Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Functions

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

Example Question #5 : How To Find Area Of A Region

Find the area between the curves y=2x and y=x^2 .

Possible Answers:

$\frac{5}{3}$

$\frac{4}{3}$

$\frac{2}{3}$

Correct answer:

$\frac{4}{3}$

Explanation:

To solve this problem, we first need to find the point where the two equations are equal. Doing this we find that

$2x=x^2\Rightarrow x^2-2x=0\Rightarrow x(x-2)=0$ .

From this, we see that the two graphs are equal at x=0 and x=2 . We also know that for 0<x<2 , y=2x is greater than y=x^2 .

So to find the area between these curves we need to evaluate the integral $\int_{0}^{2} (2x-x^2)dx$ .

The solution to the integral is

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

Evaluating this at x=2 and x=0 we get

$(2*2-\frac{1}{3}2^3)-(0-0)=\frac{12}{3}-\frac{8}{3}=\frac{4}{3}.$

Report an Error

Example Question #6 : How To Find Area Of A Region

Find the value of $\int_{0}^{\infty}xe^{-x^2}dx.$

Possible Answers:

$\frac{3}{2}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

To solve this problem, we will need to do a -substitution. Letting

$u=-x^2\Rightarrow \frac{du}{dx}=-2x\Rightarrow \frac{-du}{2}=xdx$ .

Substituting our u(x) function back into the integral, we get

$\int \frac{-1}{2}e^udu=\frac{-1}{2}e^u=\frac{-1}{2}e^{-x^2}.$

Evaluating this at $x=\infty$ and x=0 we get

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

Report an Error

Example Question #7 : How To Find Area Of A Region

Find the average value of x^3 - 2x^2 +x on the interval (0, 1).

Possible Answers:

1/6

1/3

1/12

-1/3

Correct answer:

1/12

Explanation:

The average is given by integration as:

$\text{avg}_{A, B}[ f ] = \frac{1}{B - A} \int_A^B f(x)~dx$

This means that:

$\text{avg}_{0, 1}[x \to x^3 - 2x^2 + x ] = \frac{1}{1 - 0} \int_0^1 [x^3 - 2x^2 + x]~dx$

$= \left[\frac{x^4}{4} - \frac{2 x^3}{3} + \frac{x^2}{2}\right]_0^1 = \frac{1}{4} - \frac{2}{3} + \frac{1}{2} = \frac{1}{12}$

Report an Error

Example Question #2971 : Functions

We have the function $f(x)=\sqrt{x}$ and it is used to form a three-dimensional figure by rotating it about the line $y=4$ . Find the volume of that figure from $x=0$ to $x=5$ .

Possible Answers:

56.111

Area is infinite

66.667

78.832

Correct answer:

78.832

Explanation:

Imagine a point somewhere on the function $f(x)=\sqrt{x}$ and it rotates about $y=4$ to form a circle with a circumference of $2\pi (4-\sqrt{x})$ where $(4-\sqrt{x})$ is the radius. You may have to draw a picture/graph to make sure this is clear.

Next, pretend that this circle is a circular strip with thickness $\Delta x$ . To find the area of that circular strip, we times that thickness by the circumference so that we have

$2\pi (4-\sqrt{x})\Delta x$

Now, imagine that the three dimensional figure is made up of many of these circular strips. To find the total volume, we need to sum up the areas of all of these strips. We do this by turning $2\pi (4-\sqrt{x})\Delta x$ into an integral from 0 to 5.

$Volume=2\pi \int_{0}^{5}(4-\sqrt{x})dx$

Perform your integration

$Volume=2\pi \left [ 4(5)-\frac{2}{3}(5)^{\frac{3}{2}} - 4(0)+\frac{2}{3}(0)^{\frac{3}{2}} \right ]=2\pi\left [ 20-\frac{2}{3}\sqrt{5^3} \right ]$

Volume=78.832

Report an Error

Example Question #2972 : Functions

Calculate the area between the parabola f(x)=4+5x-x^2 and the line f(x)=x-1 .

Possible Answers:

$\frac{98}{3}$

$\frac{100}{3}$

Correct answer:

Explanation:

To calculate the area between these two functions, we are going to need to set up a definite integral, so our first step is to see where our two boundaries intersect. We do this by setting our two functions equal to each other, and solving for the x values at which they intersect:

4+5x-x^2=x-1

x^2-4x-5=(x-5)(x+1)=0

x=5,x=-1

Here we can see that our two functions intersect at x=5 and x= -1, so these will be the limits for our definite integral. Now that we know our limits, we set up our integral of the function of the upper boundary minus the function for the lower boundary:

$\int_{-1}^{5}((4+5x-x^2)-(x-1))dx=\int_{-1}^{5}(-x^2+4x+5)dx$

$=(-\frac{1}{3}x^3+2x^2+5x)_{-1}^{5}$

$=(-\frac{1}{3}(5)^3+2(5)^2+5(5))-(-\frac{1}{3}(-1)^3+2(-1)^2+5(-1))$

$=(-\frac{125}{3}+50+25)-(\frac{1}{3}+2-5)=(\frac{100}{3})-(-\frac{8}{3})=36$

Report an Error

Example Question #2973 : Functions

Find the area under the curve of y=1 from [-1,1] .

Possible Answers:

Correct answer:

Explanation:

To find the area under the curve, integrate the function and evaluate at the bounds.

$\int_{-1}^{1} 1 dx= x|_{-1}^{1} = 1-(-1) = 2$

Report an Error

Example Question #2974 : Functions

Find the area bounded by y=x and $y=\sqrt x$ .

Possible Answers:

$\frac{1}{6}$

$\frac{1}{3}$

$\frac{1}{5}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{6}$

Explanation:

The top curve will be $y=\sqrt x$ , and the bottom curve will be y=x . The area is the integral of the top minus the bottom curve.

First, determine the bounds of integration by setting both equations equal to each other, and solve for x. These values are where both graphs intersect.

$x= \sqrt x$

x^2 = x

x^2-x=0

x(x-1)=0

x=0

x=1

The bounds of integration will be from 0 to 1. Setup and evaluate the integral.

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

Report an Error

Example Question #15 : Area

What is the area under the curve bounded by y=x^3 from [1,3] ?

Possible Answers:

$\frac{81}{4}$

Correct answer:

Explanation:

The area under the curve is the integral of the function evaluated at the interval given.

Write the integral to be evaluated.

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

$\frac{81}{4}-\frac{1}{4} =\frac{80}{4}=20$

Report an Error

Example Question #16 : Area

Find the area of the region bounded by the following two curves:

y=x^2

$y=\sqrt{x}$

Possible Answers:

$\frac{2}{3}$

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{3}{4}$

Correct answer:

$\frac{1}{3}$

Explanation:

In order to find the area of the region bounded by the two curves, we must first find the bounds of the region by determning where the curves intersect:

$x^2=\sqrt{x}\rightarrow x^2-\sqrt{x}=0\rightarrow x^{\frac{1}{2}}(x^{\frac{3}{2}}-1)=0$

$x^{\frac{1}{2}}=0\rightarrow x=0$

$x^{\frac{3}{2}}-1=0\rightarrow x^{\frac{3}{2}}=1\rightarrow x=1$

The curves intersect at x=0 and x=1, so these are the bounds of the region for which we want to determine the area. Our next step is to set up an integral with these bounds, where the bottom curve is subtracted from the upper curve, which can then be evaluated to find the area of the region:

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

Report an Error

Example Question #2975 : Functions

Find the area of the region bounded by the following two functions:

y=4x+16

y=2x^2+10

Possible Answers:

$\frac{17}{2}$

$\frac{54}{5}$

$\frac{64}{3}$

Correct answer:

$\frac{64}{3}$

Explanation:

Before we can find the area of the region bounded by the two curves, we must first determine the bounds of that region by finding where the two functions intersect. We do this by setting them equal to each other and solving for x:

$4x+16=2x^2+10\rightarrow 2x^2-4x-6=0\rightarrow 2(x^2-2x-3)=0$

$2(x-3)(x+1)=0\rightarrow x=-1,x=3$

The functions intersect at x=-1 and x=3, so these are the bounds of the area. Now we set up an integral with these bounds, where the lower function is subtracted from the upper function:

$A=\int_{-1}^{3}[(4x+16)-(2x^2+10)]dx=\int_{-1}^{3}(4x-2x^2+6)dx$

$=\left[2x^2-\frac{2}{3}x^3+6x\right]_{-1}^{3}=(18-18+18)-\left(2+\frac{2}{3}-6\right)=\frac{64}{3}$

Report an Error

View Calculus Tutors

Anjali
Certified Tutor

Delhi University, Bachelor of Science, Physics. Kumuan University, Master of Science, Physics.

View Calculus Tutors

Jeremy
Certified Tutor

University of Illinois at Chicago, Bachelor of Science, Physics.

View Calculus Tutors

Yuanhao
Certified Tutor

Auburn University, Bachelor of Science, Civil Engineering. Carnegie Mellon University, Master of Science, Engineering Mechani...

All Calculus 1 Resources

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

Popular Subjects

French Tutors in San Francisco-Bay Area, MCAT Tutors in Denver, Chemistry Tutors in Washington DC, Computer Science Tutors in Phoenix, SSAT Tutors in New York City, Physics Tutors in Philadelphia, Biology Tutors in Miami, Chemistry Tutors in Atlanta, Calculus Tutors in Chicago, Biology Tutors in Houston

Popular Courses & Classes

ISEE Courses & Classes in Dallas Fort Worth, GMAT Courses & Classes in Los Angeles, GRE Courses & Classes in Washington DC, ACT Courses & Classes in Los Angeles, SAT Courses & Classes in San Diego, GMAT Courses & Classes in Washington DC, GMAT Courses & Classes in San Francisco-Bay Area, LSAT Courses & Classes in Phoenix, GRE Courses & Classes in San Diego, GMAT Courses & Classes in Seattle

Popular Test Prep

GRE Test Prep in Washington DC, ISEE Test Prep in Miami, MCAT Test Prep in Washington DC, GRE Test Prep in Houston, LSAT Test Prep in San Francisco-Bay Area, ACT Test Prep in Los Angeles, LSAT Test Prep in Dallas Fort Worth, ACT Test Prep in Atlanta, GMAT Test Prep in San Francisco-Bay Area, LSAT Test Prep in Los Angeles

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 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #5 : How To Find Area Of A Region

Example Question #6 : How To Find Area Of A Region

Example Question #7 : How To Find Area Of A Region

Example Question #2971 : Functions

Example Question #2972 : Functions

Example Question #2973 : Functions

Example Question #2974 : Functions

Example Question #15 : Area

Example Question #16 : Area

Example Question #2975 : Functions

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: