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

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Example Questions

Example Question #31 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Calculate the integral in the following expression:

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

Possible Answers:

x^3 + C

3x^3 + C

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

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

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

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 #31 : Antiderivatives Following Directly From Derivatives Of Basic Functions

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 #34 : Techniques Of Antidifferentiation

Evaluate the following integral

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

Possible Answers:

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

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

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

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

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 #32 : Antiderivatives Following Directly From Derivatives Of Basic Functions

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 #33 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Evaluate the following integral

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

Possible Answers:

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

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

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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Example Question #654 : Ap Calculus Ab

Evaluate the following integral

$\int (\sin{x}+\sec^2{x)}dx$

Possible Answers:

$-\cos{x}+\tan{x}+C$

$\cos{x}+\tan{x}+C$

$-\cos^2{x}+\tan^2{x}+C$

$-\cos{x}+\tan{x}$

Correct answer:

$-\cos{x}+\tan{x}+C$

Explanation:

To evaluate the integral, we use the fact that the antiderivative of $\sin{x}$ is $-\cos{x}$ (because $\frac{\mathrm{d} }{\mathrm{d} x}(-\cos{x})=\sin{x}$ ), and the antiderivative of $\sec^2{x}$ is $\tan{x}$ (because $\frac{\mathrm{d} }{\mathrm{d} x}(\tan{x})=\sec^2x$ ). Using this information, we determine that the integral is

$-\cos{x}+\tan{x}+C$

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Example Question #41 : Techniques Of Antidifferentiation

Calculate the following integral.

$\int4sin(t)-12cos(t)+3t^2dt$

Possible Answers:

-16tan(t)+t^3+c

4cos(t)+12sin(t)+t^4+c

-4cos(t)-12sin(t)+t^3+c

$-4cos(t)+12sin(t)+\frac{t^3}{3}+c$

Correct answer:

-4cos(t)-12sin(t)+t^3+c

Explanation:

Calculate the following integral.

$\int4sin(t)-12cos(t)+3t^2dt$

To do this problem, we need to recall that integrals are also called antiderivatives. This means that we can calculate integrals by reversing our integration rules.

Thus, we can have the following rules.

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

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

$\int t^ndt=\frac{t^{n+1}}{n+1}+c$

Using these rules, we can find our answer:

$\int4sin(t)-12cos(t)+3t^2dt$

Will become:

$-4cos(t)-12sin(t)+\frac{3t^3}{3}+c=-4cos(t)-12sin(t)+t^3+c$

And so our answer is:

-4cos(t)-12sin(t)+t^3+c

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Example Question #31 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Integrate:

$\int_{0}^{1}[x^2+\frac{x}{2}]dx$

Possible Answers:

$\frac{7}{12}$

$\frac{1}{12}$

Correct answer:

$\frac{7}{12}$

Explanation:

The integral of the function is equal to

$\int_{0}^{1}[x^2+\frac{x}{2}]dx= \frac{x^3}{3}+\frac{x^2}{4}\left.\begin{matrix} 1\\ 0 \end{matrix}\right|$

and was found using the following rule:

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

Finally, we evaluate by plugging in the upper bound into the resulting function and subtracting the resulting function with the lower bound plugged in:

$\frac{1}{3}+\frac{1}{4}-(0+0)=\frac{7}{12}$

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Example Question #44 : Techniques Of Antidifferentiation

Solve:

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

Possible Answers:

$\frac{\pi^2}{2}-2$

$\frac{\pi^2}{2}+2$

$\frac{\pi^2}{2}$

Correct answer:

$\frac{\pi^2}{2}+2$

Explanation:

The integral is equal to

$\int_{0}^{\pi} (x+\sin(x))dx = \frac{x^2}{2}-\cos(x) \left.\begin{matrix} \pi \\0 \end{matrix}\right|$

The rules used to integrate are

$\int x^n dx = \frac{x^{n+1}}{n+1}+C$ , $\int \sin(x)dx = -\cos(x)+C$

Now, we solve by plugging in the upper bound of integration and then subtracting the result of plugging in the lower bound of integration:

$\frac{\pi^2}{2}-(-1)-(0-1)=\frac{\pi^2}{2}+2$

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Example Question #34 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Integrate:

$\int \frac{4}{3+x^2}dx$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The integral is equal to

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

and was found using the following rule:

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

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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 #31 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #31 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #34 : Techniques Of Antidifferentiation

Example Question #32 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #33 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #654 : Ap Calculus Ab

Example Question #41 : Techniques Of Antidifferentiation

Example Question #31 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #44 : Techniques Of Antidifferentiation

Example Question #34 : Antiderivatives Following Directly From Derivatives Of Basic Functions

All AP Calculus AB Resources

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