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AP Calculus AB : Techniques of antidifferentiation

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Example Questions

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

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:

$\pi$

$\pi +2$

$\pi + 4$

$\pi -2$

$\pi - 4$

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 #31 : Techniques Of Antidifferentiation

Evaluate the integral

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

Possible Answers:

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

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

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

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

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 #23 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Integrate:

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

Possible Answers:

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

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

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

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

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 #24 : Antiderivatives Following Directly From Derivatives Of Basic Functions

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 #25 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Solve:

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

Possible Answers:

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

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

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

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

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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Example Question #651 : Ap Calculus Ab

Calculate the integral in the following expression:

$\int(e^2)^{\ln 3x}dx$

Possible Answers:

x^3 + C

$6x(e^2)^{\ln 3x} + C$

3x^3 + C

$\frac{(e^2)^{\ln3x}}{2x} + C$

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

Correct answer:

3x^3 + C

Explanation:

Solving this integral depends on knowledge of exponent rules; mainly, that $(a^b)^c = a^{b*c}$ . Using this, we can simplify the given expression.

$\int(e^2)^{\ln3x}dx = \int e^{2 * \ln 3x}dx = \int (e^{\ln 3x})^2dx =\int (3x)^2dx =\int 9x^2dx$

From here, we just follow the power rule, raising the exponent and dividing by it.

$9\int x^2 dx = 3x^3 + C$

Giving us our final answer.

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Example Question #652 : Ap Calculus Ab

Evaluate the integral

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

Possible Answers:

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

4x^5+2x^3+3x^2

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

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

Correct answer:

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

Explanation:

To evaluate the integral, we use the following definition

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

$\int (4x^4+2x^2+3x )dx=\frac{4x^5}{5}+\frac{2x^3}{3}+\frac{3x^2}{2}+C$

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Example Question #653 : Ap Calculus Ab

Evaluate the following integral

$\int (\sin(x)+\frac{x}{4})dx$

Possible Answers:

$\cos(x)+\frac{x^2}{8}+C$

$-\cos(x)+\frac{x^2}{8}$

$-\cos(x)+\frac{x^2}{8}+C$

$-\cos(x)+\frac{x^2}{4}+C$

Correct answer:

$-\cos(x)+\frac{x^2}{8}+C$

Explanation:

To solve the problem, we apply the fact that anti-derivative of $\sin(x)=-\cos(x)$ and that $\int x^n dx=\frac{x^{n+1}}{n+1}+C$

Taking the anti-derivative of each part independently, we get

$\int \sin(x)dx+\int \frac{x}{4}dx$

Finally, our answer is

$-\cos(x)+\frac{x^2}{8}+C$

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Example Question #654 : Ap Calculus Ab

Determine the value of $\int_{0}^{2}\sqrt{9x^2+12x+4} dx$ .

Possible Answers:

Correct answer:

Explanation:

We can factor the equation inside the square root:

$\\\int_0^2\sqrt{9x^2+12x+4}dx \\\\\int_0^2\sqrt{(3x+2)^2}dx \\\\\int_0^2 (3x+2)dx$

From here, increase each term's exponent by one and divide the term by the new exponent.

$\frac{3x^2}{2}+2x\big|_0^2$

Now, substitute in the upper bound into the function and subtract the lower bound function value from it.

$\\\frac{3(2)^2}{2}+2(2)-\left(\frac{3(0)^2}{2}+2(0)\right) \\\\\frac{3(4)}{2}+4 \\\\\frac{12}{2}+4 \\\\6+4 \\\\10$

Therefore,

$\int_{0}^{2}\sqrt{(3x+2)^2}dx=\int_{0}^{2}(3x+2)dx=10$

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Example Question #655 : Ap Calculus Ab

Evaluate the following integral

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

Possible Answers:

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

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

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

Correct answer:

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

Explanation:

To evaluate the integral, we use the definition

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

$\int (3x^3+2x-5 )dx=\frac{3}{4}x^4+x^2-5x+C$

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AP Calculus AB : Techniques of antidifferentiation

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 #31 : Techniques Of Antidifferentiation

Example Question #23 : Antiderivatives Following Directly From Derivatives Of Basic Functions

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

Example Question #25 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #651 : Ap Calculus Ab

Example Question #652 : Ap Calculus Ab

Example Question #653 : Ap Calculus Ab

Example Question #654 : Ap Calculus Ab

Example Question #655 : Ap Calculus Ab

All AP Calculus AB Resources

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