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.

1 Applications of Definite Integrals

1.1 Rectilinear Motion

Suppose that an object is moving along a straight path. The position, velocity, and acceleration functions governing the motion of the object can be derived from each other in the following ways.
If we start with the position function, $s(t)$ , we can compute velocity and acceleration by differentiating: $v(t) = s'(t)\quad \text {and} \quad a(t) = v'(t)$ On the other hand, if we start with the acceleration function, $a(t)$ , we can compute velocity and position by integrating: $v(t) = \int a(t) \ dt \quad \text {and} \quad s(t) = \int v(t) \ dt$ These indefinite integrals will contain a constant of integration, $C$ , which can be determined by using the initial conditions of the problem.

In the following proposition, we see that definite integrals that can be used to compute the displacement and distance traveled by an object moving along a straight path.

Suppose that an object moving along a straight path has instantaneous velocity $v(t)$ , then

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

example 1 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 $t = 2$ .

Part (a): 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: $v(t) = \int a(t) \ dt = \int -32 \ dt = -32t + C$ 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 that $C = 48$ . Thus, $v(t) = -32t + 48.$ 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, $s(t) = -16t^2 + 48t + 16$

Part (b): The displacement from $t= 0$ to $t = 2$ , is $s(2) - s(0)$ . We can use the formula for $s(t)$ to compute this: $s(2) - s(0) = \left (-16(2)^2 + 48(2) + 16\right ) - \left (-16(0)^2 + 48(0) + 16\right ) = 32 \text { feet}$ Alternatively, we can compute the displacement using the proposition and the fact that $v(t) = -32t + 48$ . The displacement from $t= 0$ to $t = 2$ , is given by $s(2) - s(0) = \int _0^2 v(t) \, dt = \left (-16t^2 + 48t\right )\bigg |_0^2 = 32 \text { feet}$

Part (c): To find the total distance traveled from $t = 0$ to time $t = 2$ , we need to 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 $v(1.5) = -32(1.5) + 48 = 0$ , we can see that $|v(t)| = |-32t + 48| = \left \{\begin {array}{rc} -32t + 48 &\ \text {if} \ t \leq 1.5 \\ 32t - 48 &\ \text {if} \ t > 1.5 \end {array} \right .$

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

(problem 1a) 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 $\text {distance traveled} = \answer {80} \ ft.$

(problem 1b) 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 $\text {maximum height} =\answer {9} \ ft.$

1.2 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 $f_{ave} = \frac {1}{b-a}\int _a^b f(x) \ dx.$

example 2 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 \\ & = \frac 12\int _0^2 x^2 \ dx \\ & = \frac{x^3}{6} \Bigg |_0^2 \\ & = \tfrac 43 \end{align*}

(problem 2) Compute the average value of $f(x) = x(x+1) +1$ on the interval $[0,3]$ .

Multiply first, then integrate

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

example 3 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*}

(problem 3) Compute the average value of $f(x) = \cos (2x) + \sin (2x)$ on the interval $[0,\pi /2]$ .

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

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

example 4 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 \\ & = \frac 12\int _{-1}^1 e^{2x} \ dx \\ & = \frac 14 e^{2x} \Bigg |_{-1}^1 \\ & = \frac{e^2 - e^{-2}}{4} \end{align*}

(problem 4) Compute the average value of $f(x) = e^{3x}$ on the interval $[-1,1]$ . $f_{ave} = \answer {\frac {e^{3} - e^{-3}}{6}}.$

2025-04-18 17:05:57