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Calculus 2 : Average Values and Lengths of Functions

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Example Questions

Example Question #1 : Average Values And Lengths Of Functions

What is the arc length if x=4cos(t) , y=4sin(t) from $[0,2\pi]$ ?

Possible Answers:

$16\pi$

$2\pi$

$4\pi$

$10\pi$

$8\pi$

Correct answer:

$8\pi$

Explanation:

Write the formula for arc length.

$s=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$

Calculate the derivatives.

$\frac{dx}{dt}=-4sin(t)$

$\frac{dy}{dt}=4cos(t)$

Substitute the derivatives and the bounds into the integral.

$s=\int_{0}^{2\pi}\sqrt{(-4sin(t))^2+(4cos(t))^2}=\int_{0}^{2\pi}\sqrt{16sin^2(t)+16cos^2(t)}$

$s=\int_{0}^{2\pi} 4 dt= 4[t]_{0}^{2\pi} = 8\pi$

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Example Question #1 : Average Values And Lengths Of Functions

What is the average value of the function

$p(t)=\frac{4t}{t^2+1}$

from t = 0 to t = 2 ?

Possible Answers:

ln(2)-1

e^2-1

ln(5)

$\frac{4}{5}$

Correct answer:

ln(5)

Explanation:

The average value of a function p(t) from t=a to t=b is found with the integral

$\frac{1}{b-a}\int_{a}^{b}p(t)dt$ .

In this case, we must compute the value of the integral

$\frac{1}{2-0}\int_{0}^{2}\frac{4t}{t^2+1}dt=\frac{1}{2}\int_{0}^{2}\frac{4t}{t^2+1}dt$ .

A substitution makes this integral clearer. Let u=t^2+1 . Then du=2tdt . We should also rewrite the limits of integration in terms of u. When t = 0, u=1, and when t = 2, u = 5. Making these substitutions results in the integral

$\frac{1}{2}\int_{1}^{5}\frac{2du}{u}=\frac{2}{2}\int_{1}^{5}\frac{du}{u}=\int_{1}^{5}\frac{du}{u}$

Evaluating this integral using the fact that

$\int \frac{du}{u}=ln(u)+c$ yields

$\int_{1}^{5}\frac{du}{u}=ln(5)-ln(1)=ln(5)$

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Example Question #1 : Average Values And Lengths Of Functions

What is the average value of the function $f(\theta)=2cos(\theta)$ over the interval $0\le\theta\le\4\pi$ ?

Possible Answers:

$\frac{\sqrt2}{2}$

$\sqrt3$

Correct answer:

Explanation:

In general, the average value of a function f(x) over the interval [a, b] is

$\frac{1}{b-a}\int_{a}^{b}f(x)dx$

This means that the average value of $f(\theta)$ over $0\le\theta\le4\pi$ is

$\frac{1}{4\pi}\int_{0}^{4\pi}2cos(\theta)d\theta$ .

Since the antiderivative of $cos(\theta)$ is $\int cos(\theta)d\theta=sin(\theta)+c$ , the integral evaluates to

$\frac{1}{4\pi}\int_{0}^{4\pi}2cos(\theta)d\theta=\frac{2}{4\pi}sin(\theta)|_{0}^{4\pi}$

$=\frac{1}{2\pi}(sin(4\pi)-sin(0))=0$ .

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Example Question #1 : Average Values And Lengths Of Functions

What is the length of the curve $g(x)=x^{\frac{3}{2}}$ over the interval $0\le x\le4$ ?

Possible Answers:

$\frac{56}{27}$

$\frac{4\pi}{3}$

$\frac{8}{27}(10\sqrt{10}-1)$

$4\sqrt2-1$

Correct answer:

$\frac{8}{27}(10\sqrt{10}-1)$

Explanation:

The general formula for finding the length of a curve f(x) over an interval [a,b] is $\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx$

In this example, the arc length can be found by computing the integral

$\int_{0}^{4}\sqrt{1+(\frac{d}{dx}x^{\frac{3}{2}})^2}\ dx$ .

The derivative of $x^{\frac{3}{2}}$ can be found using the power rule, $\frac{d}{dx} x^n=nx^{n-1}$ , which leads to

$\int_{0}^{4}\sqrt{1+\left(\frac{3}{2}x^{\frac{1}{2}}\right)^2}\ dx=\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx$ .

At this point, a substitution is useful.

Let

$u=1+\frac{9}{4}x,\ du=\frac{9}{4}dx,\frac{4}{9}du=dx$ .

We can also express the limits of integration in terms of to simplify computation. When x=0, u=1 , and when x = 4, u=10 .

Making these substitutions leads to

$\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx=\int_{1}^{10}\frac{4}{9}\sqrt{u}du=\frac{4}{9}\int_{1}^{10}u^{\frac{1}{2}}du$ .

Now use the power rule, which in general is $\int x^n=\frac{x^{n+1}}{n+1}+c$ , to evaluate the integral.

$\frac{4}{9}\int_{1}^{10}u^{\frac{1}{2}}du=\frac{4}{9}\frac{u^{\frac{3}{2}}}{\frac{3}{2}}|_{1}^{10}$

$=\frac{8}{27}(10^{\frac{3}{2}}-1^\frac{3}{2})=\frac{8}{27}(10\sqrt{10}-1)$

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Example Question #2 : Average Values And Lengths Of Functions

Given the interval, find the average value of the following function:

f(x)=(2x+3)^2; [-1,2]

Possible Answers:

Correct answer:

Explanation:

When f is integrable on [a,b], the average value of f(x) on [a,b] is defined to be:

$f(x)_{average}=\frac{1}{b-a}\int_{a}^{b}f(x)dx$

For the problem statement, we are given f(x) and the intervals [a,b]. All that needs to be done is solving the integral over this interval and dividing the result by the difference between the two intervals.

So:

$f(x)_{average}=\frac{1}{(2)-(-1)}\int_{-1}^{2}(2x+3)^2dx=\frac{1}{3}\int_{-1}^{2}(2x+3)^2dx$

To solve this integral, we have two options. We can FOIL the terms out for (2x+3)^2 and solve the integral of the resulting polynomial, or we can use a simple u-substitution. Either way, the result will always be the same. We will try both ways, to prove that this holds true:

FOIL Method:

$\frac{1}{3}\int_{-1}^{2}(2x+3)^2dx=\frac{1}{3}\int_{-1}^{2}(4x^2+12x+9)dx$

$\frac{1}{3}[\frac{4}{3}x^3+6x^2+9x]_{-1}^{2}$

$\frac{1}{3}([\frac{4}{3}(2)^3+6(2)^2+9(2)]-[\frac{4}{3}(-1)^3+6(-1)^2+9(-1)])$

$\frac{1}{3}([\frac{32}{3}+24+18]-[-\frac{4}{3}+6-9])$

$\frac{1}{3}(57)=19$

This is one of the answer choices!

U-Substition Method:

$\frac{1}{3}\int_{-1}^{2}(2x+3)^2dx\rightarrow u=2x+3, du=2dx\rightarrow \frac{1}{2}du=dx$

Next, we must adjust the bounds of the new integral which will be in terms of u.

u(-1)=2(-1)+3=1

u(2)=2(2)+3=7

So the new integral becomes:

$\frac{1}{3}\int_{1}^{7}\frac{1}{2}u^2du=\frac{1}{6}\int_{1}^{7}u^2du$

$\frac{1}{6}[\frac{1}{3}u^3]_{1}^{7}=\frac{1}{18}[(7)^3-(1)^3]=19$

As you can see both methods results in the same answer!

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Example Question #1 : Average Values And Lengths Of Functions

What is the average of value of $y = e^{3x}$ between the intervals x = 0 and x = 2 ?

Possible Answers:

$\frac{e^6 - 1}{6}$

$\frac{e^6 + 1}{6}$

$\frac{e^6}{3}$

$\frac{e^6 + 1}{3}$

$\frac{e^6 - 1}{3}$

Correct answer:

$\frac{e^6 - 1}{6}$

Explanation:

When asked for the average value between intervals of form x = a and x = b then integral becomes

$\frac{1}{b-a}\int_{a}^{b}f(x)$ so in this case we can rewrite our average value problem as

$\frac{1}{2-0}\int_{0}^{2}e^{3x}$ .

The integral comes $\frac{1}{3}e^{3x}$ evaluated at x = 2 and x = 0 which simplifies to

$\frac{e^6 - 1}{3}$ (note that e^0 is 1).

However, since we are looking for the average value, we must divide the whole thing by b-a, which in this case is 2, so the final answer is

$\frac{e^6 - 1}{6}$ .

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Example Question #7 : Average Values And Lengths Of Functions

What is the average value of the function x^2 on the interval [0,2] ?

Possible Answers:

$\frac{8}{9}$

$\frac{4}{3}$

$\frac{8}{3}$

Correct answer:

$\frac{4}{3}$

Explanation:

The average value of a function over an interval is defined to be the integral of the function divided by the length of the interval.

The general rule for this type of integration is as follows.

$\frac{1}{b-a}\int_a^b x^n dx=\frac{1}{b-a}\left[\frac{x^{n+1}}{n+1}\right]\left |_a^b$

So we get:

$\frac{1}{2-0}\int_0^2 x^2dx=\frac{1}{2}\cdot \frac{8}{3}=\frac{4}{3}$

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Example Question #1 : Average Values And Lengths Of Functions

What is the average value of the function

f(x)=x^3-1

on the interval, [0,4] ?

Possible Answers:

$\frac{31}{2}$

Correct answer:

Explanation:

The average value of a function is found by taking the integral of the function over the interval and dividing by the length of the interval.

The general rule for this type of integration is as follows.

$\frac{1}{b-a}\int_a^b x^n dx=\frac{1}{b-a}\left[\frac{x^{n+1}}{n+1}\right]\left |_a^b$

So we have:

$\frac{1}{4-0}\int_{0}^4 (x^3-1)dx=\frac{1}{4}(x^4/4-x)|_{0}^4=\frac{1}{4}(4^4/4-4)$

$=\frac{1}{4}(64-4)=\frac{60}{4}=15$

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Example Question #9 : Average Values And Lengths Of Functions

What is the average value of the function $f(x)=\sin x$ on the interval $[0,\pi/2]$ ?

Possible Answers:

$\frac{2}{\pi}$

$\sqrt{2}/2$

$\frac{1}{\pi}$

Correct answer:

$\frac{2}{\pi}$

Explanation:

The average value of a function is found by taking the integral of the function over the interval and dividing by the length of the interval.

The general rule for this type of integration is as follows.

$\frac{1}{b-a}\int_a^b \sin x dx=\frac{1}{b-a}\left[-\cos x\right]\left |_a^b$

So we have:

$\frac{1}{\pi/2-0}\int_{0}^{\pi/2} \sin x dx=\frac{2}{\pi}(-\cos x)|_{0}^{\pi/2}=\frac{2}{\pi}$

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Example Question #2 : Average Values And Lengths Of Functions

What is the average value of the function $f(x)=e^{3x}$ on the interval [0,1] ?

Possible Answers:

$\frac{1}{3}(e^3-1)$

$\frac{1}{3}e^{3/2}$

$\frac{1}{3}e^3$

e^3-1

Correct answer:

$\frac{1}{3}(e^3-1)$

Explanation:

The average value of a function is found by taking the integral of the function over the interval and dividing by the length of the interval.

The general rule for this type of integration is as follows.

$\frac{1}{b-a}\int_a^b e^u dx=\frac{1}{b-a}\left[\frac{1}{\frac{du}{dx}}e^u\right]\left |_a^b$

So we have:

$\frac{1}{1-0}\int_{0}^1 e^{3x} dx=\left(\frac{1}{3}e^{3x}\right)|_{0}^1=\frac{1}{3}(e^{3}-1)$

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Calculus 2 : Average Values and Lengths of Functions

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Average Values And Lengths Of Functions

Example Question #1 : Average Values And Lengths Of Functions

Example Question #1 : Average Values And Lengths Of Functions

Example Question #1 : Average Values And Lengths Of Functions

Example Question #2 : Average Values And Lengths Of Functions

Given the interval, find the average value of the following function:

Example Question #1 : Average Values And Lengths Of Functions

Example Question #7 : Average Values And Lengths Of Functions

Example Question #1 : Average Values And Lengths Of Functions

Example Question #9 : Average Values And Lengths Of Functions

Example Question #2 : Average Values And Lengths Of Functions

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