sin \left(\frac{n+1}{2}\mathrm{\Delta }\right)}{\sin \left(\frac{\mathrm{\Delta }}{2}\right)}\mathrm{\Delta }$$

Substituting back $\mathrm{\Delta }=x/n$ and re-arranging,

$$U(\sin ,P_n)=\frac{\sin \left(\frac{x}{2}\right)\sin \left(\frac{n+1}{n}\frac{x}{2}\right)}{\sin \left(\frac{x}{2n}\right)}\frac{x}{n}=\frac{\sin \left(\frac{x}{2}\right)\sin \left(\frac{n+1}{n}\frac{x}{2}\right)}{\frac{\sin \left(x/2n\right)}{x/n}}$$

We evaluate the limit as $n\to \mathrm{\infty }$ using the limit laws. The numerator is immediate $$\sin \left(\frac{x}{2}\right)\sin \left(\frac{n+1}{n}\frac{x}{2}\right)\mapsto \sin \left(\frac{x}{2}\right)\sin \left(\frac{x}{2}\right)$$ using that $\frac{n+1}{n}\to 1$ and the continuity of $\sin$. For the denominator, we use the fact that $\frac{\sin x}{x}\to 1$ (?cor-sinc) to see

$$\frac{\sin \left(\frac{x}{2n}\right)}{x/n}=\frac{1}{2}\frac{\sin \left(\frac{x}{2n}\right)}{\frac{x}{2n}}\to \frac{1}{2}$$

Thus

$$limU(\sin ,P_n)=\frac{\sin \left(\frac{x}{2}\right)\sin \left(\frac{x}{2}\right)}{\frac{1}{2}}=2{\sin }^2\frac{x}{2}$$

Using the half-angle identity Exercise 20.11, we can rewrite this

$$limU(\sin ,P_n)=2\frac{1-\cos (x)}{2}=1-\cos x$$

As we’ve already shown $\sin$ to be integrable, this limit of upper sums over a sequence of shrinking partitions gives the value:

$${\int }_{[0,x]}\sin =1-\cos (x)$$

We can leverage this result and the symmetries of the sine function to calculate the integral over arbitrary intervals:

Exercise 27.4 Prove that $\sin$ is integrable on the interval $[\pi /2,\pi ]$ and for any $x\in [\pi /2,\pi ]$ $${\int }_{[\frac{\pi }{2},x]}\sin =-\cos (x)$$

Hint: proceed either (1) directly, using the fact that $\sin$ is decreasing on this interval or (2) using the above, and the symmetry $\sin (\pi /2+x)=\sin (\pi /2-x)$.

Use this and subdivision to show for any $x\in [0,\pi ]$, $${\int }_{[0,x]}\sin =1-\cos x$$

Corollary 27.2 $${\int }_{[0,\pi /2]}\sin =1\text{and}{\int }_{[0,\pi ]}\sin =2$$

Exercise 27.5 Use the fact that sine is an odd function and integrable on $[0,\pi ]$ to show $\sin$ is integrable on $[-\pi ,0]$ and for any $x\in [-\pi ,0]$ $${\int }_{[x,0]}\sin =\cos (x)-1$$

Again by subdivision we can conclude that $\sin$ is integrable on $[-\pi ,\pi ]$.

Proposition 27.4 Let $a,b\in [-\pi ,\pi ]$. Then $\sin$ is integrable on $[a,b]$ and $${\int }_{[a,b]}\sin =\cos (a)-\cos (b)$$

Proof. We proceed by cases depending on the location of $a,b$. If both are positive and lie in $[0,\pi /2]$ we evaluate using Exercise 27.4

$$\begin{array}{rl}{\int }_{[a,b]}\sin & ={\int }_{[0,b]}\sin -{\int }_{[0,a]}\sin \\ & =(1-\cos b)-(1-\cos a)\\ & =\cos a-\cos b\end{array}$$

A similar calculation applies if $a,b<0$. If $a<0$ and $b>0$ we evaluate as

$$\begin{array}{rl}{\int }_{[a,b]}\sin & ={\int }_{[a,0]}\sin +{\int }_{[0,b]}\sin \\ & =(\cos a-1)+(1-\cos b)\\ & =\cos a-\cos b\end{array}$$

Corollary 27.3 $${\int }_{[-\pi ,\pi ]}\sin =0$$

Since $\sin$ is $2\pi$ periodic this is enough to conclude that $\sin$ is in fact integrable on any interval

Theorem 27.2 (Integrating sine) Let $a<b$. Then $\sin$ is integrable on $[a,b]$ and $${\int }_{[a,b]}\sin =\cos (a)-\cos (b)$$

Exercise 27.6 Prove this.

This work has immediate payoff for integrating cosine as well, since we know it to be just a shifted version of the sine:

Theorem 27.3 (Integrating cosine) Let $a<b$. Then $\cos$ is integrable on $[a,b]$ and $${\int }_{[a,b]}\cos =\sin (b)-\sin (a)$$

Exercise 27.7 Prove Theorem 27.3 using that $\sin (x+\pi /2)=\cos (x)$ and $\cos (x+\pi /2)=-\sin (x)$ (?exr-trig-shift).