4.8 Applications of Definite Integrals

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In this section we use definite integrals to study rectilinear motion and compute average value.

Applications of Definite Integrals

Average Value

Average Value If $f(x)$ is continuous on the interval $[a,b]$ , then the average value of $f(x)$ on $[a,b]$ is given by $f_{ave} = \frac {1}{b-a}\int _a^b f(x) \ dx.$

Compute the average value of $f(x) = x^2$ on the interval $[0,2]$ .
The average value is given by $\begin{align*} f_{ave} &= \frac{1}{b-a}\int_a^b f(x) \ dx \\ & = \frac12\int_0^2 x^2 \ dx \\ & = \frac{x^3}{6} \Bigg|_0^2 \\ & = \tfrac43 \end{align*}$

Compute the average value of $f(x) = x(x+1) +1$ on the interval $[0,3]$ .

Multiply first, then integrate

$f_{ave} = \answer {11/2}.$

Compute the average value of $f(x) = \sin (x)$ on the interval $[0,\pi ]$ .
The average value is given by $\begin{align*} f_{ave} &= \frac{1}{b-a}\int_a^b f(x) \ dx \\ & = \frac{1}{\pi}\int_0^\pi \sin(x) \ dx \\ & = -\frac{1}{\pi} \cos(x) \Bigg|_0^\pi \\ & = \frac{2}{\pi} \end{align*}$

Compute the average value of $f(x) = \cos (2x) + \sin (2x)$ on the interval $[0,\pi /2]$ .

An anti-derivative of $\cos (2x) + \sin (2x)$ is $\tfrac 12\sin (2x) - \tfrac 12\cos (2x)$

$f_{ave} = \answer {2/\pi }.$

Compute the average value of $f(x) = e^{2x}$ on the interval $[-1,1]$ .
The average value is given by $\begin{align*} f_{ave} &= \frac{1}{b-a}\int_a^b f(x) \ dx \\ & = \frac12\int_{-1}^1 e^{2x} \ dx \\ & = \frac14 e^{2x} \Bigg|_{-1}^1 \\ & = \frac{e^2 - e^{-2}}{4} \end{align*}$

Compute the average value of $f(x) = e^{3x}$ on the interval $[-1,1]$ . $f_{ave} = \answer {\frac {e^{3} - e^{-3}}{6}}.$

Rectilinear Motion

Suppose that a projectile is launched vertically into the air. Then, the height of the object at time $t$ , the velocity at time $t$ , and the acceleration at time $t$ are related as follows: $v(t) = s'(t)$ and $a(t) = v'(t) = s''(t).$

We can also use indefinite integrals to express these relationships: $s(t) = \int v(t) \ dt$ and $v(t) = \int a(t) \ dt.$

Suppose that a vertical projectile has instantaneous velocity $v(t)$ , then

(a): the displacement over the interval $[a,b]$ is given by $\int _a^b v(t) \ dt$ and
(b): the distance traveled over the interval $[a,b]$ is given by $\int _a^b |v(t)| \ dt.$

An object is launched vertically from a height of $16 ft$ with an initial velocity of $48 ft/sec$ .

(a): Find a formula, $s(t)$ , for the height of the object at time $t$ seconds.
(b): Find the displacement of the object from time $t = 0$ to time $t = 2$ .
(c): Find the total distance traveled from time $t = 0$ to time $t = 2$ .

Assuming that $a(t) - -32 ft/sec^2$ , as it is on the surface of the earth, we can compute $v(t)$ by using an indefinite integral: $v(t) = \int a(t) \ dt = \int -32 \ dt = -32 + C.$ To find the value of the constant $C$ , we use the fact that the initial velocity was given as $48 ft/sec$ : $v(0) = -32(0) + C = 48,$ which implies $C = 48$ . Thus $v(t) = -32t + 48.$ Next, we repeat the process to find $s(t)$ : $s(t) = \int v(t) \ dt = \int (-32t + 48) \ dt = -16t^2 + 48t + C.$ To find $C$ , we use the given information about the initial height of the object: $s(0) = -16t^2 + 48(0) + C = 16,$ which implies $C = 16$ . Thus, $s(t) = -16t^2 + 48t + 16.$ Next, to find the displacement from $t= 0$ to $t = 2$ , we compute $s(2) - s(0) = (-64+96+16)-(16) = 32 \text {feet}.$ Note that we could have computed $s(2) - s(0)$ as a definite integral: $\int _0^2 v(t) \ dt = s(2) - s(0),$ since $s(t)$ is an anti-derivative of $v(t)$ . Finally, to find the total distance traveled from $t = 0$ to time $t = 2$ , we need integrate $|v(t)|$ . To get a handle on $|v(t)|$ , recall $|x| = \left \{ \begin {array}{rc} x & \ \text {if} \ x \geq 0 \\ -x & \ \text {if} \ x <0 \end {array} \right .$ Noting that $v(1.5) = 0$ , we can see that $|v(t)| = \left \{\begin {array}{rc} -32t + 48 &\ \text {if} \ t \leq 1.5 \\ 32t - 48 &\ \text {if} \ t > 1.5 \end {array} \right .$

Hence, the total distance traveled from $t=0$ to $t = 2$ is $\int _0^2 |v(t)| \ dt = \int _0^{1.5} (-32t + 48) \ dt + \int _{1.5}^2 (32t -48) \ dt = (36)+(4) = 40 \ \text {feet}.$

An object is launched vertically from a height of $24 ft$ with an initial velocity of $64 ft/sec$ .

(a): Find a formula, $s(t)$ , for the height of the object at time $t$ seconds.
(b): Find the maximum height reached by the object.
(c): Find the displacement of the object from time $t = 0$ to time $t = 3$ .
(d): Find the total distance traveled from time $t = 0$ to time $t = 3$ .

Assume $g = -32 \ ft/sec^2$ . $s(t) = \answer {-16t^2 + 64t + 24},$
$\text {maximum height} =\answer {88} \ ft,$
$\text {displacement} =\answer {48} \ ft,$
and $\text {distance traveled} = \answer {80} \ ft.$

An object is launched vertically from a height of $5 ft$ with an initial velocity of $16 ft/sec$ .

(a): Find a formula, $s(t)$ , for the height of the object at time $t$ seconds.
(b): Find the maximum height reached by the object.

Assume $g = -32 \ ft/sec^2$ . $s(t) = \answer {-16t^2 + 16t + 5},$
and $\text {maximum height} =\answer {9} \ ft,$