The definite integral

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Definite integrals arise as the limits of Riemann sums, and compute net areas.

In the previous sections, we’ve been approximating the area under a curve by treating the region as a collection of rectangles.

Approximate the area under the graph of $f(x)=x^2$ on the interval $[0,1]$ using right-endpoints and $n$ rectangles, for $n$ a positive integer.

We’ll start with the widths: $\Delta x = \frac {b-a}{n} = \frac {1}{n}$ .

Since we’re using right-endpoints, the sample points are given by $x_k^* = a + k \Delta x = \frac {k}{n}$ .

The height of the $k$ -th rectangle is $f(x_k^*) = \left ( \frac {k}{n} \right )^2 = \frac {\answer [given]{k^2}}{n^2}$ .

The area of the $k$ -th rectangle is $f(x_k^*)\Delta x = \left (\frac {k^2}{n^2}\right )\left ( \frac {1}{n} \right ) = \frac {\answer [given]{k^2}}{\answer [given]{n^3}}$ .

The area approximation is given by the Riemann sum.

$\begin{align*} \sum_{k=1}^{n} f(x_k^*) \Delta x &= \sum_{k=1}^n \frac{k^2}{n^3}\\ &= \frac{1}{n^3} \sum_{k=1}^{n} \answer[given]{k^2} \quad \text{(We have a formula for this!)}\\ &= \frac{1}{n^3} \left( \frac{n(n+1)(2n+1)}{6}\right)\\ &= \frac{n(n+1)(2n+1)}{6n^3}\\ &= \frac{(n+1)(2n+1)}{6n^2} \end{align*}$

The approximate area is $\frac {(n+1)(2n+1)}{6n^2}$ .

The benefit of doing this calculation with $n$ still variable, is that it’s now easy to find the approximation for every possible value of $n$ ! If we wanted to approximate the area under $f(x)=x^2$ using $4$ rectangles, we could substitute $n=4$ into this formula. $\frac {(4+1)(2\cdot 4+1)}{6(4)^2} = \frac {15}{32}$ . If we wanted to use $10$ rectangles, we can substitute $n=10$ into it. $\frac {(10+1)(2\cdot 10+1)}{6(10)^2} = \frac {77}{200}$ .

Why would we want to be able to plug in different values of $n$ ? Remember how we made the approximation better and better? By taking more and more rectangles! The actual area is given in the limit as $n\to \infty$ .

Find the actual area under the graph of $f(x)=x^2$ on the interval $[0,1]$ .

We saw that the area approximation using $n$ rectangles was given by $A(n) = \frac {(n+1)(2n+1)}{6n^2}$ The actual area is given by $\begin{align*} \lim_{n\to\infty} A(n) &= \lim_{n\to\infty} \frac{(n+1)(2n+1)}{6n^2}\\ &= \lim_{n\to\infty} \frac{2n^2+3n+1}{6n^2} \\ &= \lim_{n\to\infty} \frac{4n+3}{12n} \quad \text{L'Hopital's Rule}\\ &= \lim_{n\to\infty} \frac{4}{12} \quad \text{L'Hopital's Rule}\\ &= \answer{\frac{1}{3}} \end{align*}$

All of the functions we have been examining have been positive. What changes if $f$ is negative in the region?

Let’s look at $f(x) = \cos (x)$ on $[0, \pi ]$ with $n=4$ using midpoints. $\Delta x = \frac {\pi }{4}$ , so the regular partition is $\left [ 0 \, , \, \frac {\pi }{4} \right ]$ , $\left [ \frac {\pi }{4} \, , \, \frac {\pi }{2} \right ]$ , $\left [ \frac {\pi }{2} \, , \, \frac {3\pi }{4} \right ]$ , and $\left [ \frac {3\pi }{4} \, , \, \pi \right ]$ , giving us midpoints $x_1^* = \frac {\pi }{8}$ , $x_2^* = \frac {3\pi }{8}$ , $x_3^* = \frac {3\pi }{8}$ , and $x_4^*=\frac {7\pi }{8}$ .

In the Riemann sum $f(x_1^*)\Delta x + f(x_2^*)\Delta x + f(x_3^*)\Delta x + f(x_4^*)\Delta x,$ the first two terms $f(x_1^*)\Delta x$ and $f(x_2^*)\Delta x$ are the areas of the first two rectangles. Notice that $f(x_3^*)$ and $f(x_4^*)$ are negative! They are not the heights of their rectangles. They are the opposite of the heights. That means $f(x_3^*)\Delta x$ and $f(x_4^*)\Delta x$ are not the areas of those rectangles. They are the opposite of those areas. The Riemann sum is adding the areas of the first two rectangles, then subtracting the areas of the last two rectangles.

This same situation occurs whenever our function is negative. The areas of the corresponding rectangles are being subtracted, not added.

