We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

High School Math : Calculus II — Integrals

Study concepts, example questions & explanations for High School Math

Create An Account

All High School Math Resources

8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 4 5 6 7 8 9 10 Next →

Example Question #81 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

What is the indefinite integral of 4x+3 with respect to ?

Possible Answers:

2x^3-c

2x^2+3x+c

2x^2+3x

4+c

Correct answer:

2x^2+3x+c

Explanation:

To find the indefinite integral, we're going to use the reverse power rule: raise the exponent of the variable by one and then divide by that new exponent.

$\int 4x+3 dx=(\frac{4x^{1+1}}{1+1})+(\frac{3x^{0+1}}{0+1})+c$

Be sure to include + c to compensate for any constant!

$\int 4x+3dx=(\frac{4x^{1+1}}{1+1})+(\frac{3x^{0+1}}{0+1})+c$

$\int 4x+3dx=(\frac{4x^{2}}{2})+(\frac{3x^{1}}{1})+c$

$\int 4x+3dx=2x^2+3x+c$

Report an Error

Example Question #82 : Calculus Ii — Integrals

$\int x^4-3x^2+8\ {\mathrm{d} x}=?$

Possible Answers:

$\frac{1}{5}x^5-x^3+8x+c$

5x^5-3x^3+8x+c

x^7

$\frac{1}{5}x^5-x^3+8x$

12x^2-6x

Correct answer:

$\frac{1}{5}x^5-x^3+8x+c$

Explanation:

To find the indefinite integral, or anti-derivative, we can use the reverse power rule. We raise the exponent of each variable by one and divide by that new exponent.

$\int x^4-3x^2+8\ {\mathrm{d} x}=(\frac{x^{4+1}}{4+1})-(\frac{3x^{2+1}}{2+1})+(\frac{8x^{0+1}}{0+1})+c$

Don't forget to include a to cover any constant!

Simplify.

$\int x^4-3x^2+8\ {\mathrm{d} x}=(\frac{x^{5}}{5})-(\frac{3x^{3}}{3})+(\frac{8x^{1}}{1})+c$

$\int x^4-3x^2+8\ {\mathrm{d} x}=\frac{1}{5}x^5-x^3+8x+c$

Report an Error

Example Question #2241 : High School Math

$\int \frac{2}{3}x^3+7x^2-12x \ {\mathrm{d} x}=?$

Possible Answers:

$\frac{1}{6}x^4+\frac{7}{3}x^3-12x+c$

$\frac{1}{6}x^4+\frac{7}{3}x^3-6x^2+c$

2x^2+14x

$6x^4+\frac{7}{3}x^3-6x^2$

Correct answer:

$\frac{1}{6}x^4+\frac{7}{3}x^3-6x^2+c$

Explanation:

To find the indefinite integral of $\frac{2}{3}x^3+7x^2-12x$ , we can use the reverse power rule. To do this, we raise our exponent by one and then divide the variable by that new exponent.

$\int \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=\frac{\frac{2}{3}x^{3+1}}{3+1}+\frac{7x^{2+1}}{2+1}-\frac{12x^{1+1}}{1+1}+c$

Don't forget to include a to cover any constant!

$\int \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=\frac{\frac{2}{3}x^{4}}{4}+\frac{7x^{3}}{3}-\frac{12x^{2}}{2}+c$

$\int \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=\frac{2}{12}x^4+\frac{7}{3}x^3-6x^2+c$

$\int \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=\frac{1}{6}x^4+\frac{7}{3}x^3-6x^2+c$

Report an Error

Example Question #1 : Finding Integrals By Substitution

Determine the indefinite integral:

$\dpi{120}\int\frac{\sqrt{\log_{10} x}}{x}\; \; dx$

Possible Answers:

$\frac{ 2 \sqrt{ \log_{10} x} \cdot \log_{10} x} {3 } + C$

$\frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C$

$\frac{ 2 \log_{10} x} {3 } + C$

$\frac{ 2 \sqrt{ \log_{10} x} } {3 \ln x} + C$

$\frac{ 2 \sqrt{ \log_{10} x} } {3 } + C$

Correct answer:

$\frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C$

Explanation:

$\log_{10} x = \frac{\ln x}{\ln 10}$ , so this can be rewritten as

$\dpi{120}\int\frac{\sqrt{\log_{10} x}}{x}\; \; dx$

$\dpi{120}=\int\frac{\sqrt{\ln x}}{x \sqrt{\ln 10} }\; \; dx$

$\dpi{120}= \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }\; \; dx$

Set $u = \ln x$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$

and

$du = \frac{dx}{x}$

Substitute:

$\dpi{120} \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }\; \; dx$

$\dpi{120} =\frac{1}{\sqrt{\ln 10}} \int\sqrt{u} \; du$

$\dpi{120} =\frac{1}{\sqrt{\ln 10}} \int u ^{\frac{1}{2}} \; du$

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} + C \right )$

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2u \sqrt{u}} {3} + C \right )$

The outer factor can be absorbed into the constant, and we can substitute back:

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2 \ln x \cdot \sqrt{ \ln x}} {3} \right )+ C$

$= \frac{ 2 \sqrt{ \ln x} \cdot \ln x} {3 \cdot \sqrt{\ln 10}} + C$

$= \frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C$

Report an Error

Example Question #71 : Functions, Graphs, And Limits

Evaluate:

$\dpi{120}\int_{1}^{e^{ 3}}\frac{\sqrt{\ln x}}{x}\; \; dx$

Possible Answers:

$\frac{2}{3}$

$\frac{2\sqrt{3}}{3}$

$2 \ln 3$

$2\sqrt{3}$

$2 \sqrt{\ln 3}$

Correct answer:

$2\sqrt{3}$

Explanation:

Set $u = \ln x$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$

and

$du = \frac{dx}{x}$

Also, since $u = \ln x$ , the limits of integration change to $\ln e^{3} = 3$ and $\ln 1= 0$ .

Substitute:

$\dpi{120}\int_{1}^{e^{ 3}}\frac{\sqrt{\ln x}}{x}\; \; dx$

$\dpi{120} = \int_{0}^{3}\sqrt{u} \; du$

$= \int_{0}^{3} u ^{\frac{1}{2}} \; du$

$\dpi{150} = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} & \\ & \end{matrix}\right|$ $\begin{matrix} 3\\ \\ 0 \end{matrix}$

$\dpi{150} = \left.\begin{matrix} \\ \frac{ 2u \sqrt{u}} {3} & \\ & \end{matrix}\right|$ $\begin{matrix} 3\\ \\ 0 \end{matrix}$

$=\frac{ 2\cdot 3\cdot \sqrt{3}} {3} - \frac{ 2\cdot 0\cdot \sqrt{0}} {3} =2\sqrt{3}$

Report an Error

Example Question #71 : Asymptotic And Unbounded Behavior

Determine the indefinite integral:

$\dpi{120} \int\cos x \sqrt{\sin x }\; \; dx$

Possible Answers:

$\dpi{120} \dpi{120} \frac{ 2 \sqrt{\cos x }} {3} +C$

$\dpi{120} \dpi{120} \frac{ 2 \cos x \cdot \sqrt{\cos x }} {3} +C$

$\dpi{120} \frac{ 2 \sin x \cdot \sqrt{\cos x }} {3} +C$

$\frac{ 2 \sin x \cdot \sqrt{\sin x }} {3} +C$

$\dpi{120} \frac{ 2 \sqrt{\sin x }} {3} +C$

Correct answer:

$\frac{ 2 \sin x \cdot \sqrt{\sin x }} {3} +C$

Explanation:

Set $u = \sin x$ . Then

$\frac{\mathrm{du} }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x} \sin x= \cos x$ .

and

$\cos x \; dx = du$

The integral becomes:

$\dpi{120} \int\cos x \sqrt{\sin x }\; \; dx$

$\dpi{120} =\int \sqrt{u }\; \; du$

$= \int u ^{\frac{1}{2}} \; du$

$\dpi{120} \dpi{150} = \frac{u^{\frac{1}{2}+1}}{{\frac{1}{2}+1}} + C$

$\dpi{150} = \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} + C$

$\dpi{120} =\frac{ 2u \sqrt{u}} {3} +C$

Substitute back:

$\dpi{120} =\frac{ 2 \sin x \cdot \sqrt{\sin x }} {3} +C$

Report an Error

Example Question #4 : Finding Integrals By Substitution

$\int_{4}^{9}\frac{\sqrt{x}}{1-x}dx=$

Possible Answers:

$1+\tan^{-1}(3)-\tan^{-1}(2)$

$-2+\ln\frac{3}{2}$

$\tan^{-1}(3)-\tan^{-1}(4)$

$-2+\ln\frac{9}{4}$

$-2+\ln\frac{2}{3}$

Correct answer:

$-2+\ln\frac{2}{3}$

Explanation:

This integral will require a u-substitution.

Let $u=\sqrt{x}$ .

Then, differentiating both sides, $du=\frac{1}{2\sqrt{x}}dx$ .

We need to solve for dx in order to replace all x terms with u terms.

$dx=(2\sqrt{x})du$ .

$\int\frac{\sqrt{x}}{1-x}\cdot(2\sqrt{x})du=\int\frac{2xdu}{1-x}$

This is a little tricky because we stilll have x and u terms mixed together. We need to go back to our original substitution.

$u=\sqrt{x}\Rightarrow x=u^2$

$\int\frac{2xdu}{1-x}=\int\frac{2u^2 du}{1-u^2}$

Now we have an integral that looks more manageable. First, however, we can't forget about the bounds of the definite integral. We were asked to evaluate the integral from x=4 to x=9 . Because $u=\sqrt{x}$ , the bounds will change to $\sqrt{4}=2$ and $\sqrt{9}=3$ .

Essentially, we have made the following transformation:

$\int_{4}^{9}\frac{\sqrt{x}}{1-x}dx=\int_{2}^{3}\frac{2u^2}{1-u^2}du$ .

The latter integral is easier to evaluate.

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

$=-2\int_{2}^{3}\frac{1-u^2-1}{1-u^2}du=-2\int_{2}^{3}\left ( \frac{1-u^2}{1-u^2}+\frac{-1}{1-u^2} \right )du$

$=-2\int_{2}^{3}\left ( 1-\frac{1}{1-u^2}\right )du$

At this point, we can separate the integral into two smaller integrals.

$-2\int_{2}^{3}\left ( 1-\frac{1}{1-u^2}\right )du=-2\int_{2}^{3}1\cdot du+2\int_{2}^{3}\frac{1}{1-u^2}du$ .

The integral $-2\int_{2}^{3}1\cdot du$ evaluates to -2, so now we just need to worry about the other integral. This will require the use of partial fraction decomposition. We need to rewrite $\frac{1}{1-u^2}$ as the sum of two fractions.

$\frac{1}{1-u^2}=\frac{1}{(1-u)(1+u)}=\frac{A}{1-u}+\frac{B}{1+u}$

We need to solve for the values of A and B.

$\frac{A}{1-u}+\frac{B}{1+u}=\frac{1}{(1-u)(1+u)}$

A(1+u)+B(1-u)=1=0u+1

A+Au+B-Bu=0u+1

Au-Bu+A+B=0u+1

(A-B)u+(A+B)=0u+1

This means that A-B=0 and A+B=1 . This is a relatively simple system of equations to solve, so I won't go into detail. The end result is that $A=B=\frac{1}2{}$ .

$\frac{1}{1-u^2}=\frac{\frac{1}{2}}{1-u}+\frac{\frac{1}{2}}{1+u}$

Let's now go back to the integral $2\int_{2}^{3}\frac{1}{1-u^2}du$ .

$2\int_{2}^{3}\frac{1}{1-u^2}du=2\int_{2}^{3}\left ( \frac{\frac{1}{2}}{1-u}+\frac{\frac{1}{2}}{1+u} \right )du$

Distribute the 2 to both integrals and separate it into two integrals.

$=\int_{2}^{3}\frac{1}{1-u}du + \int_{2}^{3}\frac{1}{1+u}du$

$=-\ln|1-u| ]_{2}^{3} + \ln|1+u|]_{2}^{3}$

$=-\ln|-2|+\ln|-1|+\ln4-\ln3$

$=-\ln2+\ln1+\ln4-\ln3$

$=\ln\frac{2}{3}$ .

Remember we need to add this value back to the value of $-2\int_{2}^{3}1\cdot du$ , which we already determined to be -2.

The final answer is $-2+\ln\frac{2}{3}$ .

Report an Error

Example Question #5 : Finding Integrals By Substitution

Evaluate:

$\dpi{120} \int_{0}^{\frac{\pi }{ 2}}\sin x \sqrt{\cos x }\; \; dx$

Possible Answers:

$\frac{4}{3}$

$\frac{ 2} {3}$

Correct answer:

$\frac{ 2} {3}$

Explanation:

Set $u = \cos x$ .

Then $\frac{\mathrm{du} }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x} \cos x= -\sin x$ and $\sin x \; dx = - du$ .

Also, since $u = \cos x$ , the limits of integration change to $\cos \frac{\pi }{2} = 0$ and $\cos 0 = 1$ .

Substitute:

$\dpi{120} \int_{0}^{\frac{\pi }{ 2}}\sin x \sqrt{\cos x }\; \; dx$

$= \int_{1}^{0} \left (- \sqrt{u} \right ) \; du$

$= \int_{1}^{0} \left (-u ^{\frac{1}{2}} \right ) \; du$

$= \int_{0}^{1} u ^{\frac{1}{2}} \; du$

$\dpi{150} = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} & \\ & \end{matrix}\right|$ $\begin{matrix} 1\\ \\ 0 \end{matrix}$

$\dpi{150} = \left.\begin{matrix} \\ \frac{ 2u \sqrt{u}} {3} & \\ & \end{matrix}\right|$ $\begin{matrix} 1\\ \\ 0 \end{matrix}$

$=\frac{ 2\cdot 1\cdot \sqrt{1}} {3} - \frac{ 2\cdot 0\cdot \sqrt{0}} {3} = \frac{ 2} {3}$

Report an Error

Example Question #1 : Understanding Taylor Series

Give the $x^{2}$ term of the Maclaurin series of the function $f \left ( x\right ) = \frac{1}{x-2}$

Possible Answers:

$-\frac{1}{12} x^{2}$

$-\frac{1}{2} x^{2}$

$-\frac{1}{4} x^{2}$

$-\frac{1}{8} x^{2}$

$-\frac{1}{6} x^{2}$

Correct answer:

$-\frac{1}{8} x^{2}$

Explanation:

The $x^{2}$ term of the Maclaurin series of a function has coefficient $\frac{f '' (0) }{2!} = \frac{f '' (0) }{2}$

The second derivative of can be found as follows:

$f \left ( x\right ) = \frac{1}{x-2} = (x-2) ^{-1}$

$f' \left ( x\right ) = -1\cdot (x-2)^{-2} = -1(x-2)^{-2}$

$f'' \left ( x\right ) = (-2) ( -1(x-2)^{-3}) = 2(x-2)^{-3} = \frac{2}{(x-2)^{3}}$

$f'' \left ( 0 \right ) =\frac{2}{(0-2)^{3}} = -\frac{1}{4}$

The coeficient of $x^{2}$ in the Maclaurin series is therefore

$\frac{f '' (0) }{2} = \frac{-\frac{1}{4}}{2} = - \frac{1}{8}$

Report an Error

Example Question #1 : Understanding Taylor Series

Give the (x-1)^2 term of the Taylor series expansion of the function $f (x) = \log_5 x$ about x = 1 .

Possible Answers:

$\frac{\ln 5}{ 2 } (x-1)^{2}$

$\frac{2}{ \ln 5 } (x-1)^{2}$

$\frac{1}{ \ln 25 } (x-1)^{2}$

$- \frac{1}{ \ln 25 } (x-1)^{2}$

$\ln 25\cdot (x-1)^{2}$

Correct answer:

$- \frac{1}{ \ln 25 } (x-1)^{2}$

Explanation:

The $(x-1)^{2}$ term of a Taylor series expansion about x = 1 is

$\frac{f ' ' (1)}{2!} (x-1)^{2}= \frac{f ' ' (1)}{2}(x-1)^{2}$ .

We can find f''(x) by differentiating twice in succession:

$f (x) = \log_5 x$

$f'(x) = \frac{1}{x \ln 5}$

$f'(x) =\frac{1}{ \ln 5} \cdot x^{-1}$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{1}{ \ln 5} \cdot x^{-1} \right )$

$f''(x) =\frac{1}{ \ln 5}\cdot \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{-1} \right )$

$f''(x) =\frac{1}{ \ln 5} \cdot \left ( -1 \cdot x^{-2} \right )$

$f''(x) =- \frac{1}{ \ln 5 \cdot x^{2}}$

$f''(1) =- \frac{1}{ \ln 5 \cdot 1^{2}} = - \frac{1}{ \ln 5 }$

so the (x-1)^2 term is $\frac{f''(1) }{2} \; (x-1)^{2} = - \frac{1}{2 \ln 5 } (x-1)^{2} = - \frac{1}{ \ln 25 } (x-1)^{2}$

Report an Error

← Previous 1 2 3 4 5 6 7 8 9 10 Next →

View Tutors

Jessica
Certified Tutor

Eastern Washington University, Bachelor in Arts, Psychology. University of Phoenix-Boston Campus, Master of Arts, Education.

View Tutors

Manikya
Certified Tutor

Osmania University, Bachelor of Science, Biology Teacher Education. Osmania University, Master of Science, Biology, General.

View Tutors

Alexandra
Certified Tutor

University of Pittsburgh-Pittsburgh Campus, Bachelor in Arts, Psychology.

All High School Math Resources

8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GMAT Tutors in Houston, Physics Tutors in Seattle, Physics Tutors in Boston, Biology Tutors in Boston, GRE Tutors in Chicago, MCAT Tutors in Phoenix, Biology Tutors in Dallas Fort Worth, SAT Tutors in San Diego, GRE Tutors in Miami, Calculus Tutors in Washington DC

Popular Courses & Classes

LSAT Courses & Classes in Atlanta, ISEE Courses & Classes in Phoenix, SAT Courses & Classes in Chicago, GMAT Courses & Classes in Atlanta, LSAT Courses & Classes in Philadelphia, LSAT Courses & Classes in Denver, GMAT Courses & Classes in San Diego, GMAT Courses & Classes in Washington DC, LSAT Courses & Classes in Houston, GRE Courses & Classes in Miami

Popular Test Prep

ACT Test Prep in Houston, SAT Test Prep in Denver, ISEE Test Prep in New York City, ACT Test Prep in Seattle, GRE Test Prep in Washington DC, SAT Test Prep in San Francisco-Bay Area, LSAT Test Prep in Atlanta, SAT Test Prep in Houston, GRE Test Prep in Atlanta, GMAT Test Prep in Chicago

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

High School Math : Calculus II — Integrals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #81 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #82 : Calculus Ii — Integrals

Example Question #2241 : High School Math

Example Question #1 : Finding Integrals By Substitution

Example Question #71 : Functions, Graphs, And Limits

Example Question #71 : Asymptotic And Unbounded Behavior

Example Question #4 : Finding Integrals By Substitution

Example Question #5 : Finding Integrals By Substitution

Example Question #1 : Understanding Taylor Series

Example Question #1 : Understanding Taylor Series

All High School Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: