We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : How to find area of a region

Study concepts, example questions & explanations for Calculus 1

Create An Account

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Next →

Example Question #71 : How To Find Area Of A Region

Find the indefinite integral

$\int10sin(s)+7cos(s)ds$

Possible Answers:

42cos(s)

-10cos(s)+7sin(s)+C

-20cos(s)+14cos(s)+C

20cos(s)-14sin(s)+C

-20cos(s)+14sin(s)+C

Correct answer:

-10cos(s)+7sin(s)+C

Explanation:

Know the integrals and derivatives of trigonometric functions from memory, they occur very often.

$\frac{\mathrm{d}}{\mathrm{d} x}sin(x)=cos(x)$ and $\frac{\mathrm{d} }{\mathrm{d} x}cos(x)=-sin(x)$

Now notice the pattern of 'going backwards' from taking derivatives to integrating:

$\int sin(x)dx=-cos(x)$ and $\int cos(x)dx=sin(x)$

We get,

$\int10sin(s)+7cos(s)ds=-10cos(s)+7sin(s)+C$

Report an Error

Example Question #72 : How To Find Area Of A Region

Find the indefinite integral

$\int (1+4t)^3dt$

Possible Answers:

$\frac{(1+4t)^4}{4}+C$

$\frac{(1+4t)^4}{16}+C$

$\frac{(1+4t)^2}{16}+C$

$\frac{(1+16t)^4}{4}+Ct$

$\frac{(1+42t)^4}{16}+C$

Correct answer:

$\frac{(1+4t)^4}{16}+C$

Explanation:

We have to use what is called direct substitution. We let some quantity be represented by a new variable, and take the derivative of that. Then we replace the quantity which was in the original equation, with this variable. Commonly, mathematicians use the letter for this substitution; nicknaming it a 'u-sub'.

Using direct substitution, Let u=1+4t

Recall the power rule for differentiation:

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=n\cdot x^{n-1}$

then take the derivative of both sides of the equation,

$du=\frac{\mathrm{d} }{\mathrm{d} t}(1+4t)=4dt$

$du=4dt\rightarrow \frac{du}{4}=dt$ .

You can replace (1+4t) with and now you can replace with $\frac{du}{4}$ .

Now recall the power rule for integration:

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

Finally, take the $\frac{1}{4}$ fraction outside the integral and replace the old expression with the new variable. And then integrate and when you are done, replace the new variable back with the old quantity, as if you never even did it.

$\frac{1}{4}\int u^3du=\frac{1}{4}\cdot \frac{u^4}{4}=\frac{(1+4t)^4}{16}+C$

Report an Error

Example Question #73 : How To Find Area Of A Region

Find the indefinite integral

$\int z^3\sqrt{2+z^4}dz$

Possible Answers:

$\frac{(2+z^4)^{-3/2}}{6}+C$

$\frac{(42+z^4)^{3/2}}{6}+C$

$\frac{(2+z^4)^{-1/2}}{6}+C$

$\frac{(2+z^4)^{3/2}}{6}+C$

$\frac{(2+z^4)^{1/2}}{6}+C$

Correct answer:

$\frac{(2+z^4)^{3/2}}{6}+C$

Explanation:

Begin by making the substitution and taking the derivative, and recall the power rule for differentiation.

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=n\cdot x^{n-1}$

And so,

Let u=2+z^4

$du=4z^3dz\rightarrow \frac{du}{4}=z^3dz$ .

Now we replace 2+4z with and z^3dz with $\frac{du}{4}$

Recall the power rule for integration,

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

Finally, pull out the $\frac{1}{4}$ from the integrand and write the new integral and then solve. After you solve the integral, put the original quantity back in. That is replace with what it originally equalled.

$=\frac{1}{4}\int \sqrt{u}du=\frac{1}{4}\int u^{\frac{1}{2}}du=\frac{1}{6}u^{3/2}=\frac{1}{6}(2+z^4)^{3/2}+C$

Report an Error

Example Question #74 : How To Find Area Of A Region

y_1=x and $y_2=\sqrt{x}$

Find the area of the region created by the two functions.

Possible Answers:

$\frac{2}{3}$

$\frac{1}{2}$

$\frac{1}{6}$

$\frac{5}{6}$

Correct answer:

$\frac{1}{6}$

Explanation:

First, graph the two functions in order to identify the boundaries of the region. You will find that they are [0,1] .

Therefore, when you set up your integral, it will be from zero to one.

Next you set up your integral by subtracting the lower function from the upper function.

In this case, it would be $\sqrt{x}-x$ .

This means your integral should look like

$\int_0^1\sqrt{x}-x dx$ .

Then solve using the power rule

$\int x^ndx=\frac{x^{n+1}}{n+1}$ :

$\\ \int_0^1\sqrt{x}-x dx=\left[\frac{2x^{\frac{3}{2}}}{3}-\frac{x^2}{2}\right]_0^1\\=\left[\frac{2(1)^\frac{3}{2}}{3}-\frac{(1)^2}{2}\right]-\left[\frac{2(0)^\frac{3}{2}}{3}-\frac{(0)^2}{2}\right]\\ =\left[\frac{2}{3}-\frac{1}{2}\right]-[0-0]=\frac{1}{6}$

Report an Error

Example Question #75 : How To Find Area Of A Region

y_1=cos(x) and y_2=sin(x)

Find the region (to the nearest thousandth) created by the functions and bound by the y-axis and $x=\frac{\pi}{4}$ .

Possible Answers:

$\sqrt2-1$

0.414

0.707

Correct answer:

0.414

Explanation:

First, set up your integral, using the given bounds, by subtracting the lower function from the upper function.

In this case, it would be cos(x)-sin(x) .

Therefore the integral would look like

$\int_0^\frac{\pi}{4}cos(x)-sin(x)dx$ .

Then solve using the rules for integrals of the trig functions:

$\\ \int_0^\frac{\pi}{4}cos(x)-sin(x)dx=[sin(x)-(-cos(x))]_0^\frac{\pi}{4}\\ =\left[sin\left(\frac{\pi}{4}\right)+cos\left(\frac{\pi}{4}\right)\right]-[sin(0)+cos(0)]\\=\left[\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\right]-[0+1]\\=\left[\frac{2\sqrt2}{2}\right]-[1]\\ =\sqrt2-1\approx 0.414$ .

Report an Error

Example Question #76 : How To Find Area Of A Region

y_1=x and y_2=3x

Find the region created by the two functions bounded by y-axis and x=3 .

Possible Answers:

$\frac{27}{2}$

$\frac{18}{3}$

Correct answer:

Explanation:

First, set up your integral, using the given bounds, by subtracting the lower function from the upper function.

In this case, in the given bounds, the integral will be

$\int_0^33x-xdx$ .

Then solve using the power rule

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

$\\ \int_0^33x-xdx=\left[\frac{3x^2}{2}-\frac{x^2}{2}\right]_0^3\\ \\=\left[\frac{3(3)^2}{2}-\frac{(3)^2}{2}\right]-\left[\frac{3(0)^2}{2}-\frac{(0)^2}{2}\right]\\ \\=\left[\frac{3(9)}{2}-\frac{9}{2}\right]-[0]\\ \\=\frac{27}{2}-\frac{9}{2}\\ \\=\frac{18}{2}=9$

Report an Error

Example Question #77 : How To Find Area Of A Region

y_1=x and y_2=x^2-2

What is the area of the region created by the two functions?

Possible Answers:

4.5

-4.5

Correct answer:

4.5

Explanation:

First, set up your integral using the bounds found either by graphing or setting the two functions up to equal one another and solving.

Then subtract the lower function from the upper function.

In this case, in the given bounds, the integral will be

$\int_{-1}^2x-(x^2-2)dx$ .

Then solve using the power rule

$\int x^ndx=\frac{x^{n+1}}{n+1}$ :

$\\ \int_{-1}^2 x-(x^2-2)dx=\int_{-1}^2 x-x^2+2 dx\\ \\=\left[\frac{x^2}{2}-\frac{x^3}{3}+2x\right]_{-1}^2\\ \\=\frac{(2)^2}{2}-\frac{(2)^3}{3}+2(2)-\left[\frac{(-1)^2}{2}-\frac{(-1)^3}{3}+2(-1)\right]\\ \\=\frac{4}{2}-\frac{8}{3}+4-\frac{1}{2}-\frac{1}{3}+2\\ \\=8-3-\frac{1}{2}=4.5$

Report an Error

Example Question #78 : How To Find Area Of A Region

y_1=x and y_2=x^2

What is the area of the region created by the two functions?

Possible Answers:

$\frac{5}{6}$

$\frac{1}{6}$

$\frac{-1}{6}$

$\frac{-5}{6}$

Correct answer:

$\frac{1}{6}$

Explanation:

First, graph the two functions in order to identify the boundaries of the region. You will find that they are [0,1] .

Therefore, when you set up your integral, it will be from zero to one. Next you set up your integral by subtracting the lower function from the upper function.In this case it will be x-x^2 .