The net area of a region is the area above the $x$ -axis minus the area below the $x$ -axis.

The Riemann sum is not calculating the approximate area, but the net area.

Compute the net area bounded by the graph of $f(x) = \cos (x)$ and the $x$ -axis between $x=0$ and $x=\pi$ .

Let’s look at the graph.

The area above the $x$ -axis and the area below the $x$ -axis are exactly the same. That means the net area is $\answer [given]{0}$ .

Remember that when we setup our Riemann sum to approximate the areas, we had a choice on what sample points to use. We could use left-endpoints, right-endpoints, midpoints, a combination of the three, or any other possible points we wished. Different choices of sample points leads to different values for the Riemann sums.

A function $f$ , defined on an interval $[a,b]$ , is called integrable on $[a,b]$ if $\displaystyle \lim _{n\to \infty } \sum _{k=1}^n f(x_k^*) \Delta x$ exists and has the same value for ALL choices of sample points, $x_k^*$ . If $f$ is integrable on $[a,b]$ , this limit is called the definite integral of $f$ from $x=a$ to $x=b$ , and denoted by $\int _a^b f(x) \d x$

To say it another way, if we have an integrable function, the difference between those sample points disappears once we take the limit, and this limit is the actual net area of the region.

Whenever we are working with an integrable function, the definite integral $\int _a^b f(x) \d x$ makes sense, but which functions are integrable? The next theorem tells us that most of the functions we deal with are.

If $f$ is continuous on $[a,b]$ , then $f$ is integrable on $[a,b]$ . If $f$ is bounded on $[a,b]$ with only finitely many discontinuities, then $f$ is integrable on $[a,b]$ .

Consider the following graph of $y=f(x)$ :

Compute:

(a): $\int _0^3 f(x) \d x \begin {prompt}= \answer {1.5}\end {prompt}$
(b): $\int _3^5 f(x) \d x \begin {prompt}= \answer {-1}\end {prompt}$
(c): $\int _0^5 f(x) \d x \begin {prompt}= \answer {0.5}\end {prompt}$

Use the formula for the area of a triangle.

Remember, we are dealing with “signed” area here:

Evaluate: $\displaystyle \int _0^2 (x^3-2x) \d x$ .

$f(x)=x^3-2x$ is continuous everywhere, so it is integrable on $[0,2]$ . That means this definite integral is given by the limit of Riemann sums.

We start by computing the right Riemann sum using $n$ rectangles.

$\begin{align*} \Delta x &= \answer[given]{\frac{2}{n}} \\ \\ x_k^* &= \answer[given]{\frac{2k}{n}}\\ \\ f(x_k^*) &= \left( \frac{2k}{n}\right)^3 - 2\left(\frac{2k}{n}\right)\\ &= \frac{8}{n^3}k^3 - \frac{4}{n}k \\ \\ f(x_k^*)\Delta x &= \left(\frac{8}{n^3}k^3 - \frac{4}{n}k\right) \left( \frac{2}{n}\right)\\ &= \frac{16}{n^4}k^3 - \frac{8}{n^2}k \end{align*}$

$\begin{align*} \Delta x &= \frac{b-a}{n}\\ x_k^* &= a + k\Delta x \end{align*}$

The Riemann sum is given by: $\begin{align*} \sum_{k=1}^n f(x_k^*)\Delta x &= \sum_{k=1}^n \left(\frac{16}{n^4}k^3 - \frac{8}{n^2}k \right)\\ &= \sum_{k=1}^n \frac{16}{n^4}k^3 - \sum_{k=1}^n \frac{8}{n^2}k \\ &= \frac{16}{n^4} \sum_{k=1}^n \answer[given]{k^3} - \frac{8}{n^2}\sum_{k=1}^n \answer[given]{k} \\ &= \frac{16}{n^4} \left( \frac{n^2(n+1)^2}{4} \right) - \frac{8}{n^2}\left( \frac{n(n+1)}{2} \right) \\ &= \frac{16n^2(n+1)^2}{4n^4} - \frac{8n(n+1)}{2n^2} \\ &= \frac{4(n+1)^2}{n^2} - \frac{4(n+1)}{n} \end{align*}$

The definite integral is the limit of these.

$\begin{align*} \int_0^2 (x^3-2x) \d x &= \lim_{n\to \infty}\sum_{k=1}^n f(x_k^*)\Delta x \\ &= \lim_{n\to\infty} \left(\frac{4(n+1)^2}{n^2} - \frac{4(n+1)}{n}\right) \\ &= \lim_{n\to\infty} \frac{4(n+1)^2}{n^2} - \lim_{n\to\infty} \frac{4(n+1)}{n} \\ &= \answer[given]{4} - \answer[given]{4}\\ &= \answer[given]{0} \end{align*}$