We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Calculus II

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 #361 : Integrals

Evaluate:

$\int_{-1}^{1} x ^{555}\; \mathrm{d} x$

Possible Answers:

$\frac{1}{555}$

$\frac{2}{555}$

$\frac{1}{556}$

$\frac{1}{228}$

Correct answer:

Explanation:

$\int x ^{n} \;\mathrm{d} x = \frac{ x ^{n+ 1 }}{n+1}$ ,

$\int_{-1}^{1} x ^{555}\; \mathrm{d} x$

$=\frac{ x ^{555+ 1 } }{555+ 1} \left| \begin{matrix} 1 \\ -1 \end{matrix}\right.$

$=\frac{ x ^{556 } }{556} \left| \begin{matrix} 1 \\ -1 \end{matrix}\right.$

$=\frac{ 1 ^{556 } }{556} - \frac{ (-1) ^{556 } }{556} = \frac{ 1 }{556} - \frac{ 1 }{556} =0$

Alternatively, any monomial of the form $f(x) = x ^ {2N+ 1}$ for an integer is an odd function. Since the two bounds of integration are each other's opposite, the definite integral is 0.

Report an Error

Example Question #362 : Integrals

Which of the following functions makes the statement $\int_{-1}^{1} f(x) \; \mathrm{d}x = 0$ false?

Possible Answers:

$f(x) = x ^{2} (2^x + 2^{-x})$

$f(x) = x^{2} \left ( \sin {x} + \tan {x} \right )$

$f(x) = x^{3} \sin^{2} {x}$

$f(x)=e^{3x}-e^{-3x}$

$f(x) = x^{2} \tan {x} \cos{2x}$

Correct answer:

$f(x) = x ^{2} (2^x + 2^{-x})$

Explanation:

If f(x) is an odd function - that is, if f(-c) = -f(c) for all real - then $\int_{-A}^{A} f(x) \; \mathrm{d}x = 0$ for all real , including A = 1 .

If f(x) is positive on the interval [-1,1] , then $\int_{-1}^{1} f(x) \; \mathrm{d}x > 0$ .

We can sleuth out the correct answer by demonstrating that four of these functions are odd, and the fifth is always nonnegative and sometimes positive on the interval.

$f(x) = x^{2} \tan {x} \cos{2x}$

$f(-x) =\left ( -x \right ) ^{2} \tan {\left ( -x \right )} \cos{2\left ( -x \right )}$

$=\left ( -x \right ) ^{2} \tan {\left ( -x \right )} \cos{\left ( -2x \right )}$

$=x^{2} \left (-\tan {x} \right ) \cos{2x}$

$=-x^{2} \tan {x} \cos{2x} = -f(x)$

making this function odd.

$f(x) = x^{3} \sin^{2} {x}$

$f(-x) = (-x) ^{3} \sin^{2} {(-x) }$

$= (-x) ^{3}\left ( -\sin {x } \right )^{2}$

$= -x^{3} \sin ^{2} {x }= -f(x)$

making this function odd.

$f(x) = x^{2} \left ( \sin {x} + \tan {x} \right )$

$f(-x) = (-x)^{2} \left ( \sin {(-x)} + \tan {(-x)} \right )$

$=x^{2} \left ( -\sin {x} - \tan {x} \right )$

$=- x^{2} \left ( \sin {x} + \tan {x} \right ) = - f(x)$

making this function odd.

$f(x)=e^{3x}-e^{-3x}$

$f(-x)=e^{3(-x)}-e^{-3(-x)}$

$=e^{-3x}-e^{3x}$

$=-\left ( e^{3x} - e^{-3x} \right ) = - f(x)$

making this function odd.

However,

$f(x) = x ^{2} (2^x + 2^{-x})$ is nonnegative for all real values of :

$x^{2}, 2^x, 2^ {-x }\geq 0$ , so $f(x) = x ^{2} (2^x + 2^{-x}) \geq 0$ on [-1,1] . Also, since $f(x) = x ^{2} (2^x + 2^{-x}) > 0$ for at least one real $c \in [-1,1]$ :

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

Then $\int_{-1}^{1} f(x) \; \mathrm{d}x > 0$ .

$f(x) = x ^{2} (2^x + 2^{-x})$ is the correct choice.

Report an Error

Example Question #363 : Integrals

Evaluate:

$\int_{0}^{4} |x ^2- 3| \; \mathrm{d}x$

Possible Answers:

$\frac{64 + 12\sqrt{3}}{3}$

$\frac{64 - 12\sqrt{3}}{3}$

$4\sqrt{3}$

$\frac{28 + 12\sqrt{3}}{3}$

$\frac{28}{3}$

Correct answer:

$\frac{28 + 12\sqrt{3}}{3}$

Explanation:

x ^2- 3 > 0 when $x > \sqrt{3}$ or $x < - \sqrt{3}$ and x ^2- 3 < 0 when $- \sqrt{3}< x < \sqrt{3}$ .

The function

f(x)=|x ^2- 3 |

can then be rewritten as

$f(x)= \left\{\begin{matrix} x^2 - 3, & x > \sqrt{3} \\ -x^2 + 3, &-\sqrt{3} \leq x \leq \sqrt{3} \\x^2 - 3, & x <-\sqrt{3} \end{matrix}\right.$ .

Therefore,

$\int_{0}^{4} |x ^2- 3| \; \mathrm{d}x$

$= \int_{0}^{4}f(x) \; \mathrm{d}x$

$= \int_{0}^{\sqrt{3}}f(x) \; \mathrm{d}x + \int_{\sqrt{3}}^{4}f(x) \; \mathrm{d}x$

$= \int_{0}^{\sqrt{3}}( -x^2+3) \; \mathrm{d}x + \int_{\sqrt{3}}^{4}( x^2-3)\; \mathrm{d}x$

$=\left.\begin{matrix} -\frac{x^3}{3}+ 3x \\ \\ \end{matrix}\right|\begin{matrix} \sqrt{3}\\ \\ 0 \end{matrix}+ \left.\begin{matrix} \frac{x^3}{3}- 3x \\ \\ \end{matrix}\right|\begin{matrix} 4\\ \\ \sqrt{3} \end{matrix}$

$=\left ( -\frac{(\sqrt{3}) ^3}{3}+ 3 (\sqrt{3}) \right ) - \left ( -\frac{0^3}{3}+ 3\cdot 0 \right ) + \left ( \frac{4^3}{3}- 3 \cdot 4 \right ) - \left ( \frac{(\sqrt{3})^3}{3}- 3(\sqrt{3}) \right )$

$=\left ( - \sqrt{3} + 3 \sqrt{3} \right ) - 0 + \left ( \frac{64}{3}- 12 \right ) - \left ( \sqrt{3} - 3 \sqrt{3}\right )$

$= 2 \sqrt{3} + \frac{28}{3} - \left ( -2 \sqrt{3} \right )$

$= 4 \sqrt{3} + \frac{28}{3} = \frac{28 + 12\sqrt{3}}{3}$

Report an Error

Example Question #1 : Definite Integrals

Evaluate: $\int_{\pi}^{5\pi} sin(\theta)-cos(\theta) \:d\theta$

Possible Answers:

$\frac{1}{2}$

$-\frac{1}{2}$

Correct answer:

Explanation:

The definite integral $\int_{\pi}^{5\pi} sin(\theta)-cos(\theta) \:d\theta$ can be integrated as is:

$[-cos(\theta)-sin(\theta)]_{\pi}^{5\pi}$

Evaluate the bounds. Be aware of the sign changes.

$=[-cos(5\pi)-sin(5\pi)]-[-cos(\pi)-sin(\pi)]$

=[-(-1)-0]-[-(-1)-0]=[1]-[1]=0

Report an Error

Example Question #1 : Definite Integrals

Evaluate:

$\int_{1}^{6}(3x+4)\: dy$

Possible Answers:

$\frac{145}{2}$

15x+20

94y

Correct answer:

15x+20

Explanation:

Be careful, we are integrating with the respect to y, not x. This means that the integrand itself is treated as a constant.

$\int_{1}^{6}(3x+4)\: dy$

Since we just have a constant, to integrate means to increase the value of the variable y. Thus we get the following,

$(3x+4)\int_1^6dy=(3x+4)\frac{y^{0+1}}{0+1}=(3x+4)y$

$= (3x+4)y|_{1}^{6}$ .

From here we plug in our bounds. We take the upper bound function value and subtract lower bound function value.

= (3x+4)(6)-(3x+4)(1)

=(3x+4)(5)

=15x+20

Report an Error

Example Question #1 : Definite Integrals

Calculate the following definite integral.

$\int_{0}^{1}(e^{2x}+x+2) dx$

Possible Answers:

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

e^2 +2e

e+2

$\frac{1}{2}e^2 + \frac{5}{2}$

$\frac{e^2}{2}+2$

Correct answer:

$\frac{e^2}{2}+2$

Explanation:

To help us evalute the integral, we can split up the expression into 3 parts:

$y(x)= \int_{0}^{1}(e^{2x}+x+2) dx = \int_{0}^{1}e^{2x} dx + \int_{0}^{1}x dx+\int_{0}^{1}2 dx$ .

This allows us to evaluate the integral of each of the three parts, sum them up, and then evaluate the summed up parts from 0 to 1.

The first integral is $\int_{0}^{1}e^{2x} dx = \frac{1}{2}e^{2x}$ .

The second integral is $\int_{0}^{1}xdx=\frac{1}{2}x^2$ .

The third integral is $\int_{0}^{1}2 dx=2x$ .

Sum up all these terms in evaluate between 0 and 1.

$\left(\frac{1}{2}e^{2x}+\frac{1}{2}x^2+2x\right)|_0^1$

$\left(\frac{1}{2}e^{2}+\frac{1}{2}+2\right)-\left(\frac{1}2}\right)= \frac{1}{2}e^2 +2$

Report an Error

Example Question #1 : Definite Integrals

Evaluate the following integral:

$\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}3sin(2x)+1 dx$

Possible Answers:

$\frac{\pi}{2}$

$-3 -\pi$

$-\frac{3}{2}\pi$

$\pi-3$

$\pi$

Correct answer:

$\pi$

Explanation:

The general expression for the integral of a sine function is:

$\int sin(ax)=-\frac{1}{a}cos(ax)$

Using this, the integral above now becomes,

$-\frac{3}{2}cos(2x)+x |_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$

$\left(-\frac{3}{2}cos\left(2\left(\frac{\pi}{2}\right)\right)+\frac{\pi}{2} \right)-\left(-\frac{3}{2}cos\left(2\left(-\frac{\pi}{2}\right)\right)-\frac{\pi}{2} \right)$ , which simplifies to

$\left(\frac{\pi}{2} \right)-\left(-\frac{\pi}{2} \right)=\pi$

Report an Error

Example Question #1 : Definite Integrals

Evaluate the following integral:

$\int_{0}^{2}\frac{dx}{\sqrt{8x+9}}$

Possible Answers:

$\ln \frac{5}{3}$

$4\ln(\frac{3}{5})$

$4\ln(\frac{5}{3})$

Correct answer:

$4\ln(\frac{5}{3})$

Explanation:

To integrate, we must first make the following subsitution:

$u=(8x+9)^{\frac{1}{2}}, du=4(8x+9)^{-\frac{1}{2}}$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, rewrite the given integral, change the bounds in terms of (by plugging in the upper and lower bounds into the equation for in terms of ), and integrate:

$4 \int_{3}^{5} \frac{du}{u}=4(\ln\left | 5\right |-\ln\left | 3\right |)=4\ln(\frac{5}{3})$

The integral was performed using the following rule:

$\int \frac{dx}{x} = \ln\left | x\right |+C$

To perform the definite integration, simply plug in the upper limit of integration and subtract from the result of plugging in the lower limit of integration, as shown above. Simplifying the result, we get

$4\ln(\frac{5}{3})$

Report an Error

Example Question #4 : Definite Integrals

Calculate the value of the definite integral.

$\int_{0}^{9}x+5\,dx$

Possible Answers:

85.5

40.5

Correct answer:

85.5

Explanation:

In order to calculate the definite integral, we apply the inverse power rule which states

$\int x^n\,dx = \frac{x^{n+1}}{n+1}$

Applying this to the problem in this question term by term we get

$\int_{0}^{9}x+5\,dx = \Big(\frac{x^2}{2}+5x\Big)\Big|_{0}^{9}$

And by the corollary of the Fundamental Theorem of Calculus the definite integral becomes

$=\frac{9^2}{2}+5(9)-\Big(\frac{0^2}{2}+5(0)\Big)$

And so

$\int_{0}^{9}x+5\,dx = 85.5$

Report an Error

Example Question #7 : Definite Integrals

Calculate the value of the definite integral.

$\int_{0}^{5}4\,dx$

Possible Answers:

Correct answer:

Explanation:

In order to calculate the definite integral, we apply the inverse power rule which states

$\int x^n\,dx = \frac{x^{n+1}}{n+1}$

Applying this to the problem in this question we get

$\int_{0}^{5}4\,dx = (4x)\Big|_{0}^{5}$

And by the corollary of the Fundamental Theorem of Calculus the definite integral becomes

=4(5)-4(0)=20

And so

$\int_{0}^{5}4\,dx = 20$

Report an Error

View Calculus 2 Tutors

Muhammad
Certified Tutor

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

View Calculus 2 Tutors

Magdy
Certified Tutor

Cairo University, Egypt, Bachelor of Science, Electrical Engineering. New Mexico State University-Main Campus, Doctor of Phil...

View Calculus 2 Tutors

Ping
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Computer Science. University of North Texas, Master of Arts, Education.

All Calculus 2 Resources

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

Popular Subjects

Math Tutors in Miami, Physics Tutors in Philadelphia, Spanish Tutors in Chicago, Reading Tutors in Washington DC, LSAT Tutors in Atlanta, French Tutors in Philadelphia, Chemistry Tutors in Seattle, Biology Tutors in Boston, Chemistry Tutors in San Diego, SAT Tutors in San Francisco-Bay Area

Popular Courses & Classes

ISEE Courses & Classes in San Diego, Spanish Courses & Classes in Philadelphia, Spanish Courses & Classes in San Diego, SAT Courses & Classes in Los Angeles, GMAT Courses & Classes in Boston, SAT Courses & Classes in Seattle, SSAT Courses & Classes in New York City, SAT Courses & Classes in Houston, ACT Courses & Classes in San Francisco-Bay Area, GRE Courses & Classes in Boston

Popular Test Prep

ACT Test Prep in San Francisco-Bay Area, SAT Test Prep in Seattle, ISEE Test Prep in Houston, ISEE Test Prep in Denver, SAT Test Prep in Philadelphia, LSAT Test Prep in Miami, GMAT Test Prep in Boston, LSAT Test Prep in San Diego, ACT Test Prep in Philadelphia, GMAT Test Prep in New York City

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 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #361 : Integrals

Example Question #362 : Integrals

Example Question #363 : Integrals

Example Question #1 : Definite Integrals

Example Question #1 : Definite Integrals

Example Question #1 : Definite Integrals

Example Question #1 : Definite Integrals

Example Question #1 : Definite Integrals

Example Question #4 : Definite Integrals

Example Question #7 : Definite Integrals

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: