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 #231 : Integrals

Give the area of the region bounded by the curves of the functions

$f(x) = 4e^{x}$

and

$g(x) = e^{2x} + 3$ .

Possible Answers:

$4 - \ln 3$

$12 \ln 3$

$4 + \ln 3$

$4 + \ln 27$

$4 - \ln 27$

Correct answer:

$4 - \ln 27$

Explanation:

The two curves intersect when

g(x) = f(x) .

$e^{2x} + 3 = 4e^{x}$

$e^{2x} - 4e^{x} + 3 = 0$

$\left (e^{x} \right )^{2} - 4e^{x} + 3 = 0$

$\left ( e^{x} - 1 \right ) \left ( e^{x} - 3 \right ) = 0$

$e^{x} = 1$

x = 0

$e^{x} = 3$

$x = \ln 3$

Therefore, the area between the curves is

$\left |\int_{0}^{\ln 3} \left [g(x) - f(x) \right ] \mathrm{d}x \right |$

$= \left |\int_{0}^{\ln 3} \left (e^{2x} - 4e^{x} + 3 \right ) \mathrm{d}x \right |$

$=\left | \left.\begin{matrix} \frac{e^{2x}}{2} -4e^{x}+ 3x \\ \\ \end{matrix}\right| \begin{matrix} \ln 3\\ \\ 0 \end{matrix} \right |$

$=\left |\left ( \frac{e^{2 \ln 3}}{2} -4e^{\ln 3}+ 3 \ln 3 \right )- \left ( \frac{e^{2 \cdot 0}}{2} -4e^{0}+ 3 \cdot 0 \right ) \right |$

$=\left |\left ( \frac{e^{ \ln 9}}{2} -4e^{\ln 3}+ 3 \ln 3 \right )- \left ( \frac{e^{ 0}}{2} -4e^{0}+ 0 \right ) \right |$

$\; =\left | \left ( \frac{9}{2} - 4 \cdot 3 + \ln 27 \right) - \left ( \frac{1}{2} -4 \cdot 1 + 0 \right ) \right |$ $\; =\left | \left ( \frac{9}{2} - 4 \cdot 3 + \ln 27 \right) - \left ( \frac{1}{2} -4 \cdot 1 + 0 \right ) \right |$

$\; = \left |\left ( \frac{9}{2} -12 + \ln 27 \right) - \left ( \frac{1}{2} -4 \right ) \right |$

$=\left | \frac{9}{2} -12 + \ln 27 - \frac{1}{2} + 4 \right |$

$=\left | -4 + \ln 27 \right | = 4 - \ln 27$

Report an Error

Example Question #232 : Integrals

Give the area of the region bounded by the curves of the functions

f(x) = x^3 + x - 5

and

g(x) = 17x - 5 .

Possible Answers:

128

256

Correct answer:

128

Explanation:

The two curves intersect when

f(x) = g(x) .

x^3 + x - 5 = 17x - 5

x^3 + x - 5 - 17x +5 =0

x^3 -16x =0

$x\left ( x^2 -16 \right ) =0$

$x\left ( x+4 \right )\left ( x-4 \right ) =0$

x = -4,0,4

So the curves intersect at three points: x = -4,0,4 .

The area of the region bound by the curves is

$\int_{-4}^{0}\left |f(x) - g(x) \right | \mathrm{d}x+ \int_{0}^{4}\left |f(x) - g(x) \right | \mathrm{d}x$

$=\left |\int_{-4}^{0}\left ( x^3 -16x \right )\mathrm{d}x \right |+\left | \int_{0}^{4} \left ( x^3 -16x \right ) \; \mathrm{d}x \right |$

$=\left | \left.\begin{matrix} \left ( \frac{x^{4}}{4} -\frac{16x^{2}}{2} \right ) \\ \\ \end{matrix}\right|\begin{matrix} 0\\ \\ -4 \end{matrix} \right | + \left | \left.\begin{matrix} \left ( \frac{x^{4}}{4} -\frac{16x^{2}}{2} \right ) \\ \\ \end{matrix}\right|\begin{matrix} 4\\ \\ 0 \end{matrix} \right |$

