Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Average Values and Lengths of Functions

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

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:

$8\pi$

$16\pi$

$10\pi$

$2\pi$

$4\pi$

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.

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

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

Report an Error

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

$\frac{4}{5}$

ln(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

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

Report an Error

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

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

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

Report an Error

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:

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

$\frac{56}{27}$

$4\sqrt2-1$

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

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.

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

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

Report an Error

Example Question #1 : 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!

Report an Error

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

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

Report an Error

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:

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

Report an Error

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:

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

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

Report an Error

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:

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

Report an Error

Example Question #10 : 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:

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

e^3-1

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

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

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:

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

Report an Error

View Calculus 2 Tutors

Satweka
Certified Tutor

The University of Texas at Dallas, Master of Science, Biomedical Engineering.

View Calculus 2 Tutors

Jeremy
Certified Tutor

University of Illinois at Chicago, Bachelor of Science, Physics.

View Calculus 2 Tutors

Reuben
Certified Tutor

University of Alberta, Bachelor of Science, Mathematics.

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Calculus Tutors in San Diego, Reading Tutors in New York City, ISEE Tutors in Dallas Fort Worth, LSAT Tutors in Chicago, GRE Tutors in New York City, Computer Science Tutors in New York City, MCAT Tutors in Miami, MCAT Tutors in Washington DC, Statistics Tutors in Dallas Fort Worth, Computer Science Tutors in Dallas Fort Worth

Popular Courses & Classes

Spanish Courses & Classes in Washington DC, ISEE Courses & Classes in Philadelphia, SSAT Courses & Classes in Phoenix, SSAT Courses & Classes in Philadelphia, ISEE Courses & Classes in Seattle, ACT Courses & Classes in Dallas Fort Worth, GMAT Courses & Classes in Philadelphia, MCAT Courses & Classes in Boston, ISEE Courses & Classes in Chicago, SSAT Courses & Classes in New York City

Popular Test Prep

SSAT Test Prep in Philadelphia, MCAT Test Prep in Denver, GMAT Test Prep in San Francisco-Bay Area, LSAT Test Prep in Phoenix, ACT Test Prep in Dallas Fort Worth, GMAT Test Prep in New York City, ISEE Test Prep in Philadelphia, LSAT Test Prep in San Francisco-Bay Area, LSAT Test Prep in Miami, LSAT Test Prep in Boston

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

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 #1 : 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 #10 : Average Values And Lengths Of Functions

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: