The definite integral

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Definite integrals compute net area.

The process of approximating areas under curves led to the notion of a Riemann sum

$A\approx \sum _{k=1}^n f(x_k^*)\Delta x,$ where $f$ is a nonnegative, continuous function on the interval $[a,b]$ , and

$x_k^*$ is a sample point for the $k^{th}$ rectangle, $k=1,2,..., n$ .

The limit of Riemann sums, as $n\to \infty$ , gives the exact area between the curve $y=f(x)$ and the interval on the $x-$ axis:

$A=\lim _{n\to \infty }\sum _{k=1}^n f(x_k^*)\Delta x.$ It does not matter whether we consider only right Riemann sums, or left Riemann sums, or midpoint Riemann sums, or others: the limit of any kind of Riemann sum as $n\to \infty$ is equal to the area, as long as $f$ is nonnegative and continuous on $[a,b]$ .

What happens when a continuous function $f$ assumes negative values on the interval $[a,b]$ ?

We can still form Riemann sums and take the limit.

The question is: What is the meaning of a Riemann sum in that case?

The graph of the function $f$ on the interval $[0,10]$ is given in the figure below.

The figure illustrates a right Riemann sum with $n=5$ rectangles for $f$ on $[0,10]$ . $\sum _{k=1}^5 f(x_k^*)\Delta x= f(2)\Delta x+ f(4)\Delta x+ f(6)\Delta x+f(8)\Delta x+ f(10)\Delta x$

So, the Riemann sum is the sum of signed areas of rectangles: rectangles that lie above the $x$ -axis contribute positive values, and rectangles that lie below the $x$ -axis contribute negative values to the Riemann sum.

$\begin{align*} \sum_{k=1}^5 f(x_k^*)\Delta x= &\left(\underbrace{f(2)\Delta x}_{\text{nonnegative}}+ \underbrace{f(4)\Delta x}_{\text{nonnegative}}+ \underbrace{f(6)\Delta x}_{\text{nonnegative}}\right)\\ &+ \left(\underbrace{f(8)\Delta x}_{\text{negative}}+ \underbrace{f(10)\Delta x}_{\text{negative}}\right) \end{align*}$

When we take the limit of Riemann sums, it seems that we should get that

$\lim _{n\to \infty }\sum _{k=1}^n f(x_k^*)\Delta x= A_1+(-A_2),$ where the areas $A_1$ and $A_2$ are depicted in the figure below.

In other words, the limit of Riemann sums seems to be equal to the sum of signed areas of the regions that lie entirely above or below the $x$ -axis. The signed area of regions that lie above the $x$ -axis is positive, and the signed area of regions that lie below the $x$ -axis is negative.

This sum of signed areas is called the net area of the region between the graph of $f$ and the interval on the $x$ - axis .

The limit of Riemann sums will exist for any continuous functions on the interval $[a,b]$ , even if $f$ assumes negative values on $[a,b]$ . The limit of Riemann sums gives the net area of the region between the graph of $f$ and an interval on the $x$ - axis.

This leads to the following definition.

Let $f$ be a function which is continuous on the interval $[a,b]$ . We define the definite integral of $f$ on $[a,b]$ by $\int _a^b f(x) \d x=\lim _{n\to \infty }\sum _{k=1}^n f(x_k^*)\Delta x.$

The definite integral is a number that gives the net area of the region between the curve $y=f(x)$ and the $x$ -axis on the interval $[a,b]$ .

The graph a function $f$ on the interval $[0,9]$ is given in the figure. The areas of four regions that lie either above or below the $x$ -axis are labeled in the figure.

Consider the integral $\int _0^9 f(x) \d x$ Express the integral in terms of areas $A_1$ , $A_2$ , $A_3$ and $A_4$ .

$\begin{align*} \int_0^9 f(x) \d x&= -A_1+A_2-A_3+A_4\\ \end{align*}$

Consider the integral $\int _0^{10}-x \d x.$ Compute the integral in two ways:

(a): by interpreting the integral as the net area of the region between the curve $y=-x$ and the interval $[0,10]$ on the $x$ -axis;
(b): using the definition of the definite integral, i.e. by computing the limit of Riemann sums.

(a)

The area between the $x$ -axis and the curve can be easily computed, since it is the area of a triangle.

$A=\frac {1}{2}\cdot 10\cdot 10=\answer [given]{50}.$

Then, it follows that

$\int _0^{10}-x \d x=-A$

(b)

We use the definition of the definite integral and write $\int _0^{10}-x \d x=\lim _{n\to \infty }\sum _{k=1}^n f(x_k^*)\Delta x=\lim _{n\to \infty }\sum _{k=1}^n -x_k^*\Delta x$ It does not matter what type of a Riemann sum we use, so we choose a right Riemann sum.

$\int _0^{10}-x \d x=\lim _{n\to \infty }\sum _{k=1}^n -x_k\Delta x$ A right Riemann sum with $n=5$ is illustrated in the figure below.

We can apply the constant multiple rule for sums and limits:

$\begin{align*} \int_0^{10}-x \d x &=\lim_{n\to\infty}\sum_{k=1}^n -x_k\Delta x\\ &=\lim_{n\to\infty}\left(-\sum_{k=1}^n x_k\Delta x\right)\\ &=-\lim_{n\to\infty}\sum_{k=1}^n x_k\Delta x \end{align*}$

We can now finish our computation of the limit of right Riemann sums.

$\begin{align*} \int_0^{10}-x \d x&=-\lim_{n\to\infty}\sum_{k=1}^n x_k\Delta x\\ &=-\lim_{n\to\infty}\sum_{k=1}^n k\cdot\frac{10}{n}\cdot\frac{10}{n}\\ &=-\lim_{n\to\infty}\left(\frac{10}{n}\right)^2\sum_{k=1}^n k\\ &=-\lim_{n\to\infty}\left(\frac{10}{n}\right)^2\frac{n(n+1)}{2}\\ &=-\lim_{n\to\infty}50\frac{n+1}{n}\\ &=\answer[given]{-50}\\ \end{align*}$

(a)

Express the limit as a definite integral. $\lim _{n\to \infty } \sum _{k=1}^n \left (\sqrt {1-\left (-1+\frac {2k}{n}\right )^2}\right ) \left (\frac {2}{n}\right )$

(b)

Compute this limit: $\lim _{n\to \infty } \sum _{k=1}^n \left (\sqrt {1-\left (-1+\frac {2k}{n}\right )^2}\right ) \left (\frac {2}{n}\right )$

(i): This is a limit of Riemann sums! Specifically, it is a limit of Riemann sums of $n$ rectangles, where $\Delta x = \answer [given]{\frac {2}{n}},$
$x_k^* = -1+\frac {2k}{n},$ $a= x_0^* = \answer [given]{-1},$ and $b=x_n^* = \answer [given]{1}.$ Hence, we may rewrite this as $\lim _{n\to \infty } \sum _{k=1}^n \left (\sqrt {1-(x_k^*)^2}\right )=\int _{\answer [given]{-1}}^{\answer [given]{1}}\sqrt {1-x^2} \d x.$
(ii): The limit computes the area between the $x$ -axis and the curve
$y = \sqrt {1-x^2}$ . Let’s see it:

By geometry, we know that this semicircle has area $\answer [given]{\pi /2}$ . Hence $\lim _{n\to \infty } \sum _{k=1}^n \left (\sqrt {1-(x_k^*)^2}\right )\Delta x =\answer [given]{\pi /2}.$

The definite integral computes the net area (sum of signed areas) between $y=f(x)$ and the $x$ -axis on the interval $[a,b]$ .

Consider the graph of a function $f$ on the interval $[0,5]$ .

Compute the definite integral $\int _0^5 f(x) \d x.$

The net area in the figure above is given by $\int _0^5 f(x) \d x=A_1-A_2$

The areas $A_1$ and $A_2$ are easily computed, since both are the areas of triangles. Therefore, $\int _0^5 f(x) \d x=\answer [given]{1.25}$

There is more to this example, than just a value of the integral.

Notice that $\int _0^3 f(x) \d x= A_1,$ $\int _3^5 f(x) \d x=-A_2,$ and $\int _0^5 f(x) \d x=A_1-A_2.$ It follows that $\int _0^5 f(x) \d x= \int _0^3 f(x) \d x+ \int _3^5 f(x) \d x.$

This looks like a property of the definite integral. Are there other properties?

Properties of the definite integral Let $f$ and $g$ be defined on a closed interval $[a,b]$ that contains the value $c$ , and let $k$ be a constant. The following hold:

(a): $\int _a^a f(x)\d x = 0$
(b): $\int _a^c f(x)\d x + \int _c^b f(x)\d x = \int _a^b f(x)\d x$
(c): $\int _a^bf(x)\d x = -\int _b^a f(x)\d x$
(d): $\int _a^bk\cdot f(x)\d x = k\cdot \int _a^bf(x)\d x$
(e): $\int _a^b \left (f(x)\pm g(x)\right )\d x = \int _a^bf(x)\d x \pm \int _a^bg(x)\d x$

We will address each property in turn:

(a): Here, there is no “area under the curve” when the region has no width; hence this definite integral is $0$ .
(b): This states that total area is the sum of the areas of subregions. Here a picture is worth a thousand words:

It is important to note that this still holds true even if $a<b<c$ . We discuss this in the next point.
(c): For now, this property can be viewed as merely a convention to make other properties work well. However, later we will see how this property has a justification all its own.
(d): This states that when one scales a function by, for instance, $7$ , the area of the enclosed region also is scaled by a factor of $7$ .
(e): This states that the integral of the sum is the sum of the integrals.

