Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Integral Applications

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 #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

Example Question #321 : Integrals

What is the average value of the function $f(x)=e^{2x}$ on the interval [0,5] ?

Possible Answers:

$e^{5}$

$\frac{1}{10}(e^{10}-1)$

$\frac{1}{10}e^{10}$

$\frac{1}{5}(e^{10}-1)$

Correct answer:

$\frac{1}{10}(e^{10}-1)$

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:

$\frac{1}{5-0}\int_{0}^5 e^{2x} dx=\frac{1}{5}\left(\frac{1}{2}e^{2x}\right)|_{0}^5=\frac{1}{10}(e^{10}-1)$

Report an Error

Example Question #12 : Average Values And Lengths Of Functions

What is the average value of the function f(x)=2x+3x^2 on the interval [0,2] ?

Possible Answers:

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:

$\frac{1}{2-0}\int_{0}^2 (3x^2+2x) dx=\frac{1}{2}(x^3+x^2)|_{0}^2=\frac{1}{2}(2^2+2^3)=\frac{12}{2}=6$

Report an Error

Example Question #321 : Integrals

Find the length L of the graph of the function f(x) on the given interval.

$f(x)=2\sqrt{2}x+7\sqrt{3}$

$3 \leq x \leq 5$

Possible Answers:

L=1

L=312

L=6

$L=4\sqrt{39}$

L=18

Correct answer:

L=6

Explanation:

To find the length of a line along a given function f(x) from a to b, utilize the equation

$L=\int^{b}_{a}\sqrt{1+[f'(x)]^{2}} dx$

Applying the equation for this problem, we first find the first derivative of the function:

$f'(x)=2\sqrt{2}$

Then, we use this expression in the above formula:

$L=\int^{5}_{3}\sqrt{1+(2\sqrt{2})^{2}} dx=L=\int^{5}_{3}3dx=3x\left.\begin{matrix} \\ \end{matrix}\right|^{5}_{3}=6$

Report an Error

Example Question #14 : Average Values And Lengths Of Functions

Let f(x) bet a function continuous on the defined below and having the derivative

$f'(x)=\sqrt{x^{4}-1}$

Find the length L of the graph of f(x) on the interval $1 \leq x \leq 4$

Possible Answers:

L=7

L=21

L=17

L=63

L=13

Correct answer:

L=21

Explanation:

To find the length of a line along a given function f(x) from a to b, utilize the equation

$L=\int^{b}_{a}\sqrt{1+[f'(x)]^{2}} dx$

Applying the equation for this problem and using the given expression for f'(x):

$L=\int^{4}_{1}\sqrt{1+(\sqrt{x^{4}-1})^{2}} dx=\int^{4}_{1}x^{2}dx=\frac{x^{3}}{3}\left.\begin{matrix} \\ \end{matrix}\right|^{4}_{1}=\frac{63}{3}=21$

Report an Error

Example Question #151 : Integral Applications

Find the average value of the function $y = x\sin(x)$ over the interval $[\frac{\pi}{2}, \pi]$ .

Possible Answers:

None of the other answers

$2\pi-3$

$\frac{\pi}{2}$

Correct answer:

None of the other answers

Explanation:

The correct answer is $\frac{2(\pi-1)}{\pi}$ .

To find the average value of the function, we use

Average value $= \frac{1}{b-a}\int_{a}^{b} x\sin(x) dx$ ,

with $a = \frac{\pi}{2}, b=\pi$ in this case.

So we have

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

$= \frac{2}{\pi}\int_{\pi/2}^{\pi} x\sin(x) dx$

Using integration by parts with $u = x, dv = \sin(x)dx$ and hence $du=dx, v = -\cos(x)$ , we have

$=\frac{2}{\pi}\{[-x\cos(x)]_{\pi/2}^{\pi}-\int_{\pi/2}^{\pi}-\cos(x)dx\}$

$=\frac{2}{\pi}\{(\pi)- [-\sin(x)]_{\pi/2}^{\pi}\}$

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

$=\frac{2(\pi-1)}{\pi}$

Report an Error

Example Question #16 : Average Values And Lengths Of Functions

Find the arc length of the graph $f(x)=\ln(\sec(x))$ on the interval $[-\frac{\pi}{4},\frac{\pi}{4}]$ .

Possible Answers:

$\frac{1}{2}$

$\frac{\sqrt2}{2}$

$\ln\biggr(\frac{\sqrt2-1}{\sqrt2+1}\biggr)$

$\ln\biggr(\frac{\sqrt2+1}{\sqrt2-1}\biggr)$

Correct answer:

$\ln\biggr(\frac{\sqrt2+1}{\sqrt2-1}\biggr)$

Explanation:

The arc length for a function f(x) on the interval [a,b] is given by the following integral:

$L[a,b]=\int_a^b\sqrt{1+(f^{'}(x))^2}dx$ .

Thus we must compute the value of $f^{'}(x)$ for the function f(x)=ln(sec(x)) .

To do so, we use the chain rule:

$f^{'}(x)=\frac{1}{sec(x)}(\sec(x))^{'}=\frac{1}{\sec(x)}\sec(x)\tan(x)=\tan(x)$ .

Now plugging $f^{'}(x)$ to the arc length formula, and using the interval $[-\frac{\pi}{4},\frac{\pi}{4}]$ , we obtain,

$L[-\frac{\pi}{4},\frac{\pi}{4}]=\int_\frac{-\pi}{4}^\frac{\pi}{4}\sqrt{1+\tan^2(x)}dx$ .

Using the trig identity $1+tan^{2}(x)=\sec^{2}(x)$ , our integral becomes:

$\int_\frac{-\pi}{4}^\frac{\pi}{4}\sqrt{\sec^2(x)}dx=\int_\frac{-\pi}{4}^\frac{\pi}{4}\sec(x)dx$

$=\ln\left | \sec(x)+tan(x)) \right |\biggr|_\frac{-\pi}{4}^\frac{\pi}{4}=\ln\left | \sec(\frac{\pi}{4})+tan(\frac{\pi}{4})) \right |-\ln\left | \sec(\frac{-\pi}{4})+tan(\frac{-\pi}{4})) \right |$

$=\ln\left | \sqrt2+1 \right |-\ln\left | \sqrt2-1\right |=\ln\left |\frac{\sqrt2+1}{\sqrt2-1} \right |$ .

In the last step above, we invoked the property

$\ln(a)-\ln(b)=\ln\frac{a}{b}$ .

Since the expression in the absolute value is positive, we can get rid of the absolute value brackets.

Thus we obtain

$L[-\frac{\pi}{4},\frac{\pi}{4}]=\ln\biggr(\frac{\sqrt2+1}{\sqrt2-1}\biggr)$ .

Report an Error

Example Question #11 : Average Values And Lengths Of Functions

What is the average value of f(x)=x^3+1 on the interval [0,10] ?

Possible Answers:

1255

1250

502

2510

251

Correct answer:

251

Explanation:

To find the average value of a function, you must use the following equation:

$\frac{1}{b-a}\int_{a}^{b}f(x)dx$ . Now, we simply substitute our values, and integrate.

$\frac{1}{10}\int_{0}^{10}x^3+1dx=\frac{1}{10}[\frac{x^4}{4}+x]$ . Now, we plug in our bounds (upper - lower).

$\frac{1}{10}[\frac{(10^4}{4}+10)-(\frac{0^4}{4}+0)]=\frac{1}{10}[2510]=251.$

Report an Error

Example Question #151 : Integral Applications

Find the average value of the function $f(x)=\frac{3}{x}$ from to .

Possible Answers:

$\frac{9}{e-1}$

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

$\frac{9}{1-e}$

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

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

Correct answer:

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

Explanation:

The average value equation is $\frac{1}{b-a}\int_{a}^{b}f(x)dx=\frac{1}{e-1}\int_{1}^{e}\frac{3}{x}dx.$

$=\frac{1}{e-1}[3ln(x)].$ Now, all we have to do is plug in our limits.

$\frac{1}{e-1}[3ln(e)-3ln(1)]=\frac{3}{e-1}.$

The second term drops out because the natural log of equals zero. In the first term, natural log and cancel each other out, just leaving the out front.

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

Filiberto
Certified Tutor

University of Arizona, Bachelor of Science, Electrical Engineering.

View Calculus 2 Tutors

Chelsea
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Physics. The University of Texas at Dallas, Master of Science, Bioinf...

All Calculus 2 Resources

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

Popular Subjects

GRE Tutors in Washington DC, SSAT Tutors in Atlanta, Biology Tutors in Atlanta, English Tutors in Atlanta, ISEE Tutors in Boston, Statistics Tutors in New York City, SSAT Tutors in Denver, Spanish Tutors in Houston, Chemistry Tutors in Dallas Fort Worth, GRE Tutors in Seattle

Popular Courses & Classes

SAT Courses & Classes in Houston, Spanish Courses & Classes in Washington DC, ACT Courses & Classes in Los Angeles, ACT Courses & Classes in Dallas Fort Worth, ISEE Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in Washington DC, GRE Courses & Classes in Phoenix, GMAT Courses & Classes in Houston, LSAT Courses & Classes in Dallas Fort Worth, Spanish Courses & Classes in San Francisco-Bay Area

Popular Test Prep

ISEE Test Prep in Atlanta, GRE Test Prep in New York City, GRE Test Prep in Chicago, LSAT Test Prep in Dallas Fort Worth, GRE Test Prep in Philadelphia, LSAT Test Prep in New York City, GRE Test Prep in San Diego, ISEE Test Prep in Los Angeles, GRE Test Prep in Atlanta, LSAT Test Prep in Miami

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 : Integral Applications

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #9 : Average Values And Lengths Of Functions

Example Question #10 : Average Values And Lengths Of Functions

Example Question #321 : Integrals

Example Question #12 : Average Values And Lengths Of Functions

Example Question #321 : Integrals

Example Question #14 : Average Values And Lengths Of Functions

Example Question #151 : Integral Applications

Example Question #16 : Average Values And Lengths Of Functions

Example Question #11 : Average Values And Lengths Of Functions

Example Question #151 : Integral Applications

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: