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. $s(t) = s(a)+\int _a^t v(z) dz.$
$s(b) -s(a)$ 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.
$\int _a^b |v(t)| dt$ 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: $\int _0^\pi \sin (t) dt \begin {prompt}= \answer [given]{2}\end {prompt}$ What is the displacement of the particle from $t=0$ to $t=2\pi$ ? That is, compute: $\int _0^{2\pi } \sin (t) dt \begin {prompt}= \answer [given]{0}\end {prompt}$ What is the distance traveled by the particle from $t=0$ to $t=\pi$ ? That is, compute: $\int _0^\pi \left | \sin (t) \right | dt \begin {prompt}= \answer [given]{2}\end {prompt}$ What is the distance traveled by the particle from $t=0$ to $t=2\pi$ ? That is, compute: $\int _0^{2\pi } \left | \sin (t) \right | dt \begin {prompt}=\answer [given]{4}\end {prompt}$

Change in the amount

We can apply the Fundamental Theorems of Calculus to a variety of problems where both accumulation and rate of change play important roles. For example, we can consider a tank that is being filled with fuel at some rate. Given the rate, we can ask what is the amount of fuel in the tank at a certain time. Or, a tank that is being emptied at a given rate, or a culture of bacteria growing in a Petri dish, or a population of a city, etc. This brings us to our next theorem.

Let $A(t)$ denote the amount of some substance/population at the time $t$ .

Assume that the function $A$ is continuous, differentiable and its derivative, $A'$ , continuous on some time interval $[a,b]$ .

Then, the change in the amount $A$ over the time interval $[a,b]$ is given by $A(b)-A(a)=\int _a^bA'(t) dt.$

The equation follows directly from the Second Theorem of Calculus. The left hand side obviously gives the change in the amount $A$ over the time interval $[a,b]$ , which is the difference between the amount at the end of the time interval and the amount at the beginning of the time interval.

The right hand side of the equation gives the “accumulation of a rate”. You can think of this definite integral as the limit of Riemann sums, where each Riemann sum is the“sum of the changes of the amount over small intervals of time”.

An experiment is conducted in which a culture of bacteria is grown in a Petri dish. The initial population was estimated at 1200 cells. The growth rate of the population is estimated at $P'(t)=100e^{-0.2 t}$ cells/day.

(a): By how much has the population grown during the first three days of the experiment?
The growth of population during the first three days of the experiment is given by: $\begin{align*} P(3)-P(0) &=\int_0^3 P'(t) dt\\ &=\int_0^3 100 e^{-0.2 t} dt\\ &=\bigg[ \frac{100}{-0.2}e^{-0.2 t} \bigg]_0^3\approx 226 \end{align*}$ During the first three days, the population grew approximately by 226 cells.
(b): Compute the right Riemann sum of the function $P'$ and the interval $[0,3]$ with $n=3$ . What does this Riemann sum approximate? Is this approximation an underestimate or an overestimate and why?
This Riemann sum approximates the change in population during the first three days of the experiment. We expect it to be an underestimate, because the function $P'$ is decreasing and this is a right Riemann sum.
Since $n=3$ , $\Delta t=1$ .
$\begin{align*} P(3)-P(0) &\approx\sum_{k=1}^3P'(k)\\ &=P'(1)+P'(2)+P'(3)\\ &=100e^{-0.2 }+100e^{-0.4}+100e^{-0.6}\\ &\approx 204 \end{align*}$
(c): Find the population at any time $t\ge 0$ .
$\begin{align*} P(t)-P(0)&=\int_0^t P'(z) dz\\ &=\int_0^t 100 e^{-0.2 z} dz\\ &=\bigg[ \frac{100}{-0.2}e^{-0.2 z} \bigg]_0^t\\ &=500\left(1-e^{-0.2 t}\right) \end{align*}$ Therefore, for $t\ge 0$ , $P(t)=P(0)+500\left (1-e^{-0.2 t}\right )=1700 -500e^{-0.2 t}$

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: $\frac {f_1+f_2+\dots +f_n}{n} = \frac {1}{n}\sum _{k=1}^n f_k$ If we want to know $\overline {f}$ , the average value of a function on the interval $[a,b]$ , a naive approach might be to introduce $n+1$ equally spaced grid points on the interval $[a,b]$ $a=x_0 < x_1 < \dots < x_{n}=b$ and choose a sample point $x_k^*$ in each interval $[x_k,x_{k+1}]$ , $k=1,\dots ,n$ .

We will approximate the average value of $f$ on the interval $[a,b]$ with the average of $f(x_1^*)$ , $f(x_2^*)$ , …, and $f(x_n^*)$ : $\overline {f}\approx \frac {f(x_1^*) + f(x_2^*) + \dots + f(x_n^*)}{n} = \frac 1n\sum _{k=1}^n f(x_k^*).$ 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!

What will happen as $n\to \infty$ ?

We take the limit as $n\to \infty$ : $\overline {f}=\lim _{n\to \infty } \frac {1}{b-a} \sum _{k=1}^n f(x_k^*)\Delta x = \frac {1}{b-a} \int _a^b f(x) dx.$ 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 $\overline {f}= \frac {1}{b-a}\int _a^b f(x) dx.$

Multiplying this equation by $(b-a)$ , we obtain that $\overline {f}(b-a)= \int _a^b f(x) dx.$ If $f$ is positive, the average value of a function gives the height of a single rectangle whose area is equal to $\int _a^b f(x) dx.$

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 $v(t)$ in ft/s.

What is the average velocity of the object?

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

Choose all the correct expressions for $\overline {v}$ , the average velocity of an object moving along a straight line over the time interval $[a,b]$ .

(Reminder: $s$ is the position function, and $a$ the acceleration).

$\overline {v}=\frac {1}{b-a}\int _a^b s(t) dt$ $\overline {v}=\frac {1}{b-a}\int _a^b v(t) dt$ $\overline {v}=\frac {1}{b-a}\int _a^b a(t) dt$ $\overline {v}=\frac {a(b)-a(a)}{b-a}$ $\overline {v}=\frac {v(b)-v(a)}{b-a}$ $\overline {v}=\frac {s(b)-s(a)}{b-a}$

We know from before that $\overline {v}=\frac {s(b)-s(a)}{b-a}.$ We also know that the position function is given by $s(t)=s(a)+\int _a^tv(z) dz.$ Therefore, $\overline {v}=\frac {s(b)-s(a)}{b-a}=\frac {s(a)+\int _a^bv(t) dt-s(a)}{b-a}=\frac {1}{b-a}\int _a^bv(t) dt.$ Therefore, we can compute the average velocity using either the position function or the velocity function.

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 $\frac {1}{2}\int _{-1}^1 \sqrt {1-x^2} dx.$ 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 $\int _a^bf(x) dx = f(c)(b-a).$

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.

Proof: Define an accumulation (area) function, $F$ , $F(x)= \int _a^x f(t) dt$ Since $F$ is continuous on the interval $[a,b]$ and differentiable on the interval $(a,b)$ , we can apply the Mean Value Theorem to the function $F$ on the interval $[a,b]$ . Therefore, there exist a number $c$ in $(a,b)$ such that $F'(c)=\frac {F(b)-F(a)}{b-a}$ But we know that $F'(c)=f(c)$ , and that $F(b)-F(a)=\int _a^bf(t) dt =\int _a^bf(x) dx$ . Therefore, $f(c)=\frac {1}{b-a}\int _a^bf(x) dx$ $\blacksquare$

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

Let $f(x)=\sin {x}$ . Consider $\int _0^\pi \sin x dx$ . Find a value $c$ guaranteed by the Mean Value Theorem.

We first need to evaluate $\int _0^\pi \sin x dx$ . $\int _0^\pi \sin x\ dx =\bigg [ -\cos x \bigg ]_0^\pi = 2.$ Thus we seek a value $c$ in $[0,\pi ]$ such that $\pi \sin c =2$ . $\pi \sin c = 2\ \ \Rightarrow \ \ \sin c = 2/\pi \ \ \Rightarrow \ \ c_1 = \arcsin (2/\pi ) \approx 0.69.$ There is another such point: $c_2=\pi -0.69$ . Indeed, $f( \pi -0.69)=\sin {(\pi -0.69)}=\sin {(0.69)}=\frac {2}{\pi }=\overline {f}.$ A graph of $\sin x$ , along with a rectangle with height $\sin (0.69)$ , is pictured below. The area of the rectangle is the same as the area under $\sin x$ on $[0,\pi ]$ . The two points $c_1$ and $c_2$ are also shown in the figure. Recall, $f(c_1)=f(c_2)=\overline {f}$ .