This means your integral should look like

$\int_0^1 x-x^2 dx$ .
Then solve using the power rule

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

$\\ \int_0^1 x-x^2 dx=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1\\ \\=\frac{(1)^2}{2}-\frac{(1)^3}{3}-\left[\frac{(0)^2}{2}-\frac{(0)^3}{3}\right]\\ \\=\frac{1}{2}-\frac{1}{3}\\ \\=\frac{1}{6}$

Report an Error

Example Question #4071 : Calculus

y_1=x and y_2=-x^2

What is the area of the region created by the two functions?

Possible Answers:

$\frac{5}{6}$

$\frac{-5}{6}$

$\frac{1}{6}$

$\frac{-1}{6}$

Correct answer:

$\frac{1}{6}$

Explanation:

First, graph the two functions in order to identify the boundaries of the region. You will find that they are [-1,0] . Therefore, when you set up your integral, it will be from zero to one.

Next you set up your integral by subtracting the lower function from the upper function. In this case, it would be x^2-x .

This means your integral should look like

$\int_{-1}^0x^2-x dx$ .

Then solve using the power rule

$\int x^ndx=\frac{x^{n+1}}{n+1}$ :

$\\ \int_{-1}^0 x^2-xdx=-\int_{-1}^0x^2+xdx\\ \\=-\left[\frac{x^3}{3}+\frac{x^2}{2}\right]_{-1}^0\\ \\=-\left[\frac{(0^3)}{3}+\frac{(0)^2}{2}-\left[\frac{(-1)^3}{3}+\frac{(-1)^2}{2}\right]\right]\\ \\=-\left[-\left[\frac{-1}{3}+\frac{1}{2}\right]\right]\\ \\=-\left[\frac{1}{3}-\frac{1}{2}\right]\\ \\=-\left[-\frac{1}{6}\right]\\ \\=\frac{1}{6}$

Report an Error

Example Question #151 : Regions

y=sin(x)

What is the area of the region created by the function and the given bounds y-axis and $\pi$ ?

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

Set up your integral using the given bounds, then solve:

Remember the rules of trigonometric functions,

$\int sin(x)=-cos(x)$ .

Therefore our equation becomes,

$\\ \int_0^\pi sin(x) dx=-cos(x)]_0^\pi\\ \\=-cos(\pi)-(-cos(0))\\ \\=-(-1)-(-(1))\\ \\=1+1=2$ .

Report an Error

← Previous 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Next →

View Calculus Tutors

Heather
Certified Tutor

University of Wisconsin-River Falls, Bachelor of Science, Mathematics. University of Iowa, Master of Science, Mathematics.

View Calculus Tutors

Anthony
Certified Tutor

George Washington University, Bachelor of Science, Biology, General. Columbia University in the City of New York, Master of S...

View Calculus Tutors

Gregory
Certified Tutor

Pennsylvania State University-Main Campus, Bachelor of Science, Chemical Engineering. Carnegie Mellon University, Doctor of P...

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

MCAT Tutors in Phoenix, ACT Tutors in Boston, Algebra Tutors in San Francisco-Bay Area, Chemistry Tutors in Seattle, Reading Tutors in Boston, SSAT Tutors in Boston, French Tutors in Atlanta, GRE Tutors in San Francisco-Bay Area, Math Tutors in Atlanta, LSAT Tutors in Houston

Popular Courses & Classes

ISEE Courses & Classes in Los Angeles, LSAT Courses & Classes in Houston, GMAT Courses & Classes in Chicago, Spanish Courses & Classes in Boston, MCAT Courses & Classes in Los Angeles, GMAT Courses & Classes in Washington DC, GRE Courses & Classes in Seattle, ISEE Courses & Classes in Miami, Spanish Courses & Classes in Dallas Fort Worth, ISEE Courses & Classes in Seattle

Popular Test Prep

SSAT Test Prep in Phoenix, MCAT Test Prep in Washington DC, ACT Test Prep in Houston, LSAT Test Prep in New York City, SAT Test Prep in Washington DC, GRE Test Prep in Chicago, SSAT Test Prep in Atlanta, SSAT Test Prep in San Francisco-Bay Area, GRE Test Prep in Dallas Fort Worth, SAT Test Prep in Atlanta

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 1 : How to find area of a region

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #71 : How To Find Area Of A Region

Example Question #72 : How To Find Area Of A Region

Example Question #73 : How To Find Area Of A Region

Example Question #74 : How To Find Area Of A Region

Example Question #75 : How To Find Area Of A Region

Example Question #76 : How To Find Area Of A Region

Example Question #77 : How To Find Area Of A Region

Example Question #78 : How To Find Area Of A Region

Example Question #4071 : Calculus

Example Question #151 : Regions

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: