Call Now to Set Up Tutoring:

(888) 888-0446

AP Calculus AB : Comparing relative magnitudes of functions and their rates of change

Study concepts, example questions & explanations for AP Calculus AB

Create An Account

All AP Calculus AB Resources

3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #34 : Integrals

Evaluate the integral below:

$f(x)=\int \frac{dx}{x^2-16}$

Possible Answers:

$\frac{1}{4} ln\left | \frac{x+2}{x-2} \right |+c$

$\frac{1}{8} ln\left | \frac{x+4}{x-4} \right |+c$

$\frac{1}{4} ln\left | \frac{x-2}{x+2} \right |+c$

$\frac{1}{8} ln\left | \frac{x-4}{x+4} \right |+c$

Correct answer:

$\frac{1}{8} ln\left | \frac{x-4}{x+4} \right |+c$

Explanation:

In this case we have a rational function as $\frac{N(x)}{D(x)}$ , where

N(x)=1

and

$D(x)=\frac{1}{x^2-16}$

D(x) can be written as a product of linear factors:

$\frac{1}{x^2-16}=\frac{1}{(x-4)(x+4)}\equiv \frac{A}{x-4}+\frac{B}{x+4}$

It is assumed that A and B are certain constants to be evaluated. Denominators can be cleared by multiplying both sides by (x - 4)(x + 4). So we get:

1=A(x+4)+B(x-4)

First we substitute x = -4 into the produced equation:

$1=-8B\Rightarrow B=-\frac{1}{8}$

Then we substitute x = 4 into the equation:

$1=8A\Rightarrow A=\frac{1}{8}$

Thus:

$\frac{1}{(x-4)(x+4)}=\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4}$

Hence:

$\int \frac{dx}{x^2-16}=\int(\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4})dx$

$=\frac{1}{8}ln\left | x-4 \right |-\frac{1}{8}ln\left |x+4 \right |+c$

$=\frac{1}{8}(ln\left | x-4 \right |-ln\left |x+4 \right |)+c$

$=\frac{1}{8}ln\left | \frac{x-4}{x+4}\right |+c$

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:

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

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

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

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

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

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 #2 : Finding Integrals By Substitution

Evaluate:

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

Possible Answers:

$\frac{2}{3}$

$2 \ln 3$

$2 \sqrt{\ln 3}$

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

$2\sqrt{3}$

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 #2 : Finding Integrals

$\int_{2}^{4}x^2{\mathrm{d} x}=?$

Possible Answers:

$18\frac{2}{3}$

$28\tfrac{2}{3}$

$5\tfrac{1}{2}$

$11\tfrac{3}{4}$

$22\tfrac{1}{3}$

Correct answer:

$18\frac{2}{3}$

Explanation:

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our f(x)=x^2 , we can use the reverse power rule to find the indefinite integral or anti-derivative of our function:

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

Remember to include the + c for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(\frac{1}{3}b^3+c)-(\frac{1}{3}a^3+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{2}^{4}f(x){\mathrm{d} x}=(\frac{1}{3}\cdot 4^3)-(\frac{1}{3}\cdot 8^3)$

$\int_{2}^{4}f(x){\mathrm{d} x}=(\frac{64}{3})-(\frac{8}{3})$

$\int_{2}^{4}f(x){\mathrm{d} x}=\frac{56}{3}=18\frac{2}{3}$

Report an Error

Example Question #43 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

$\int_{1}^{2}\frac{1}{x}{\mathrm{d} x}=?$

Possible Answers:

0.81

1.2

0.30

0.75

0.69

Correct answer:

0.69

Explanation:

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

As it turns out, since our $f(x)=\frac{1}{x}$ , the power rule really doesn't help us. $\frac{1}{x}$ has a special anti derivative: $\ln{x}$ .

$\int \frac{1}{x}{\mathrm{d} x}=\ln{x}+c$

Remember to include the + c for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(\ln{b}+c)-(\ln{a}+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{1}^{2}f(x){\mathrm{d} x}=(\ln(2))-(\ln(1))$

$\int_{1}^{2}f(x){\mathrm{d} x}=(\ln(2))-(0)$

$\int_{1}^{2}f(x){\mathrm{d} x}\approx 0.69$

Report an Error

Example Question #44 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

$\int_{3}^{40}e^x{\mathrm{d} x}=?$

Possible Answers:

$2.35\cdot 10^{17}$

$1.05\cdot 10^{17}$

$2.35\cdot 10^{19}$

$2.35\cdot 10^{15}$

$2.35\cdot 10^{9}$

Correct answer:

$2.35\cdot 10^{17}$

Explanation:

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

As it turns out, since our f(x)=e^x , the power rule really doesn't help us. e^x is the only function that is it's OWN anti-derivative. That means we're still going to be working with e^x .

