# 1.2 The definite integral  (Page 2/16)

 Page 2 / 16

Previously, we discussed the fact that if $f\left(x\right)$ is continuous on $\left[a,b\right],$ then the limit $\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}f\left({x}_{i}^{*}\right)\text{Δ}x$ exists and is unique. This leads to the following theorem, which we state without proof.

## Continuous functions are integrable

If $f\left(x\right)$ is continuous on $\left[a,b\right],$ then f is integrable on $\left[a,b\right].$

Functions that are not continuous on $\left[a,b\right]$ may still be integrable, depending on the nature of the discontinuities. For example, functions with a finite number of jump discontinuities on a closed interval are integrable.

It is also worth noting here that we have retained the use of a regular partition in the Riemann sums. This restriction is not strictly necessary. Any partition can be used to form a Riemann sum. However, if a nonregular partition is used to define the definite integral, it is not sufficient to take the limit as the number of subintervals goes to infinity. Instead, we must take the limit as the width of the largest subinterval goes to zero. This introduces a little more complex notation in our limits and makes the calculations more difficult without really gaining much additional insight, so we stick with regular partitions for the Riemann sums.

## Evaluating an integral using the definition

Use the definition of the definite integral to evaluate ${\int }_{0}^{2}{x}^{2}dx.$ Use a right-endpoint approximation to generate the Riemann sum.

We first want to set up a Riemann sum. Based on the limits of integration, we have $a=0$ and $b=2.$ For $i=0,1,2\text{,…,}\phantom{\rule{0.2em}{0ex}}n,$ let $P=\left\{{x}_{i}\right\}$ be a regular partition of $\left[0,2\right].$ Then

$\text{Δ}x=\frac{b-a}{n}=\frac{2}{n}.$

Since we are using a right-endpoint approximation to generate Riemann sums, for each i , we need to calculate the function value at the right endpoint of the interval $\left[{x}_{i-1},{x}_{i}\right].$ The right endpoint of the interval is ${x}_{i},$ and since P is a regular partition,

${x}_{i}={x}_{0}+i\text{Δ}x=0+i\left[\frac{2}{n}\right]=\frac{2i}{n}.$

Thus, the function value at the right endpoint of the interval is

$f\left({x}_{i}\right)={x}_{i}^{2}={\left(\frac{2i}{n}\right)}^{2}=\frac{4{i}^{2}}{{n}^{2}}.$

Then the Riemann sum takes the form

$\sum _{i=1}^{n}f\left({x}_{i}\right)\text{Δ}x=\sum _{i=1}^{n}\left(\frac{4{i}^{2}}{{n}^{2}}\right)\frac{2}{n}=\sum _{i=1}^{n}\frac{8{i}^{2}}{{n}^{3}}=\frac{8}{{n}^{3}}\sum _{i=1}^{n}{i}^{2}.$

Using the summation formula for $\sum _{i=1}^{n}{i}^{2},$ we have

$\begin{array}{cc}\sum _{i=1}^{n}f\left({x}_{i}\right)\text{Δ}x\hfill & =\frac{8}{{n}^{3}}\sum _{i=1}^{n}{i}^{2}\hfill \\ \\ \\ \\ & =\frac{8}{{n}^{3}}\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right]\hfill \\ & =\frac{8}{{n}^{3}}\left[\frac{2{n}^{3}+3{n}^{2}+n}{6}\right]\hfill \\ & =\frac{16{n}^{3}+24{n}^{2}+n}{6{n}^{3}}\hfill \\ & =\frac{8}{3}+\frac{4}{n}+\frac{1}{6{n}^{2}}.\hfill \end{array}$

Now, to calculate the definite integral, we need to take the limit as $n\to \infty .$ We get

$\begin{array}{cc}{\int }_{0}^{2}{x}^{2}dx\hfill & =\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}f\left({x}_{i}\right)\text{Δ}x\hfill \\ \\ \\ & =\underset{n\to \infty }{\text{lim}}\left(\frac{8}{3}+\frac{4}{n}+\frac{1}{6{n}^{2}}\right)\hfill \\ & =\underset{n\to \infty }{\text{lim}}\left(\frac{8}{3}\right)+\underset{n\to \infty }{\text{lim}}\left(\frac{4}{n}\right)+\underset{n\to \infty }{\text{lim}}\left(\frac{1}{6{n}^{2}}\right)\hfill \\ & =\frac{8}{3}+0+0=\frac{8}{3}.\hfill \end{array}$

Use the definition of the definite integral to evaluate ${\int }_{0}^{3}\left(2x-1\right)dx.$ Use a right-endpoint approximation to generate the Riemann sum.

6

## Evaluating definite integrals

Evaluating definite integrals this way can be quite tedious because of the complexity of the calculations. Later in this chapter we develop techniques for evaluating definite integrals without taking limits of Riemann sums. However, for now, we can rely on the fact that definite integrals represent the area under the curve, and we can evaluate definite integrals by using geometric formulas to calculate that area. We do this to confirm that definite integrals do, indeed, represent areas, so we can then discuss what to do in the case of a curve of a function dropping below the x -axis.

## Using geometric formulas to calculate definite integrals

Use the formula for the area of a circle to evaluate ${\int }_{3}^{6}\sqrt{9-{\left(x-3\right)}^{2}}dx.$

The function describes a semicircle with radius 3. To find

${\int }_{3}^{6}\sqrt{9-{\left(x-3\right)}^{2}}dx,$

we want to find the area under the curve over the interval $\left[3,6\right].$ The formula for the area of a circle is $A=\pi {r}^{2}.$ The area of a semicircle is just one-half the area of a circle, or $A=\left(\frac{1}{2}\right)\pi {r}^{2}.$ The shaded area in [link] covers one-half of the semicircle, or $A=\left(\frac{1}{4}\right)\pi {r}^{2}.$ Thus,

$\begin{array}{}\\ \\ {\int }_{3}^{6}\sqrt{9-{\left(x-3\right)}^{2}}\hfill & =\frac{1}{4}\pi {\left(3\right)}^{2}\hfill \\ & =\frac{9}{4}\pi \hfill \\ & \approx 7.069.\hfill \end{array}$

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