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 #81 : Integral Applications

Find the area under the curve for f(x)=|2x^3-3x-10| from x=-2 to x=4

Possible Answers:

138

114

122

Correct answer:

122

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:

$\int_{-2}^{4}|2x^3-3x-10|dx$

Solution:

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

In other words, find the sum of these two integrals.

$\int_{-2}^{2}10+3x-2x^3dx + \int_{2}^{4}2x^3-3x-10dx$

First piece:

$\int_{-2}^{2}10+3x-2x^3dx = 10x+\frac{3}{2}x^2-\frac{2}{4}x^4\Big|_{-2}^{2}$

$=10\cdot2+\frac{3}{2}2^2-\frac{2}{4}2^4 - (10\cdot-2+\frac{3}{2}(-2)^2-\frac{2}{4}(-2)^4)$

=20+6-8-(-20+6-8) = 40

Second piece:

$\int_{2}^{4}2x^3-3x-10dx = \frac{2}{4}x^4-\frac{3}{2}x^2-10x\Big|_{2}^{4}$

$\frac{2}{4}\cdot256-\frac{3}{2}\cdot16 - 40 - (8-6-20) = 128-24-40+18 = 82$

Sum:

Add the 2 integrals together.

40+82 = 122

Report an Error

Example Question #251 : Integrals

Find the area under the curve for f(x)=sin(x)+x^3 from x=0 to $x=\pi/2$

Possible Answers:

$\frac{\pi^4}{64}+1$

$\frac{\pi^2}{32}$

$\frac{\pi^2}{64}-1$

$\frac{\pi^4}{32}+1$

Correct answer:

$\frac{\pi^4}{64}+1$

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:

$\int_{0}^{\pi/2}sin(x)+x^3dx$

Solution:

$\int sin(x)+x^3dx$

$-cos(x)+\frac{1}{4}x^4\Big|_{0}^{\frac{\pi}{2}}$

$\frac{1}{4}\cdot\frac{\pi^4}{16}+cos(0)$

Answer:

$\frac{\pi^4}{64}+1$

Report an Error

Example Question #21 : Area Under A Curve

Find the area under the curve for f(x)=(x-3)^2+6 from x=0 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:

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

Solution:

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

$=\int_{0}^{2} x^2 + 9 -6x + 6 dx$

$=\frac{1}{3}x^3 -3x^2 + 15x \Big|_{0}^{2}$

$=\frac{8}{3} - 3\cdot4 + 15\cdot2$

$=\frac{8}{3} + 18 = \frac{62}{3} = 20.6666$

When rounded, it is equal to

Report an Error

Example Question #251 : Integrals

Find the area under the curve for f(x)=(x-7)^2+3x^2 from x=1 to x=4

Possible Answers:

126

132

107

119

Correct answer:

126

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:

$\int_{1}^{4}(x-7)^2+3x^2dx$

Solution:

$\int_{1}^{4}(x-7)^2+3x^2dx$

$=\int_{1}^{4} x^2-14x+49+3x^2 dx$

$=\int_{1}^{4} 4x^2 - 14x + 49 dx$

$=\frac{4}{3} x^3 - 7x^2 + 49x\Big|_{1}^{4}$

$=\frac{4}{3} 4^3 - 7\cdot4^2 + 49\cdot4 - (\frac{4}{3} - 7 + 49)$

After simplifying, the answer is 126

Report an Error

Example Question #2002 : Calculus Ii

Find the area under the curve for f(x)=21x^2-40x from x=-2 to x=1 , rounded to the nearest integer.

Possible Answers:

107

115

123

132

Correct answer:

123

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:

$\int_{-2}^{1}21x^2-40xdx$

Solution:

$\int21x^2-40xdx$

$\frac{21}{3}x^3-\frac{40}{2}x^2\Big|_{-2}^{1}$

$(7\cdot1^3-20\cdot1^2)-(7\cdot(-2)^3-20\cdot(-2)^2)$

$(7-20)-(7\cdot-8-(20\cdot4))$

-13-(-56-80)

-13-(-136)=-13+136=123

The area under the curve is 123

Report an Error

Example Question #261 : Integrals

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 #2011 : Calculus Ii

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 #2012 : Calculus Ii

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

View Calculus 2 Tutors

Steven
Certified Tutor

University of Waterloo, Bachelor of Science, Chemical Engineering. McMaster University, Master of Science, Chemical Engineeri...

View Calculus 2 Tutors

Philipp
Certified Tutor

University of Nevada-Las Vegas, Bachelor of Science, Mechanical Engineering.

View Calculus 2 Tutors

Andrew
Certified Tutor

Louisiana State University Eunice, Bachelor of Science, Computer Software Engineering.

All Calculus 2 Resources

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

Popular Subjects

English Tutors in Philadelphia, Statistics Tutors in Phoenix, ACT Tutors in Dallas Fort Worth, Algebra Tutors in Chicago, Calculus Tutors in Denver, GMAT Tutors in Chicago, Spanish Tutors in Atlanta, Spanish Tutors in New York City, Algebra Tutors in Atlanta, Chemistry Tutors in Atlanta

Popular Courses & Classes

ISEE Courses & Classes in Washington DC, ACT Courses & Classes in Phoenix, ACT Courses & Classes in Seattle, SSAT Courses & Classes in Philadelphia, SAT Courses & Classes in Denver, Spanish Courses & Classes in Atlanta, MCAT Courses & Classes in San Francisco-Bay Area, SAT Courses & Classes in Washington DC, GRE Courses & Classes in Atlanta, LSAT Courses & Classes in Philadelphia

Popular Test Prep

ACT Test Prep in Boston, LSAT Test Prep in Boston, MCAT Test Prep in Miami, ACT Test Prep in New York City, LSAT Test Prep in Phoenix, SSAT Test Prep in Phoenix, SAT Test Prep in Atlanta, SSAT Test Prep in Boston, MCAT Test Prep in San Francisco-Bay Area, ACT 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 : Integral Applications

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #81 : Integral Applications

Example Question #251 : Integrals

Example Question #21 : Area Under A Curve

Example Question #251 : Integrals

Example Question #2002 : Calculus Ii

Example Question #261 : Integrals

Example Question #2011 : Calculus Ii

Example Question #88 : Integral Applications

Example Question #2012 : Calculus Ii

Example Question #89 : Integral Applications

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: