Applications of integrals

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We give more contexts to understand integrals.

Velocity and displacement, speed and distance

Some values include ‘‘direction’’ that is relative to some fixed point.

$v(t)$ is the velocity of an object at time $t$ . This represents the ‘‘change in position’’ at time $t$ .
$s(t)$ is the position of an object at time $t$ . This gives location with respect to the origin. If we can assume that $s(a) = 0$ , then $s(t) = \int _a^t v(x) \d x.$
$s(b) -s(a)$ is the displacement, the distance between the starting and finishing locations.

On the other hand speed and distance are values without ‘‘direction.’’

$|v(t)|$ is the speed.
$\int _a^b |v(t)| \d t$ is the distance traveled.

Consider a particle whose velocity at time $t$ is given by $v(t) = \sin (t)$ .

What is the displacement of the particle from $t=0$ to $t=\pi$ ? That is, compute: $\int _0^\pi \sin (t) \d t \begin {prompt}= \answer [given]{2}\end {prompt}$ What is the displacement of the particle from $t=0$ to $t=2\pi$ ? That is, compute: $\int _0^{2\pi } \sin (t) \d t \begin {prompt}= \answer [given]{0}\end {prompt}$ What is the distance traveled by the particle from $t=0$ to $t=\pi$ ? That is, compute: $\int _0^\pi \left | \sin (t) \right | \d t \begin {prompt}= \answer [given]{2}\end {prompt}$ What is the distance traveled by the particle from $t=0$ to $t=2\pi$ ? That is, compute: $\int _0^{2\pi } \left | \sin (t) \right | \d t \begin {prompt}=\answer [given]{4}\end {prompt}$

Average value

Conceptualizing definite integrals as ‘‘signed area’’ works great as long as one can actually visualize the ‘‘area.’’ In some cases, a better metaphor for integrals comes from the idea of average value. Looking back to your days as an even younger mathematician, you may recall the idea of an average: $\frac {f_1+f_2+\dots +f_n}{n} = \frac {1}{n}\sum _{k=1}^n f_i$ If we want to know the average value of a function, a naive approach might be to partition the interval $[a,b]$ into $n$ equally spaced subintervals, $a=x_0 < x_1 < \dots < x_{n}=b$ and choose any $x_k^*$ in $[x_i,x_{i+1}]$ . The average of $f(x_1^*)$ , $f(x_2^*)$ , …, $f(x_n^*)$ is: $\frac {f(x_1^*) + f(x_2^*) + \dots + f(x_n^*)}{n} = \frac 1n\sum _{k=1}^n f(x_k^*).$ Multiply this last expression by $1 = \frac {(b-a)}{(b-a)}$ :

$\begin{align*} \frac1n \sum_{k=1}^n f(x_k^*)\frac{(b-a)}{(b-a)} &= \sum_{k=1}^n f(x_k^*)\frac1n \frac{(b-a)}{(b-a)} \\ &= \frac{1}{b-a} \sum_{k=1}^n f(x_k^*)\frac{b-a}n \\ &=\frac{1}{b-a} \sum_{k=1}^n f(x_k^*)\Delta x \end{align*}$ where $\Delta x = (b-a)/n$ . Ah! On the right we have a Riemann Sum! Now take the limit as $n\to \infty$ : $\lim _{n\to \infty } \frac {1}{b-a} \sum _{k=1}^n f(x_k^*)\Delta x = \frac {1}{b-a} \int _a^b f(x)\d x.$ This leads us to our next definition:

Let $f$ be continuous on $[a,b]$ . The average value of $f$ on $[a,b]$ is given by $\frac {1}{b-a}\int _a^b f(x)\d x.$

The average value of a function gives the height of a single rectangle whose area is equal to $\int _a^b f(x) \d x$

An application of this definition is given in the next example.

An object moves back and forth along a straight line with a velocity given by $v(t) = (t-1)^2$ on $[0,3]$ , where $t$ is measured in seconds and $v(t)$ is measured in ft/s.

What is the average velocity of the object?

By our definition, the average velocity is: $\begin{align*} \frac{1}{3-0}\int_0^3 (t-1)^2\d t &=\frac13 \int_0^3 t^2-2t+1\d t\\ &= \frac13\eval{\frac{t^3}{3}-t^2+t}_0^3\\ &= 1\text{ ft/s}. \end{align*}$

When we take the average of a finite set of values, it does not matter how we order those values. When we are taking the average value of a function, however, we need to be more careful.

For instance, there are at least two different ways to make sense of a vague phrase like ‘‘The average height of a point on the unit semi circle’’

One way we can make sense of ‘‘The average height of a point on the unit semi circle’’ is to compute the average value of the function $f(x) =\sqrt {1-x^2}$ on the interval $[-1,1]$ .

Compute the average value of the function $f(x) =\sqrt {1-x^2}$ on the interval $[-1,1]$ .

By definition, we wish to compute $\frac {1}{2}\int _{-1}^1 \sqrt {1-x^2} \d x.$ Computing this integral geometrically, we find the average value to be $\answer [given]{\frac {\pi }{4}}$ .

Another way we can make sense of ‘‘The average height of a point on the unit semi circle’’ is the average value of the function $g(\theta ) =\sin (\theta )$ on $[0,\pi ]$ , since $\sin (\theta )$ is the height of the point on the unit circle at the angle $\theta$ .

Compute the average value of the function $g(\theta ) =\sin (\theta )$ on the interval $[0,\pi ]$ .

By definition, we wish to compute $\frac {1}{\pi } \int _0^\pi \sin (\theta ) \d \theta .$ Computing this integral geometrically, we find the average value to be $\answer [given]{\frac {2}{\pi }}$ .

See if you can understand intuitively why the average using $f$ should be larger than the average using $g$ .

Mean value theorem for integrals

Just as we have a Mean Value Theorem for Derivatives, we also have a Mean Value Theorem for Integrals.

The Mean Value Theorem for integrals Let $f$ be continuous on $[a,b]$ . There exists a value $c$ in $[a,b]$ such that $\int _a^bf(x)\d x = f(c)(b-a).$

This is an existential statement. The Mean Value Theorem for Integrals tells us:

The average value of a continuous function is in the range of the function.

We demonstrate the principles involved in this version of the Mean Value Theorem in the following example.

Consider $\int _0^\pi \sin x \d x$ . Find a value $c$ guaranteed by the Mean Value Theorem.

We first need to evaluate $\int _0^\pi \sin x\d x$ . $\int _0^\pi \sin x\d x =\eval {-\cos x}_0^\pi = 2.$ Thus we seek a value $c$ in $[0,\pi ]$ such that $\pi \sin c =2$ . $\pi \sin c = 2\ \ \Rightarrow \ \ \sin c = 2/\pi \ \ \Rightarrow \ \ c = \arcsin (2/\pi ) \approx 0.69.$

A graph of $\sin x$ is sketched along with a rectangle with height $\sin (0.69)$ are pictured below. The area of the rectangle is the same as the area under $\sin x$ on $[0,\pi ]$ .