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AP Calculus AB : Asymptotic and Unbounded Behavior

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

Example Questions

Example Question #51 : Asymptotic And Unbounded Behavior

Evaluate the integral:

$\int (x+2) dx$

Possible Answers:

$\frac{x^2}{2}$

$\frac{x^2}{2}+2x+C$

x^2+2

x^2+2x

Correct answer:

$\frac{x^2}{2}+2x+C$

Explanation:

In order to find the antiderivative, add 1 to the exponent and divide by the exponent.

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

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Example Question #52 : Asymptotic And Unbounded Behavior

Evaluate:

$\int \frac{1}{x^4} dx$

Possible Answers:

$\frac{-3}{x^3}$

$\frac{-5}{x^5}$

$\frac{-1}{3x^3}$

$\frac{-1}{5x^5}$

Correct answer:

$\frac{-1}{3x^3}$

Explanation:

$\int x^{-4}= \frac{x^{-4+1}}{-3}= \frac{x^{-3}}{-3}=\frac{-1}{3x^3}$

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Example Question #24 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Evaluate:

$\int (sin^2x*cosx) dx$

Possible Answers:

$(\frac{1}{3})sin^3x+C$

$\frac{cos^3x}{3}$

$\frac{sin^3x}{3}-\frac{cos^2x}{2}$

cos^3x

Correct answer:

$(\frac{1}{3})sin^3x+C$

Explanation:

You should first know that the derivative of sin(x)=cos(x) .

Therefore, looking at the equation you can see that the antiderivative should involve something close to: $\square sin^3x+C$

Now to figure out what value represents the square take the derivative of sin^3x and set it equal to what the original integral contained.

3sin^2x*cosx = sin^2x*cosx

Since the derivative of sin^3x contains a 3 that the integral does not show, we know that the square is equal to 1/3 . Thus, the answer is $\frac{1}{3}sin^3x+C$ .

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Example Question #54 : Functions, Graphs, And Limits

Evaluate:

$\int cos(2x)dx$

Possible Answers:

$-\frac{1}{2}sin(2x)$

2cos(2x)

$(\frac{1}{2})sin(2x)+C$

2sin(2x)

-2sin(2x)

Correct answer:

$(\frac{1}{2})sin(2x)+C$

Explanation:

The antiderivative of cos(x) = sin(x) . The derivative of sin(2x) = 2cos(2x) . However, since there is no 2 in the original integral, we must divide sin(2x) by 2. Therefore, the answer is (1/2)sin(2x)+C.

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Example Question #53 : Asymptotic And Unbounded Behavior

Evaluate the integral:

$\int (x^3-2x^2+5x-1) dx$

Possible Answers:

x^4-6x^3+10x^2-x +C

${4x^4}-6x^3+5x+C$

3x^2-4x+5+C

$\frac{x^4}{4}-\frac{2x^3}{3}+\frac{5x^2}{2}-x+C$

Correct answer:

$\frac{x^4}{4}-\frac{2x^3}{3}+\frac{5x^2}{2}-x+C$

Explanation:

When taking the antiderivative add one to the exponent and then divide by the exponent.

$\frac{x^{3+1}}{4}-\frac{2*x^{2+1}}{3}+\frac{5x^{1+1}}{2}-\frac{1*x^{0+1}}{1}$

$\frac{x^4}{4}-\frac{2x^3}{3}+\frac{5x^2}{2}-x+C$

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Example Question #54 : Asymptotic And Unbounded Behavior

Evaluate the integral:

$\int (e^x)dx$

Possible Answers:

e^0

$e^{x^2}+C$

e^x +C

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

Cannot be evaluated

Correct answer:

e^x +C

Explanation:

The derivative of e^x=e^x . Therefore, the antiderivative of e^x is equal to itself.

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Example Question #57 : Asymptotic And Unbounded Behavior

Evaluate:

$\int_{-4}^{4} (x^{9} + tan(x) - x^{3}cos(x))dx$

Possible Answers:

$5\pi$

Can't be determined from the information given.

Correct answer:

Explanation:

$f(x) = x^{9} + tan(x) - x^{3}cos(x)$ and

$f(-x) = (-x)^{9} + tan(-x) - (-x)^{3}cos(-x) =$

$-x^{9} - tan(x) + x^{3}(cos(x)) =$

$-[x^{9} + tan(x) - x^{3}(cos(x))] = -f(x)$

Recall that tan(x) is an odd function and cos(x) is an even function.

Thus, since f(x) is an odd function, the integral of this function from to will be zero.

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Example Question #55 : Asymptotic And Unbounded Behavior

Evaluate this indefinite integral:

$\int sin^{4}(x)cos^{3}(x)dx$

Possible Answers:

$\frac{cos^{7}(x)}{7} - \frac{cos^{5}(x)}{5}$

$\frac{sin^{5}(x)}{5} - \frac{sin^{7}(x)}{7} + C$

$\frac{sin^{5}(x)}{5} - \frac{sin^{7}(x)}{7}$

$\frac{cos^{5}(x)}{5} - \frac{cos^{7}(x)}{7} + C$

$\frac{sin^{5}(x)}{5} + \frac{sin^{7}(x)}{7} + C$

Correct answer:

$\frac{sin^{5}(x)}{5} - \frac{sin^{7}(x)}{7} + C$

Explanation:

To approach this problem, first rewrite the integral expression as shown below:

$\int sin^{4}(x)cos^{2}(x)cos(x)dx$ .

Then, recognize that cos^2(x) = 1 - sin^2(x) , and substitute this into the integral expression:

$\int sin^{4}(x)(1-sin^{2}x)cos(x)dx$

Use substitution, letting u = sin(x) and du = cos(x)dx . The integral can then be rewritten as

$\int u^{4}(1-u^{2})du = \int (u^{4} - u^{6})du.$

Evaluating this integral gives

$\frac{u^{5}}{5} - \frac{u^{7}}{7} + C$ .

Finally, substituting u = sin(x) back into this expression gives the final answer:

$\frac{sin^{5}(x)}{5} - \frac{sin^{7}(x)}{7} + C$

(As this is an indefinite integral, must be included).

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Example Question #2182 : High School Math

Evaluate:

$\int_{1}^{e^{100}} \frac{1}{x}dx$

Possible Answers:

$\frac{100}{e}$

100e

$\ln 100$

$\frac{1}{100}$

100

Correct answer:

100

Explanation:

$\int_{1}^{e^{100}} \frac{dx}{x} = \ln e^{100} - \ln 1 = 100-0 = 100$

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Example Question #2183 : High School Math

Find $\int_{0}^{\pi} (\cos^{2}x+\sin^{2}x)dx$

Possible Answers:

$\pi$

$2\pi$

Correct answer:

$\pi$

Explanation:

This is most easily solved by recognizing that $\sin^{2}x+\cos^{2}x=1$ .

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AP Calculus AB : Asymptotic and Unbounded Behavior

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #51 : Asymptotic And Unbounded Behavior

Example Question #52 : Asymptotic And Unbounded Behavior

Example Question #24 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #54 : Functions, Graphs, And Limits

Example Question #53 : Asymptotic And Unbounded Behavior

Example Question #54 : Asymptotic And Unbounded Behavior

Example Question #57 : Asymptotic And Unbounded Behavior

Example Question #55 : Asymptotic And Unbounded Behavior

Example Question #2182 : High School Math

Example Question #2183 : High School Math

All AP Calculus AB Resources

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