$\int e^x{\mathrm{d} x}=e^x+c$

Remember to include the + c for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(e^b+c)-(e^a+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{3}^{40}f(x){\mathrm{d} x}=(e^{40})-(e^3)$

$\int_{3}^{40}f(x){\mathrm{d} x}=(2.35\cdot 10^{17})-(20.09)$

Because e^3 is so small in comparison to the value we got for $e^{40}$ , our answer will end up being $2.35\cdot 10^{17}$

Report an Error

Example Question #42 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

What is the indefinite integral of x^2+5x ?

Possible Answers:

10x^2+c

x^4+5x^2+c

3x+5

2x+5

$\frac{1}{3}x^3+\frac{5}{2}x^2+c$

Correct answer:

$\frac{1}{3}x^3+\frac{5}{2}x^2+c$

Explanation:

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent. For this problem, that would look like:

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

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

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

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

Report an Error

Example Question #43 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

What is the indefinite integral of 5x+8 ?

Possible Answers:

16x^3+c

$\frac{5}{6}x^3+4x^2+cx+b$

$\frac{5}{2}x^2+8x +c$

Correct answer:

$\frac{5}{2}x^2+8x +c$

Explanation:

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent. For this problem, that would look like:

$\int 5x+8{\mathrm{d} x}=\frac{5x^{1+1}}{1+1}+\frac{8x^{0+1}}{0+1}+c$

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

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

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

Report an Error

Example Question #44 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

What is the indefinite integral of x^3+2x+5 ?

Possible Answers:

4x^4+x^2+5x+c

3x^2+2

$\frac{1}{4}x^4+x^2+5x+c$

$\frac{1}{4}x^4+2x^2+5x+c$

Correct answer:

$\frac{1}{4}x^4+x^2+5x+c$

Explanation:

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent.

We're going to treat as 5x^0 , as anything to the zero power is one.

For this problem, that would look like:

$\int x^3+2x+5{\mathrm{d} x}=\frac{x^{3+1}}{3+1}+\frac{2x^{1+1}}{1+1}+\frac{5x^{0+1}}{0+1}+c$

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

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

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

Report an Error

Example Question #61 : Finding Integrals

Determine the indefinite integral:

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

Possible Answers:

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

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

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

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

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

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

View AP Calculus AB Tutors

Muhammad
Certified Tutor

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

View AP Calculus AB Tutors

Austin
Certified Tutor

University of Arizona, Bachelor of Science, Physiology.

View AP Calculus AB Tutors

Shakeel
Certified Tutor

All AP Calculus AB Resources

3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Spanish Tutors in Houston, Reading Tutors in Atlanta, Reading Tutors in Philadelphia, English Tutors in Los Angeles, Algebra Tutors in Seattle, Spanish Tutors in San Francisco-Bay Area, ISEE Tutors in Phoenix, GRE Tutors in Philadelphia, Reading Tutors in Miami, Biology Tutors in San Diego

Popular Courses & Classes

ACT Courses & Classes in Chicago, ISEE Courses & Classes in San Francisco-Bay Area, SAT Courses & Classes in Dallas Fort Worth, GMAT Courses & Classes in Phoenix, SAT Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Seattle, GRE Courses & Classes in Seattle, LSAT Courses & Classes in Los Angeles, LSAT Courses & Classes in Boston, SSAT Courses & Classes in Dallas Fort Worth

Popular Test Prep

MCAT Test Prep in Phoenix, GRE Test Prep in Seattle, GMAT Test Prep in Phoenix, ISEE Test Prep in Dallas Fort Worth, SAT Test Prep in Phoenix, GMAT Test Prep in Washington DC, ISEE Test Prep in San Diego, GMAT Test Prep in Atlanta, ISEE Test Prep in Los Angeles, ACT 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

AP Calculus AB : Comparing relative magnitudes of functions and their rates of change

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #34 : Integrals

Example Question #1 : Finding Integrals By Substitution

Example Question #2 : Finding Integrals By Substitution

Example Question #2 : Finding Integrals

Example Question #43 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #44 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #42 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #43 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #44 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #61 : Finding Integrals

All AP Calculus AB Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: