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AP Calculus AB : Techniques of antidifferentiation

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Example Questions

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 #41 : Techniques Of Antidifferentiation

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 #41 : Techniques Of Antidifferentiation

Integrate:

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

Possible Answers:

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

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

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

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

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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Example Question #112 : Integrals

Integrate:

$\int_{0}^{4}(x-x^2)dx$

Possible Answers:

$\frac{40}{3}$

$\frac{88}{3}$

$-\frac{40}{3}$

Correct answer:

$-\frac{40}{3}$

Explanation:

The integral is equal to

$\int_{0}^{4}(x-x^2)dx = \frac{x^2}{2}-\frac{x^3}{3}\left.\begin{matrix} 4\\0 \end{matrix}\right|$

The following rule was used to integrate:

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

Now, we find the numerical answer by plugging in the upper bound of integration and subtracting what we get from plugging in the lower bound of integration:

$8-\frac{64}{3}-(0-0)=-\frac{40}{3}$

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Example Question #42 : Techniques Of Antidifferentiation

Integrate:

$\int_{0}^{2\pi}5\sec^2(x)dx$

Possible Answers:

$\pi$

Undefined

Correct answer:

Explanation:

The integral is equal to

$\int_{0}^{2\pi}5\sec^2(x)dx=5\tan(x)\left.\begin{matrix} 2\pi\\0 \end{matrix}\right|$

and was found using the identical rule.

Evaluating by plugging in the upper bound and subtracting from what we get from plugging in the lower bound, we get

$5(0)-(5 \cdot 0)=0$

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Example Question #1 : Antiderivatives By Substitution Of Variables

Use a change of variable (aka a u-substitution) to evaluate the integral,

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

Possible Answers:

$\frac{1}{6}(2x+1)^{\frac{2}{3}}+C$

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

$\frac{1}{3}x^{\frac{3}{2}}+C$

$\frac{1}{6}(x^2+1)^{\frac{2}{3}}+\sqrt{x}+C$

$\frac{3}{4}(x^2+1)^{\frac{3}{2}}+C$

Correct answer:

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

Explanation:

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

Integrals such as this are seen very commonly in introductory calculus courses. It is often useful to look for patterns such as the fact that the polynomial under the radical in our example, 1+x^2 , happens to be one order higher than the factor outside the radical, You know that if you take a derivative of a second order polynomial you will get a first order polynomial, so let's define the variable:

u = 1+x^2 (1)

Now differentiate with respect to to write the differential for ,

du = 2x dx (2)

Looking at equation (2), we can solve for xdx , to obtain $xdx =\frac{1}{2}du$ . Now if we look at the original integral we can rewrite in terms of

$\int x\sqrt{1+x^2}dx = \int\sqrt{1+x^2}\underbrace{xdx} = \int\sqrt{u}\left(\frac{1}{2}du \right )$

Now proceed with the integration with respect to .

$\int\sqrt{u}\left(\frac{1}{2}du \right ) = \frac{1}{2}\int u^{\frac{1}{2}}du$

$= \frac{1}{2}\left[\frac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1} \right ]+C$

$=\frac{1}{2} \left(\frac{2}{3}u^{\frac{3}{2}} \right )+C$

$=\frac{1}{3}u^{\frac{3}{2}}+C$

Now write the result in terms of using equation (1), we conclude,

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

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Example Question #2 : Antiderivatives By Substitution Of Variables

Use u-subtitution to fine

$\int \sin^2(x)\cos (x)dx$

Possible Answers:

$\frac{1}{6}sin^3(x)cos^2(x)$

$\frac{1}{4}sin^4(x)$

$\frac{-1}{6}cos^3(x)sin^2(x)$

$\frac{1}{6}sin^2(x)$

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

Correct answer:

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

Explanation:

Let u=sin(x)

Then du=cos(x)dx

Now we can subtitute

$\int sin^2(x) \cos(x) dx=\int u^2du= \frac{1}{3}u^3$

Now we substitute back

$\frac{1}{3}u^3=\frac{1}{3} \sin^3(x)$

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Example Question #3 : Antiderivatives By Substitution Of Variables

Evaluate $\int_0^1 \frac{6x}{1+x^2} dx$

Possible Answers:

$3\ln2$

$\ln3$

$2\ln3$

$\ln2$

Correct answer:

$3\ln2$

Explanation:

We can use substitution for this integral.

Let u = 1+x^2 ,

then du = 2x dx .

Multiplying this last equation by , we get 3du = 6x dx .

Now we can make our substitutions

$\int_0^1 \frac{6x}{1+x^2} dx$ . Start

$=\int_1^2 \frac{3}{u}du$ . Swap out 1+x^2 with , and 6x dx with 3du . Make sure you also plug the bounds on the integral into u = 1+x^2 for to get the new bounds.

$=3\int_1^2 \frac{1}{u}du$ . Factor out the .

$=3 \ln(u)|_1^2$ . Integrate (absolute value signs are not needed since $1\le u \le2$ .)

$=3(\ln(2)-\ln(1))$ . Evaluate

$=3\ln2 -0 = 3\ln2$ .

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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 #654 : Ap Calculus Ab

Example Question #41 : Techniques Of Antidifferentiation

Example Question #41 : Techniques Of Antidifferentiation

Example Question #44 : Techniques Of Antidifferentiation

Example Question #41 : Techniques Of Antidifferentiation

Example Question #112 : Integrals

Example Question #42 : Techniques Of Antidifferentiation

Example Question #1 : Antiderivatives By Substitution Of Variables

Example Question #2 : Antiderivatives By Substitution Of Variables

Example Question #3 : Antiderivatives By Substitution Of Variables

All AP Calculus AB Resources

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