Due to the geometric nature of integration, geometric properties of functions can help us compute integrals.

Compute: $\int _0^6 |x-3| \d x$

This may seem difficult at first. Perhaps the first thing to do is look at a graph of $y=x-3$ :

Now we can graph $y=|x-3|$ :

Recall, $|x-3|$ is in fact a piece-wise defined function

$|x-3| = \begin {cases} x-3 & \text {if $3\le x $}\\ -(x-3)&\text {if $x<3$}. \end {cases}$

Now we see that we really have two triangles, each with base $3$ and height $3$ . Hence

$\begin{align*} \int_0^6 |x-3| \d x &= \int_0^3 \answer[given]{3-x} \d x + \int_3^6 \answer[given]{x-3} \d x\\ &= \frac{3\cdot 3}{2} + \frac{3\cdot 3}{2}\\ &=\answer[given]{9}. \end{align*}$

A function $f$ is an odd function if $f(-x) = -f(x),$ and a function $g$ is an even function if $g(-x) = g(x).$

The names odd and even come from the fact that these properties are shared by functions of the form $x^n$ where $n$ is either odd or even. For example, if $f(x) = x^3$ , then $f(-7) = -f(7),$ and if $g(x) = x^4$ , then $g(-7) = g(7).$ Geometrically, even functions have symmetry with respect to the $y-$ axis . Cosine is an even function:

On the other hand, odd functions have $180^\circ$ rotational symmetry around the origin. Sine is an odd function:

Let $f$ be an odd function defined for all real numbers. Compute: $\int _{-2}^2 f(x) \d x \begin {prompt}=\answer {0}\end {prompt}$

Since our function is odd, it must look something like:

The integral above computes the following net area:

Let $f$ be an odd function defined for all real numbers. Which of the following are equal to $\int _2^4 f(x) \d x ?$

$\int _{4}^{2} f(x) \d x$ $\int _{-4}^{-2} f(x) \d x$ $\int _{-2}^{-4} f(x) \d x$ $\int _{-2}^{4} f(x) \d x$ $\int _{4}^{-2} f(x) \d x$ $\int _{2}^{-4} f(x) \d x$ $\int _{-4}^{2} f(x) \d x$ $-\int _{-4}^{2} f(x) \d x$ $-\int _{-4}^{-2} f(x) \d x$

Net area versus geometric area

We know that the net area of the region between a curve $y=f(x)$ and the $x$ -axis on $[a,b]$ is given by $\int _a^b f(x) \d x.$ On the other hand, if we want to know the geometric area, meaning the “actual” area, we compute $\int _a^b |f(x)| \d x.$

The graph of a function $f$ is given in the figure below.

(a)

Express the geometric area of the region between the curve $y=f(x)$ and the $x$ -axis on the interval $[0,9]$ as a definite integral.

(b)

Express the geometric area of the region between the curve $y=f(x)$ and the $x$ -axis on the interval $[0,9]$ in terms of definite integrals of $f$ .

(c)

Express the geometric area of the region between the curve $y=f(x)$ and the $x$ -axis on the interval $[0,9]$ in terms of areas $A_1$ , $A_2$ , $A_3$ and $A_4$ .

(i): The geometric area is given by the integral $\int _0^9 |f(x)| \d x$ The figure below depicts the graph of $|f|$ .
(ii): Using the properties of definite integrals, we can write $\begin{align*} \int_0^9 |f(x)| \d x = \int_0^1 |f(x) |\d x &+ \int_1^5 |f(x)| \d x\\ &+ \int_5^8 |f(x)| \d x\\ &+ \int_8^9 |f(x)| \d x \end{align*}$ Rewriting the absolute value signs: $\begin{align*} = \int_0^1 (-f(x)) \d x &+ \int_1^5 f(x) \d x\\ &+ \int_5^8 (-f(x)) \d x\\ &+ \int_8^9 f(x) \d x \end{align*}$ Pulling out negative signs: $\begin{align*} = -\int_0^1 f(x) \d x &+ \int_1^5 f(x) \d x\\ &- \int_5^8 f(x) \d x\\ &+ \int_8^9 f(x) \d x \end{align*}$
(iii): $\int _0^9 |f(x)| \d x =A_1 +A_2+A_3+A_4$

True or false: $\int _a^b |f(x)| \d x = \left |\int _a^b f(x) \d x\right |$

true false

Consider $f(x) = x$ on the interval $[-1,1]$ . Here $\int _a^b |f(x)| \d x = 1 \qquad \text {but}\qquad \left |\int _a^b f(x) \d x\right | = 0.$