Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Integrals

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 #86 : Integral Applications

Find the area under the curve for f(x)=14x^3+5x^2 from x=-2 to x=2 , rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{-2}^{2}14x^3+5x^2dx$

Solution:

$\int14x^3+5x^2dx$

$\frac{14}{4}x^4+\frac{5}{3}x^3\Big|_{-2}^{2}$

$(\frac{7}{2}2^4 + \frac{5}{3}2^3)-(\frac{7}{2}(-2)^4+\frac{5}{3}(-2)^3)$

$(\frac{7}{2}\cdot16+\frac{5}{3}\cdot8)-(\frac{7}{2}\cdot16-\frac{5}{3}\cdot8)$

$(56+\frac{40}{3})-(56-\frac{40}{3}) = \frac{80}{3} = 26.67$

When rounded, the area under the curve is

Report an Error

Example Question #87 : Integral Applications

Find the area under the curve for f(x)=3x^2+e^x from x=0 to x=2

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{0}^{2}3x^2+e^xdx$

Solution:

$\int3x^2+e^xdx$

$\frac{3}{3}x^3+e^x\Big|_{0}^{2}$

$x^3+e^x\Big|_{0}^{2}$

(2^3+e^2)-(e^0)

8+e^2-1 = e^2 + 7 = 14.389

When rounded to the nearest integer, the area under the curve is

Report an Error

Example Question #88 : Integral Applications

Find the area under the curve for $f(x)=7x+e^{2x}$ from x=0 to x=2 , when rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{0}^{2}7x+e^{2x}dx$

Solution:

$\int7x+e^{2x}dx$

$\frac{7}{2}x^2+\frac{1}{2}e^{2x}\Big|_{0}^{2}$

$(\frac{7}{2}\cdot2^2+\frac{1}{2}e^{2\cdot2})-(\frac{1}{2})$

$14+\frac{1}{2}e^4-\frac{1}{2} = \frac{27}{2}+\frac{1}{2}e^4 = 40.799$

When rounded, the area under the curve is

Report an Error

Example Question #31 : Area Under A Curve

Find the area under the curve for f(x)=|4x+x^2| from x=-2 to x=2 , rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{-2}^{2}|4x+x^2|dx$

Solution:

This function is negative from x=[-2,0], and positve everywhere else. Split this integral up into 2 pieces, multiplying x=[-2,0] region by -1, and sum everything up.

1st Piece:

$\int_{-2}^{0}-4x-x^2dx$

$-2x^2-\frac{1}{3}x^3\Big|_{-2}^{0}$

$0-(-2\cdot4-\frac{1}{3}\cdot(-8)) = -(-8+\frac{8}{3}) = 8 - \frac{8}{3} = \frac{16}{3}$

2nd piece:

$\int_{0}^{2}4x+x^2dx$

$2x^2+\frac{1}{3}x^3\Big|_{0}^{2}$

$2\cdot4+\frac{8}{3} = 8+\frac{8}{3} = \frac{32}{3}$

Sum:

$\frac{16}{3} + \frac{32}{3} = \frac{48}{3} = 16$

The area under the curve is

Report an Error

Example Question #89 : Integral Applications

Find the area under the curve for f(x)=|-6x+2x^2| from x=-3 to x=2 , rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

inding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{-2}^{2}|-6x+2x^2|dx$

Solution:

This function is negative from x=[0,2], and positve everywhere else. Split this integral up into 2 pieces, multiplying x=[0,2] region by -1, and sum everything up.

1st piece:

$\int_{-3}^{0}|-6x+2x^2|dx$

$-3x^2+\frac{2}{3}x^3\Big|_{-3}^{0}$

$0-(-3\cdot(-3)^2+\frac{2}{3}\cdot(-3)^3)$

$9+\frac{2}{3}\cdot27 = 9+18=27$

2nd piece:

$\int_{0}^{2}|-6x+2x^2|dx$

$3x^2-\frac{2}{3}x^3\Big|_{0}^{2}$

$3*4-\frac{2}{3}\cdot8 = 12-\frac{16}{3}$

sum:

$27+12-\frac{16}{3} = \frac{155}{3} = 51.667$

When rounded to the nearest integer, the area under the curve is

Report an Error

Example Question #33 : Area Under A Curve

Find the area under the curve for f(x)=|-3x+2x^2+x^3| from x=-1 to x=1 , rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{-1}^{1}|-3x+2x^2+x^3|dx$

Solution:

This function is negative from x=[0,1] , and positve everywhere else. Split this integral up into 2 pieces, multiplying x=[0,1] region by , and sum everything up.

First piece:

$\int_{-1}^{0}|-3x+2x^2+x^3|dx$

$-\frac{3}{2}x^2+\frac{2}{3}x^3+\frac{1}{4}x^4\Big|_{-1}^{0}$

$0-(-\frac{3}{2}-\frac{2}{3}+\frac{1}{4}) = \frac{3}{2}+\frac{2}{3}-\frac{1}{4} = \frac{23}{12}$

Second piece:

$\int_{0}^{1}|-3x+2x^2+x^3|dx$

$\frac{3}{2}x^2-\frac{2}{3}x^3-\frac{1}{4}x^4\Big|_{0}^{1}$

$\frac{3}{2}-\frac{2}{3}-\frac{1}{4} = \frac{7}{12}$

Sum:

$\frac{23}{12}+\frac{7}{12} = \frac{30}{12} = \frac{5}{2}=2.5$

When rounded to the nearest integer, the area under the curve is

Report an Error

Example Question #261 : Integrals

Find the area under the curve for f(x)=sin(x)+1 from $x=\frac{\pi}{4}$ to $x=\frac{\pi}{2}$

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}sin(x)+1dx$

Solution:

$\int sin(x)+1dx$

$x-cos(x)\Big|_{\frac{\pi}{4}}^{\frac{\pi}{2}}$

$({\frac{\pi}{2}}-cos({\frac{\pi}{2}}))-({\frac{\pi}{4}} - cos({\frac{\pi}{4}}))$

$\frac{\pi}{4} + \frac{1}{\sqrt2} = 1.4925$

Rounded to the nearest integer, the area under the curve is

Report an Error

Example Question #91 : Integral Applications

Find the area under the curve for f(x)=2sin(x)+x from x=0 to $x=\pi$

Possible Answers:

$\frac{1}{2}\pi^2 + 4$

$\frac{1}{2}\pi + 2$

$\pi^2 + 2$

$\frac{1}{2}\pi^2 + 6$

Correct answer:

$\frac{1}{2}\pi^2 + 4$

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{0}^{\pi}2sin(x)+xdx$

Solution:

$\int2sin(x)+xdx$

$\frac{1}{2}x^2-2cos(x)\Big|_{0}^{\pi}$

$\frac{1}{2}\pi^2-2\cdot cos(\pi)-(-2cos(0)) = \frac{1}{2}\pi^2-2\cdot cos(\pi) + 2cos(0)$

$\frac{1}{2}\pi^2-2\cdot-1 + 2 = \frac{1}{2}\pi^2 + 4$

The area under the curve is $\frac{1}{2}\pi^2 + 4$

Report an Error

Example Question #36 : Area Under A Curve

Find the area under the curve for f(x)=|3x^2-4x| from x=-2 to x=2

Possible Answers:

Correct answer:

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{-2}^{2}|3x^2-4x|dx$

Solution:

This function is negative from $x=[0,\frac{4}{3}],$ and positve everywhere else. Split this integral up into 3 pieces, multiplying x=[0,\frac{4}{3}] region by -1, and sum everything up.

1st piece:

$\int_{-2}^{0}3x^2-4xdx$

$3\cdot\frac{1}{3}x^3-\frac{4}{2}x^2\Big|_{-2}^{0}$

$x^3-2x^2\Big|_{-2}^{0}$

$0-((-2)^3-2\cdot4) = 8+8 = 16$

2nd piece:

$\int_{0}^{\frac{4}{3}}|3x^2-4x|dx$

$2x^2 - x^3\Big|_{0}^{\frac{4}{3}}$

$2\cdot(\frac{4}{3})^2 - (\frac{4}{3})^3 = \frac{32}{9}-\frac{64}{27} = \frac{32}{27}$

3rd piece:

$\int_{\frac{4}{3}}^{2}|3x^2-4x|dx$

$x^3 - 2x^2\Big|_{\frac{4}{3}}^{2}$

$(8-2\cdot4)-((\frac{4}{3})^3-2\cdot(\frac{4}{3})^2) = \frac{32}{9}-\frac{64}{27} = \frac{32}{27}$

Sum:

$16+\frac{32}{27} + \frac{32}{27} = \frac{496}{27} = 18.37$

When rounded to the nearest integer, the area under the curve is

Report an Error

Example Question #37 : Area Under A Curve

Find the area under the curve for f(x)=(x+6)^2-36 from x=1 to x=4

Possible Answers:

111

123

107

127

Correct answer:

111

Explanation:

Finding the area of a region is the same as integrating over the range of the function and it can be rewritten into the following:

$\int_{1}^{4} (x+6)^2-36dx$

Solution:

First, simplify the function and then evaluate the integral.

1. Simplify the function

(x+6)^2-36

x^2+36+12x-36

x^2+12x

2. Evaluate the integral

$\int_{1}^{4} x^2+12xdx$

$\frac{1}{3}x^3+6x^2\Big|_{1}^{4}$

$\frac{64}{3}+6\cdot16-(\frac{1}{3}+6) = 111$

The area under the curve is 111

Report an Error

View Calculus 2 Tutors

Rodolfo
Certified Tutor

The University of Texas at San Antonio, Bachelor of Science, Mathematics.

View Calculus 2 Tutors

Muhammad
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Mechanical Engineering.

View Calculus 2 Tutors

Hirdeshwar
Certified Tutor

Indian Institute of Technology, Powai, Bombay, India, Doctor of Philosophy, Mathematics.

All Calculus 2 Resources

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

Popular Subjects

Statistics Tutors in Miami, French Tutors in Chicago, Physics Tutors in Seattle, GRE Tutors in Chicago, Reading Tutors in Atlanta, Reading Tutors in New York City, SSAT Tutors in New York City, SAT Tutors in San Francisco-Bay Area, Math Tutors in Washington DC, Chemistry Tutors in Houston

Popular Courses & Classes

SSAT Courses & Classes in Miami, SAT Courses & Classes in Boston, SSAT Courses & Classes in Washington DC, Spanish Courses & Classes in Chicago, GMAT Courses & Classes in Miami, LSAT Courses & Classes in Houston, SSAT Courses & Classes in Seattle, Spanish Courses & Classes in Washington DC, SSAT Courses & Classes in Chicago, ISEE Courses & Classes in New York City

Popular Test Prep

GRE Test Prep in San Francisco-Bay Area, SAT Test Prep in San Francisco-Bay Area, SAT Test Prep in Washington DC, MCAT Test Prep in Miami, LSAT Test Prep in Chicago, LSAT Test Prep in Dallas Fort Worth, SAT Test Prep in Denver, GMAT Test Prep in Phoenix, ISEE Test Prep in Denver, SAT Test Prep in San Diego

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 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #86 : Integral Applications

Example Question #87 : Integral Applications

Example Question #88 : Integral Applications

Example Question #31 : Area Under A Curve

Example Question #89 : Integral Applications

Example Question #33 : Area Under A Curve

Example Question #261 : Integrals

Example Question #91 : Integral Applications

Example Question #36 : Area Under A Curve

Example Question #37 : Area Under A Curve

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: