Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 7

Applications of the Definite Integral

Objectives

When you have completed this unit, you should be able to

1. use integration to solve problems of
   1. rectilinear motion.
   2. net change.
   3. areas between curves.
   4. work.
   5. average of a function.

In this unit, we apply the definite integral to solve problems with applications in physics (rectilinear motion, net change, work and average of a function), and problems with a geometric application (area between curves).

Rectilinear Motion

If ${\displaystyle a\left(t\right)}$ is an acceleration function, then its antiderivative is the velocity function ${\displaystyle v\left(t\right)\text{,}}$ and the antiderivative of the velocity function ${\displaystyle v\left(t\right)}$ is the displacement function ${\displaystyle s\left(t\right)\text{.}}$ That is,

${\displaystyle v\left(t\right)=\int a\left(t\right)dt\text{ and }s\left(t\right)=\int v\left(t\right)dt.}$

So, the change of the velocity over the time interval ${\displaystyle \left[t_0,t_1\right]}$ is given by the definite integral

${\displaystyle {\int }_{t_0}^{t_1}a\left(t\right)dt;}$

and the displacement over the time interval ${\displaystyle \left[t_0,t_1\right]}$ is given by the definite integral

${\displaystyle {\int }_{t_0}^{t_1}v\left(t\right)dt={\int }_{t_0}^{t_1}{s}^{\prime }\left(t\right)dt=s\left(t_2\right)-s\left(t_1\right).}$

Since ${\displaystyle \left|v\left(t\right)\right|}$ is the speed, the distance traveled over the time interval ${\displaystyle \left[t_0,t_1\right]}$ is the definite integral

${\displaystyle {\int }_{t_0}^{t_1}\left|v\left(t\right)\right|dt.}$

Example 7.1. Spotting a police car, a driver hits the brakes to reduce his speed from $90\text{mi/h}$ to $60\text{mi/h}$ at a constant rate over a distance of $200\text{ft}$. What is the acceleration in ${\text{ft/s}}^2$? How long would it take for the driver to bring his car to a complete stop from $90\text{mi/h}$?

Let ${\displaystyle t_0}$ be the time the breaks are applied, and ${\displaystyle t_1}$ be the time the car’s velocity reaches 60 mi/h. Hence, $s(0)=0$, $v(t_0)=90\text{mi/h}=132\text{ft/s}$, and $v(t_1)=60\text{mi/h}=88\text{ft/s}$.

${\displaystyle v\left(t_1\right)-v\left(t_0\right)=88-132={\int }_0^{t_1}adt=a{t}_1.}$

Hence, ${\displaystyle a{t}_1=-44\text{,}}$ and

${\displaystyle s\left(t_1\right)-s\left(t_0\right)=200={\int }_0^{t_1}v\left(t\right)dt=v_0{t}_1+\frac{a}{2}{t}_1^2.}$

Solving for ${\displaystyle t_1\text{,}}$ we get

${\displaystyle 200=132t_1+\frac{-44}{2}{t}_1\text{ and }{t}_1=\frac{100}{55};}$

and therefore, ${\displaystyle a=-\frac{44}{t_1}=-\frac{121}{5}}$ ft/s ${\displaystyle {}^2\text{.}}$

The car comes to a complete stop when ${\displaystyle v\left(t\right)=0\text{,}}$ by Equation 7.1, above,

${\displaystyle 0=v\left(t\right)=132-\frac{121}{5}t.}$

Solving for ${\displaystyle t\text{,}}$ we get ${\displaystyle t=5.45\text{.}}$

The constant of deceleration is

$a=-{\displaystyle \frac{121}{5}}{\text{ft/s}}^2$,

and the time to a complete stop is $5.45\text{seconds}$.

Example 7.2. [See Exercise 68 on page 271 of the textbook.]

A car is traveling at $50\text{mi/h}$ when the brakes are fully applied, producing a constant deceleration of $22{\text{ft/s}}^2$. What is the distance covered before the car comes to a stop?

The deceleration is negative (the acceleration decreases); hence, ${\displaystyle a\left(t\right)=-22\text{.}}$ The distance traveled is ${\displaystyle s\left(t_2\right)-s\left(t_1\right)\text{,}}$ where ${\displaystyle t_2}$ is the instant when the car stops, and ${\displaystyle t_1=0}$ is the initial time when the brakes are applied.

We are looking for

${\displaystyle s\left(t_2\right)-s\left(t_1\right)={\int }_{t_1}^{t_2}v\left(t\right)dt.}$

Then, $v(t_1)=50\text{mi/h}$ and $v(t_1)=50(5280)/3600=220/3\text{ft/s}$. The car stops at the time ${\displaystyle t_2}$; thus, ${\displaystyle v\left(t_2\right)=0}$ and we have

${\displaystyle 0-\frac{220}{3}=v\left(t_2\right)-v\left(t_1\right)={\displaystyle {\int }_{t_1}^{t_2}a\left(t\right)dt={\displaystyle {\int }_0^{t_2}-22dt}}=-22\left(t_2-0\right)=-22t_2\text{.}}$

Solving for ${\displaystyle t_2\text{,}}$ we find that $t_2=10/3$. We need the velocity function, so

${\displaystyle v\left(t\right)=v_0-22t=\frac{220}{3}-22t,}$

and

${\displaystyle {\int }_0^{10∕3}-22t+\frac{220}{3}dt=-11t^2+\frac{220t}{3}{|}_0^{10∕3}=-11{\left(\frac{10}{3}\right)}^2+\frac{2200}{9}=\frac{1100}{9}\text{.}}$

The distance is approximatively $122.22\text{ft}$.

An object that is falling freely is also an example of uniformly accelerated motion; in this case, the acceleration ${\displaystyle g}$ due to gravity is approximately $9.8{\text{m/s}}^2$ or $32{\text{ft/s}}^2$. Furthermore, since we have chosen the up direction to be positive and the down direction to be negative, ${\displaystyle a\left(t\right)=-g\text{.}}$ It follows from Equations 7.1 and 7.2 that

${\displaystyle v\left(t\right)=v_0-gt}$ (7.3)

and

${\displaystyle s\left(t\right)=s_0+v_{0}t-\frac{g}{2}{t}^2.}$ (7.4)

Example 7.3. A penny is released from the CN tower at a point 300 m above the ground. Assuming that the free-fall model applies, how long does it take for the penny to hit the ground, and what is its speed at the time of impact?

We take ${\displaystyle g=9.8}$ m/s${\displaystyle {}^2\text{.}}$ Initially, we have ${\displaystyle s_0=300}$ and ${\displaystyle v_0=0\text{,}}$ so from Equation 7.4, above, ${\displaystyle s\left(t\right)=300-9.8\frac{t^2}{2}=300-4.9t^2\text{.}}$

The impact occurs when ${\displaystyle s\left(t\right)=0\text{.}}$ Solving the equation ${\displaystyle 300-4.9t^2=0}$ for ${\displaystyle t\text{,}}$ we find that ${\displaystyle t\approx 7.82}$ seconds; it takes the penny about ${\displaystyle 7.82}$ seconds to hit the ground.

By Equation 7.3, the velocity at this time is ${\displaystyle v\left(7.82\right)=-9.8\left(7.82\right)=-76.64}$ and the speed at the time of impact is ${\displaystyle \left|v\left(7.82\right)\right|=76.64}$ m/s.