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Wize University Calculus 2 Textbook > Applications of Integrals

Basic Applications of Integrals

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Average Value of a Function

The average value of a function f(x)f(x) on the interval [a,b] is defined by 1b−a∫abf(x) dx\displaystyle \frac{1}{b-a}\int_a^bf\left(x\right)\ dx

Example
Find the average value of y=x2y=x^2 over [1,3][1,3]
Avg=13−1∫13x2dx\displaystyle \text{Avg}=\frac{1}{3-1}\int_1^3x^2dx
Avg=12[x33]13\displaystyle \text{Avg}=\frac{1}{2}\left[\frac{x^3}{3}\right]_1^3
Avg=12[9−13]\displaystyle \text{Avg}=\frac{1}{2}\left[9-\frac{1}{3}\right]
Avg=133\displaystyle \text{Avg}=\frac{13}{3}
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Example: Average Value

Compute the average value of f(x)=3x2+6x+3\displaystyle f(x)=3x^2+6x+3 on the interval [2,3][2,3].

Average value =13−2∫23f(x)dx=∫23(3x2+6x+3)dx=[x3+3x2+3x]23=(27+27+9)−(8+12+6)=(54+9)−(26)=37\begin{aligned} \text{Average value } &= \dfrac{1}{3-2}\int_2^3f(x)dx\\ &= \int_2^3(3x^2+6x+3)dx\\ &= \left[x^3+3x^2+3x\right]_2^3\\ &= (27+27+9) - (8+12+6)\\ &= (54+9)-(26)\\ &= 37\\ \end{aligned}

Find the average value of the function f(x)=1x2f\left(x\right)=\frac{1}{x^2} over the interval [1, 3]\left[1,\ 3\right].

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Displacement, Velocity, and Acceleration


Example

A vehicle's acceleration is given by the function a(t)=3t2−ta\left(t\right)=3t^2-t, where tt is given in seconds.
a.) If the vehicle's initial velocity is 7m/s, find a function that represents the velocity of the vehicle.
v(t)=∫a(t) dtv\left(t\right)=\int_{ }^{ }a\left(t\right)\ dt
v(t)=∫(3t2−t) dtv\left(t\right)=\int_{ }^{ }\left(3t^2-t\right)\ dt
v(t)=t3−t22+cv\left(t\right)=t^3-\frac{t^2}{2}+c
We sub in the initial velocity to solve for c:
7=03−022+c   →   c=77=0^3-\frac{0^2}{2}+c\ \ \ \to\ \ \ c=7
Therefore, the velocity of the vehicle is given by the function v(t)=t3−t22+7v\left(t\right)=t^3-\frac{t^2}{2}+7

b.) What is the vehicle's displacement between t=0t=0 and t=2t=2s?
d(t)=∫v(t) dtd\left(t\right)=\int_{ }^{ }v\left(t\right)\ dt
So, the displacement between 0 and 2 seconds is given by the definite integral
∫02v(t) dt\int_0^2v\left(t\right)\ dt
=∫02(t3−t22+7)dt=\int_0^2\left(t^3-\frac{t^2}{2}+7\right)dt
=[t44−t36+7t]02=\left[\frac{t^4}{4}-\frac{t^3}{6}+7t\right]_{_0}^{^2}
=[244−236+14]−[0−0+0]=\left[\frac{2^4}{4}-\frac{2^3}{6}+14\right]-\left[0-0+0\right]
=[4−43+14]=\left[4-\frac{4}{3}+14\right]
=16 23=16\ \frac{2}{3}m
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Rates of Change

In general, when we are given a function representing the rate of change of xx, we integrate to find the function representing xx.

Example

Air is entering a chamber at a rate of et2 L/s\frac{e^t}{2}\ L/s. The chamber initially holds 1L1L of air at t=0t=0, how long will it take for the air in the chamber to reach 5L?

To find the function representing the volume of air in the chamber, we integrate the rate of change function.
v(t)=∫et2dtv\left(t\right)=\int_{ }^{ }\frac{e^t}{2}dt
v(t)=et2+cv\left(t\right)=\frac{e^t}{2}+c

Substitute the initial volume to solve for c:
1=e02+c   →   c=121=\frac{e^0}{2}+c\ \ \ \to\ \ \ c=\frac{1}{2}
So, the volume function is v(t)=et2+12v\left(t\right)=\frac{e^t}{2}+\frac{1}{2}.

Set the volume equal to 5:
5=et2+125=\frac{e^t}{2}+\frac{1}{2}
92=et2\frac{9}{2}=\frac{e^t}{2}
9=et9=e^t
t=ln⁡9t=\ln9

Therefore, the air in the chamber will reach5L5L at t=ln⁡9 st=\ln9\ s.