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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

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

Example Questions

Example Question #21 : Techniques Of Antidifferentiation

Compute the following integral:

$\int 6t^5-12t+6e^t-sin(t)+14cos(t)dt$

Possible Answers:

t^6-6t^2+6e^t+cos(t)+14sin(t)+c

t^6-6t^2+6e^t+cos(t)+14sin(t)+5c

t^6-6t^2+6e^t+cos(t)+14sin(t)

t^6-6t^2+6e^t-cos(t)-14sin(t)+c

Correct answer:

t^6-6t^2+6e^t+cos(t)+14sin(t)+c

Explanation:

Compute the following integral:

$\int 6t^5-12t+6e^t-sin(t)+14cos(t)dt$

Now, we need to recall a few rules.

$\int e^xdx=e^x+c$

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

$\int sin(x)dx=-cos(x)+c$

$\int cos(x)dx=sin(x)+c$

We can use all these rules to change our original function into its anti-derivative.

$\int 6t^5-12t+6e^t-sin(t)+14cos(t)dt$

We can break this up into separate integrals for each term, and apply our rules individually.

$\int 6t^5dt-\int 12tdt+\int 6e^tdt-\int sin(t)dt+\int 14cos(t)dt$

The first two integrals can be found using rule 2

$\int 6t^5dt-\int 12tdt=\frac{6t^6}{6}+c-\frac{12t^2}{2}+c=t^6-6t^2+c$

Next, let's tackle the middle integral:

$\int 6e^tdt=6e^t +c$

Then the "sine" integral

$-\int sin(t)=- (-cos(t))+c=cos(t)+c$

And finally, the cosine integral.

$\int 14cos(t)dt=14sin(t)+c$

Now, we can put all of this together to get:

t^6-6t^2+6e^t+cos(t)+14sin(t)+c

Note that we only have 1 c, because the c is just a constant.

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Example Question #641 : Ap Calculus Ab

Solve:

$\int (5\sec^2(x)+\sec(x)\tan(x))dx$

Possible Answers:

$5\tan(x)+\ln(\sec(x))+C$

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

$5\tan(x)+\sec(x)+C$

$5\tan(x)+\sec(x)$

$5(\tan(x)+\sec(x))+C$

Correct answer:

$5\tan(x)+\sec(x)+C$

Explanation:

The integral can be solved knowing the derivatives of the following functions:

$\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$

Given that the integrand is simply the sum of these two derivatives, we find that our integral is equal to

$\int (5\sec^2(x)+\sec(x)\tan(x))dx=5\tan(x)+\sec(x)+C$

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Example Question #13 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Solve:

$\int \frac{5}{x^3}dx$

Possible Answers:

$\frac{5}{2x^2}+C$

$5\ln\left | x^2\right |+C$

None of the other answers

$-\frac{5}{2x^2}$

Correct answer:

None of the other answers

Explanation:

The integral is equal to

$\int \frac{5}{x^3}dx = -\frac{5}{2x^2}+C$

and was given by the following rule:

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

Using this rule becomes more clear when we rewrite the integral as

$\int 5x^{-3}dx$

Note that because none of the answer choices had the integration constant C along with the proper integral result, the correct choice was "None of the other answers." Always check after solving an indefinite integral for C!

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Example Question #22 : Techniques Of Antidifferentiation

Integrate:

$\int_{1}^{e}(x^2+x+\frac{5}{x})dx$

Possible Answers:

$\frac{e^3}{3}+\frac{e^2}{2}+\frac{25}{6}$

$2{e^3}+3{e^2}+\frac{25}{6}$

$\frac{e^3}{3}+\frac{e^2}{2}+\frac{35}{6}$

${e^3}+e^2+\frac{25}{6}$

Correct answer:

$\frac{e^3}{3}+\frac{e^2}{2}+\frac{25}{6}$

Explanation:

The integral of the function is equal to

$\int_{1}^{e}(x^2+x+\frac{5}{x})dx=\frac{x^3}{3}+\frac{x^2}{2}+5\ln\left | x\right |\left.\begin{matrix} e\\1 \end{matrix}\right|$

The rules used for integration were

$\int x^n dx= \frac{x^{n+1}}{n+1}+C$ , $\int \frac{dx}{x}=\ln\left | x\right |+C$

For the definite component of the integration, we plug in the upper limit of integration, and subtract the result of plugging in the lower limit of integration:

$\frac{e^3}{3}+\frac{e^2}{2}+5-(\frac{1}{3}+\frac{1}{2}+0)= \frac{e^3}{3}+\frac{e^2}{2}+\frac{25}{6}$

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Example Question #641 : Ap Calculus Ab

Evaluate the integral

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

Possible Answers:

$5x+\frac{1}{2}x^2+C$

$\frac{5}{3}x^3+\frac{1}{4}x^4$

$\frac{5}{3}x^3+\frac{1}{4}x^4+C$

Correct answer:

$\frac{5}{3}x^3+\frac{1}{4}x^4+C$

Explanation:

To find the derivative of the expression, we use the following rule

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

Applying to the integrand from the problem statement, we get

$\frac{5}{3}x^3+\frac{1}{4}x^4+C$

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Example Question #642 : Ap Calculus Ab

Find the antiderivative of the following.

x^2

Possible Answers:

2x^3

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

Correct answer:

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

Explanation:

Follow the following formula to find the antiderivatives of exponential functions:

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

Thus, the antiderivative of x^2 is $\frac{x^3}{3}$ .

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Example Question #643 : Ap Calculus Ab

Find the antiderivative of the following.

$\cos (x)$

Possible Answers:

$\sin (x)$

$\tan (x)$

$-\cos(x)$

$-\sin (x)$

Correct answer:

$\sin (x)$

Explanation:

$\cos (x)$ is the derivative of $\sin (x)$ . Thus, the antiderivative of $\cos (x)$ is $\sin (x)$ .

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Example Question #24 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Find the antiderivative of the following.

$\sec^2(x)$

Possible Answers:

$2\sec (x)$

$2\tan (x)$

$\tan (x)$

$\frac{\tan (x)}{2}$

Correct answer:

$\tan (x)$

Explanation:

$\sec^2(x)$ is the derivative of $\tan (x)$ . Thus, the antiderivative of $\sec^2(x)$ is $\tan (x)$ .

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Example Question #31 : Techniques Of Antidifferentiation

Define $f(x) = \left\{\begin{matrix} 1 -\cos x,& x< 0\\ \sin x ,& x \ge 0 \end{matrix}\right.$

Evaluate $\int_{-\pi}^{\pi} f(x) \textup{d}x$ .

Possible Answers:

$\pi +2$

$\pi$

$\pi + 4$

$\pi - 4$

$\pi -2$

Correct answer:

$\pi +2$

Explanation:

f(x) has different definitions on $[-\pi, 0)$ and $[0,\pi]$ , so the integral must be rewritten as the sum of two separate integrals:

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0} f(x) \textup{d}x + \int_{-0}^{\pi} f(x) \textup{d}x$

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x$

Calculate the integrals separately, then add:

$\int_{-\pi}^{0}(1-\cos x) \textup{d}x = \int_{-\pi}^{0}1 \textup{ d}x - \int_{-\pi}^{0} \cos x \textup{ d}x$

$=\left (\begin{matrix} x\\ \\ \end{matrix} \left|\begin{matrix} 0 \\ -\pi \end{matrix}\right \right )- \left (\begin{matrix} \sin x\\ \\ \end{matrix} \left|\begin{matrix} 0 \\ -\pi \end{matrix}\right \right )$

$= (0-(-\pi))- (\sin 0 - \sin (-\pi))$

$= \pi- (0 -0)$

$= \pi$

$\int_{0}^{\pi} \sin x \textup{ d}x$

$= \left.\begin{matrix} -\cos x \\ \\ \end{matrix}\right| \begin{matrix} \pi \\ 0 \end{matrix}$

$= -\cos \pi - (- \cos 0)$

=-( -1) - (- 1)

= 2

$\int_{-\pi}^{\pi} f(x) \textup{d}x = \int_{-\pi}^{0}(1-\cos x) \textup{d}x + \int_{0}^{\pi} \sin x \textup{ d}x = \pi +2$

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Example Question #32 : Techniques Of Antidifferentiation

Evaluate the integral

$\int (4x^5+3x^2+1 )dx$

Possible Answers:

$\frac{4}{6}x^6+x^3+C$

$\frac{4}{6}x^6+x^3+x$

$\frac{4}{6}x^6+x^3+x+C$

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

Correct answer:

$\frac{4}{6}x^6+x^3+x+C$

Explanation:

To evaluate the integral, we use the rules for integration which tell us

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

Applying to the integral from the problem statement, we get

$\frac{4}{6}x^6+x^3+x+C$

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #21 : Techniques Of Antidifferentiation

Example Question #641 : Ap Calculus Ab

Example Question #13 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #22 : Techniques Of Antidifferentiation

Example Question #641 : Ap Calculus Ab

Example Question #642 : Ap Calculus Ab

Example Question #643 : Ap Calculus Ab

Example Question #24 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #31 : Techniques Of Antidifferentiation

Example Question #32 : Techniques Of Antidifferentiation

All AP Calculus AB Resources

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