$=\left | \left.\begin{matrix} \left ( \frac{x^{4}}{4} -8x^{2} \right ) \\ \\ \end{matrix}\right|\begin{matrix} 0\\ \\ -4 \end{matrix} \right | + \left | \left.\begin{matrix} \left \left ( \frac{x^{4}}{4} -8x^{2} \right ) \\ \\ \end{matrix}\right|\begin{matrix} 4\\ \\ 0 \end{matrix} \right |$

$=\left | \left ( \frac{0^{4}}{4} -8 \cdot 0^{2} \right ) - \left ( \frac{(-4)^{4}}{4} -8 \cdot (-4)^{2} \right ) \right | + \left | \left ( \frac{4^{4}}{4} -8 \cdot 4^{2} \right ) - \left ( \frac{0^{4}}{4} -8 \cdot 0^{2} \right ) \right |$

$=\left | 0 - \left ( \frac{256}{4} -8 \cdot 16 \right ) \right | + \left | \left ( \frac{256}{4} -8 \cdot 16 \right ) - 0 \right |$

$=\left | - \left (64 -128\right ) \right | + \left | \left ( 64 -128 \right ) \right |$

= 64 + 64 = 128

Report an Error

Example Question #233 : Integrals

Give the area of the region bounded by the curves of the functions

$f(x) = \frac{16}{x}$

and

g(x) = 10 -x .

Possible Answers:

$16 \ln 4$

$30 - 16 \ln 4$

$30 - 16 \ln 16$

$56 - 16 \ln 16$

$56 - 16 \ln 4$

Correct answer:

$30 - 16 \ln 4$

Explanation:

The two curves intersect when

g(x) = f(x) .

$\frac{16}{x} =10 -x$

$x - 10 + \frac{16}{x} = 0$

Since it can be assumed that $x \neq 0$ ,

$x \left ( x - 10 + \frac{16}{x} \right )=x\cdot 0$

$x ^{2}- 10x + 16 = 0$

(x-2) (x-8)= 0

Therefore, either x-2 = 0 , in which case x = 2 , or x - 8 = 0 , in which case x = 8 .

Therefore, the area between the curves is

$\int_{2}^{8}\left |g(x) - f(x) \right | \mathrm{d}x$

$=\left |\int_{2}^{8} \left ( x - 10 + \frac{16}{x} \right ) \mathrm{d}x \right |$

$=\left | \left.\begin{matrix} \frac{x^2}{2} -10x + 16 \ln x\\ \\ \end{matrix}\right| \begin{matrix} 8\\ \\ 2 \end{matrix} \right |$

$=\left | \left (\frac{8^2}{2} -10 \cdot 8 + 16 \ln 8 \right ) - \left ( \frac{2^2}{2} -10 \cdot 2 + 16 \ln 2 \right ) \right |$

$=\left | \left (32-80 + 16 \ln 8 \right ) - \left ( 2-20 + 16 \ln 2 \right ) \right |$

$=\left | \left (-48 + 16 \ln 8 \right ) - \left ( -18 + 16 \ln 2 \right ) \right |$

$=\left | -48 + 18 + 16 \ln 8 - 16 \ln 2 \right |$

$= \left | -30 + 16( \ln 8 - \ln 2 ) \right | =\left | -30 + 16 \ln 4 \right |= 30 - 16 \ln 4$

Report an Error

Example Question #1 : Area Under A Curve

Find the area under the curve f(x) = e^x from x = -2 to x = 2 .

Possible Answers:

$e^2 + 2e^{-2}$

$1-2e^{-2}$

2e^2

$e^2 - e^{-2}$

2e^2 - 1

Correct answer:

$e^2 - e^{-2}$

Explanation:

The area under a curve f(x) between two x-values $x_{1},x_2$ corresponds to the integral

$\int_{x_1}^{x_2} f(x)dx$ .

In this case, f(x) = e^x, x_1 = -2, x_2 = 2 , and we can find by the fundamental theorem of calculus that...

$\int_{-2}^{2} e^xdx$

= [e^x] evaluated from to

$= e^2 - e^{-2}$

Report an Error

Example Question #2 : Area Under A Curve

Find the area between f(x) and g(x) , such that $f(x) = \sqrt{x}$ and g(x) = x^3

Possible Answers:

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{5}{12}$

$\frac{1}{3}$

$\frac{7}{12}$

Correct answer:

$\frac{5}{12}$

Explanation:

The first step to find the area between two curves is to set them equal to each other and find their intersection points.

$\\f(x) = g(x)\\ \sqrt{x} = x^3\\ x = x^6\\ x = 0, x = 1$

The next step is to find which curve is "on the top" and which is "on the bottom" inside the interval (0,1) . We find that f(x) is on the top and g(x) is on the bottom.

Now, to find the area between the two curves, we take the integral from to of the top curve minus the bottom curve.

$\\\int_{0}^{1}(f(x) - g(x))dx\\ \int_{0}^{1}(\sqrt{x} - x^3)dx\\ \left [ \frac{2}{3}x^{3/2} - \frac{1}{4}x^4 \right ]$ evaluated from to

$\\= \left(\frac{2}{3} - \frac{1}{4}\right) - (0 - 0)\\ = \frac{5}{12}$

Report an Error

Example Question #236 : Integrals

Find the total area under the curve $f(x)=2xe^{x^2}$ from $x = 0\ to\ x = 5.$

Possible Answers:

$\frac{25e^{25}}{2}$

$10e^{25}$

$e^{25}$

$e^{25}-1$

Correct answer:

$e^{25}-1$

Explanation:

The definite integral of f(x) over the interval [0, 5] is the area under the curve.

$\int_{0}^{5}2xe^{x^2}dx$

A substitution makes this integral easier to evaluate. Let $u=e^{x^2}$ . Then $du=2xe^{x^2}dx$ . We can also change the limits of integration so that they are in terms of u. When x=5, $u=e^{25}$ . When x=0, $u=e^{0}=1$ . Making these substitutions, the integral becomes

$\int_{1}^{e^{25}}du=u \left |_{1}^{25}=e^{25}-1$

Report an Error

Example Question #1 : Area Under A Curve

What is the area under the curve $f(x)=\frac{x}{x^2+1}$ over the interval $1\le x\le 2$ ?

Possible Answers:

$\frac{1}{2}ln\left(\frac{5}{2}\right)$

$\frac{ln5}{2}$

$\frac{21}{5}$

e^2-1

Correct answer:

$\frac{1}{2}ln\left(\frac{5}{2}\right)$

Explanation:

The area under a curve f(x) over the interval [a, b] is $\int_{a}^{b}f(x)dx$ .

In this example, this leads to the definite integral

$\int_{1}^{2}\frac{x}{x^2+1}dx$ .

A substitution makes the antiderivative of this function more obvious.

Let

$u=x^2+1,\ du=2xdx, \frac{du}{2}=xdx.$ .

We can also convert the limits of integration to be in terms of to simplify evaluation. When $x=1,\ u=2$ , and when $x=2, \ u=5$ .

Making these substitutions results in

$\int_{2}^{5}\frac{\frac{du}{2}}{u}=\frac{1}{2}\int_{2}^{5}\frac{du}{u}$ .

Recall that in general $\int\frac{dx}{x}=ln(x)+c$ , so evaluating the integral leads to

$\frac{1}{2}\int_{2}^{5}\frac{du}{u}=\frac{1}{2}ln(u)|_{2}^{5}=\frac{1}{2}(ln(5)-ln(2))$

$=\frac{1}{2}ln\left(\frac{5}{2}\right)$

Report an Error

Example Question #238 : Integrals

Find the following indefinite integral:

$\int \frac{1}{x^2+x-6} \ dx$

Possible Answers:

$\ln|x^2+x-6|+C$

$\frac{1}{5}}\ln|x-2|-\frac{1}{5}\ln|x+3|$

$-\frac{1}{5}}\ln|x-2|+\frac{1}{5}\ln|x+3| + C$

$\ln|x^2+x-6|$

$\frac{1}{5}}\ln|x-2|-\frac{1}{5}\ln|x+3| + C$

Correct answer:

$\frac{1}{5}}\ln|x-2|-\frac{1}{5}\ln|x+3| + C$

Explanation:

In order to find the anti-derivative, lets first factor the denominator.

$\int \frac{1}{x^2+x-6} \ dx$

$= \int \frac{1}{(x+3)(x-2)} \ dx$

Now we use partial-fraction decomposition to get this into a form we know how to take the anti-derivative.

$\int \frac{A}{(x+3)} +\frac{B}{(x-2)} \ dx$

Now we need to find the values of A and B. First lets write out the equation.

1=A(x-2)+B(x+3)

1=Ax-2A+Bx+3B

1=x(A+B)+(3B-2A)

Now we can get our system of equations in order to find A and B.

$\\ 1: A+B=0 \\ 2: 3B-2A=1$

From equation 1, we get

A=-B

Now we can substitute this into equation 2.

3B-2(-B)=1

$5B=1 \rightarrow B=\frac{1}{5}$

Now we substitue B back into equation 1.

$A=-\frac{1}{5}$

Now we can put the values of A, and B into our problem.

$\int \frac{-\frac{1}{5}}{(x+3)} +\frac{\frac{1}{5}}{(x-2)} \ dx$

Now we can simply take the anti-derivative.

$\int \frac{-\frac{1}{5}}{(x+3)} +\frac{\frac{1}{5}}{(x-2)} \ dx$

$=\frac{1}{5}}\ln|x-2|-\frac{1}{5}\ln|x+3| + C$

Report an Error

Example Question #239 : Integrals

Evaluate:

$\int_{0}^{1} t \ dt$

Possible Answers:

$\frac{1}{4}$

$\frac{1}{3}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

To evaluate the integral, we use inverse power law.

Remember that the inverse power law is

$\int x^a =\frac{1}{a+1}x^{a+1}$ .

Lets apply this to our problem.

$\int_{0}^{1} t \ dt$

$= \frac{1}{2}t^2\left |_{0}^{1}$

From here we plug in the bounds and subtract the lower bound function value from the upper bound function value.

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

$= \frac{1}{2}-0=\frac{1}{2}$

Report an Error

Example Question #240 : Integrals

Find the area under the curve of the function y = x^2 - 2x between x = 2 and x = 4 .

Possible Answers:

$\frac{25}{3}$

$\frac{20}{3}$

$\frac{32}{3}$

$\frac{64}{3}$

Correct answer:

$\frac{20}{3}$

Explanation:

Asking for the area under the curve means you are going to be doing an integral of the function provided. The bounds are given for you. The integral thus looks like this: $\int_{2}^{4} x^{2} - 2x dx$ .

When integrated, you get

$\frac{x^3}{3} - x^2$ evaluated at x = 4 and x = 2 .

At x = 4 the expression evaluates to $\frac{64}{3} - 16$ and at x = 2 it evalutes to $\frac{8}{3} - 4$ .

Subtracting the two gives you your final answer of $\frac{20}{3}$ .

Report an Error

View Calculus 2 Tutors

Bobby
Certified Tutor

University of Houston - Downtown, Bachelor of Science, Applied Mathematics.

View Calculus 2 Tutors

Riley
Certified Tutor

Reed College, Bachelor in Arts, Mathematics.

View Calculus 2 Tutors

Aaron
Certified Tutor

Tulsa Community College, Associate in Science, Mechanical Engineering.

All Calculus 2 Resources

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

Popular Subjects

GRE Tutors in Miami, Computer Science Tutors in Houston, Spanish Tutors in Houston, Statistics Tutors in Seattle, Math Tutors in Miami, SSAT Tutors in Philadelphia, LSAT Tutors in Miami, Algebra Tutors in Boston, MCAT Tutors in Seattle, LSAT Tutors in Chicago

Popular Courses & Classes

GMAT Courses & Classes in Houston, SAT Courses & Classes in Miami, GMAT Courses & Classes in Los Angeles, SSAT Courses & Classes in San Diego, SSAT Courses & Classes in Boston, ISEE Courses & Classes in Washington DC, MCAT Courses & Classes in Phoenix, SAT Courses & Classes in Seattle, MCAT Courses & Classes in Chicago, SAT Courses & Classes in Dallas Fort Worth

Popular Test Prep

GMAT Test Prep in Los Angeles, GRE Test Prep in Denver, ISEE Test Prep in San Francisco-Bay Area, SSAT Test Prep in Boston, SAT Test Prep in Houston, GRE Test Prep in San Diego, LSAT Test Prep in Miami, SAT Test Prep in Dallas Fort Worth, LSAT Test Prep in Atlanta, SAT 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

Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #231 : Integrals

Example Question #232 : Integrals

Example Question #233 : Integrals

Example Question #1 : Area Under A Curve

Example Question #2 : Area Under A Curve

Example Question #236 : Integrals

Example Question #1 : Area Under A Curve

Example Question #238 : Integrals

Example Question #239 : Integrals

Example Question #240 : Integrals

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: