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AP Calculus AB : Antiderivatives following directly from derivatives of basic functions

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Example Questions

Example Question #21 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Integrate:

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

Possible Answers:

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

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

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

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

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:

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

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

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

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:

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

2x^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 #24 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Find the antiderivative of the following.

$\cos (x)$

Possible Answers:

$-\sin (x)$

$-\cos(x)$

$\tan (x)$

$\sin (x)$

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:

$2\sec (x)$

$\tan (x)$

$2\tan (x)$

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

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 #644 : Ap Calculus Ab

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:

$\pi - 4$

$\pi +2$

$\pi$

$\pi -2$

$\pi + 4$

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 #643 : Ap Calculus Ab

Evaluate the integral

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

Possible Answers:

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

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

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

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

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

Integrate:

$\int (x^2 +\frac{1}{2x})dx$

Possible Answers:

$\frac{x^3}{3}+\frac{1}{2}\ln\left | x\right |$

$x^3+\frac{1}{2}\ln\left | x\right |+C$

$\frac{x^3}{3}+\frac{2}{\ln(x)}+C$

$\frac{x^3}{3}+\frac{1}{2}\ln\left | x\right |+C$

Correct answer:

$\frac{x^3}{3}+\frac{1}{2}\ln\left | x\right |+C$

Explanation:

To evaluate the integral, we can split it into two integrals:

$\int x^2 dx+ \frac{1}{2}\int \frac{dx}{x}$

After integrating, we get

$\frac{x^3}{3}+\frac{1}{2}\ln\left | x\right |+C$

where a single constant of integration comes from the sum of the two integration constants from the two individual integrals, added together.

The rules used to integrate are

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

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Example Question #651 : Ap Calculus Ab

Solve:

$\int \frac{dx}{12+9x^2}$

Possible Answers:

$\frac{1}{4\sqrt{3}}\arctan(\frac{3x}{\sqrt{3}})+C$

$\frac{1}{2\sqrt{3}}\arctan(\frac{x}{2\sqrt{3}})+C$

$\arctan(\frac{3x}{2\sqrt{3}})+C$

$\frac{1}{2\sqrt{3}}\arctan(\frac{3x}{2\sqrt{3}})+C$

Correct answer:

$\frac{1}{2\sqrt{3}}\arctan(\frac{3x}{2\sqrt{3}})+C$

Explanation:

The integral is equal to

$\frac{1}{2\sqrt{3}}\arctan(\frac{3x}{2\sqrt{3}})+C$

and was found using the following rule:

$\int \frac{dx}{a^2+x^2}=\frac{1}{a}\arctan{\frac{x}{a}}+C$

where $a=\sqrt{12}=2\sqrt{3}$

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

Solve:

$\int (\frac{1}{x^2}+\frac{1}{3x})dx$

Possible Answers:

$\frac{1}{x}+\frac{1}{3}\ln \left | x\right |+C$

$-\frac{1}{x}+\frac{1}{3}\ln \left | 3x\right |+C$

$-\frac{1}{x}+\frac{1}{3}\ln \left | x\right |$

$-\frac{1}{x}+\frac{1}{3}\ln \left | x\right |+C$

Correct answer:

$-\frac{1}{x}+\frac{1}{3}\ln \left | x\right |+C$

Explanation:

To integrate, we can split the integral into the sum of two separate integrals:

$\int \frac{dx}{x^2}+\frac{1}{3}\int \frac{dx}{x}$

Integrating, we get

$-\frac{1}{x}+\frac{1}{3}\ln\left | x\right |+C$

which was found using the following rules:

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

Note that the constants of integration were combined to make a single integration constant in the final answer.

(The first integral can be rewritten as $\int x^{-2}dx$ for clarity.)

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AP Calculus AB : Antiderivatives following directly from derivatives of basic functions

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #21 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #641 : Ap Calculus Ab

Example Question #642 : Ap Calculus Ab

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

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

Example Question #644 : Ap Calculus Ab

Example Question #643 : Ap Calculus Ab

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

Example Question #651 : Ap Calculus Ab

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

All AP Calculus AB Resources

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