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AP Calculus AB : 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 #33 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Evaluate this indefinite integral:

$\int sin^{4}(x)cos^{3}(x)dx$

Possible Answers:

$\frac{sin^{5}(x)}{5} - \frac{sin^{7}(x)}{7} + C$

$\frac{sin^{5}(x)}{5} + \frac{sin^{7}(x)}{7} + C$

$\frac{cos^{7}(x)}{7} - \frac{cos^{5}(x)}{5}$

$\frac{sin^{5}(x)}{5} - \frac{sin^{7}(x)}{7}$

$\frac{cos^{5}(x)}{5} - \frac{cos^{7}(x)}{7} + C$

Correct answer:

$\frac{sin^{5}(x)}{5} - \frac{sin^{7}(x)}{7} + C$

Explanation:

To approach this problem, first rewrite the integral expression as shown below:

$\int sin^{4}(x)cos^{2}(x)cos(x)dx$ .

Then, recognize that cos^2(x) = 1 - sin^2(x) , and substitute this into the integral expression:

$\int sin^{4}(x)(1-sin^{2}x)cos(x)dx$

Use substitution, letting u = sin(x) and du = cos(x)dx . The integral can then be rewritten as

$\int u^{4}(1-u^{2})du = \int (u^{4} - u^{6})du.$

Evaluating this integral gives

$\frac{u^{5}}{5} - \frac{u^{7}}{7} + C$ .

Finally, substituting u = sin(x) back into this expression gives the final answer:

$\frac{sin^{5}(x)}{5} - \frac{sin^{7}(x)}{7} + C$

(As this is an indefinite integral, must be included).

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Example Question #1 : Integrals

Evaluate:

$\int_{1}^{e^{100}} \frac{1}{x}dx$

Possible Answers:

$\frac{100}{e}$

$\ln 100$

100e

$\frac{1}{100}$

100

Correct answer:

100

Explanation:

$\int_{1}^{e^{100}} \frac{dx}{x} = \ln e^{100} - \ln 1 = 100-0 = 100$

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Example Question #1 : Integrals

Find $\int_{0}^{\pi} (\cos^{2}x+\sin^{2}x)dx$

Possible Answers:

$\pi$

$2\pi$

Correct answer:

$\pi$

Explanation:

This is most easily solved by recognizing that $\sin^{2}x+\cos^{2}x=1$ .

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Example Question #1 : Finding Integrals

$\int_{\pi}^{2\pi}\sin(2x){\mathrm{d} x}=?$

Possible Answers:

$\frac{1}{2}$

$\pi$

Correct answer:

Explanation:

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $f(x)=\sin(2x)$ , we can't use the power rule. Instead we end up with:

$\int f(x){\mathrm{d} x}=\int \sin(2x){\mathrm{d} x}=\frac{-\cos(2x)}{2}+c$

Remember to include the + c for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(\frac{-\cos(2b)}{2}+c)-(\frac{-\cos(2a)}{2}+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{\pi}^{2\pi}f(x){\mathrm{d} x}=(\frac{-\cos(4\pi)}{2})-(\frac{-\cos(2\pi)}{2})$

$\int_{\pi}^{2\pi}f(x){\mathrm{d} x}=(\frac{-1}{2})-(\frac{-1}{2})$

$\int_{\pi}^{2\pi}f(x){\mathrm{d} x}=0$

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Example Question #1 : Finding Indefinite Integrals

$\int \frac{3}{x}=$

Possible Answers:

$3x^{-1}+C$

$3\ln \left | x \right |+C$

$3x^{-2}+C$

$-3e^{-x}$

$-3x^{-2}+C$

Correct answer:

$3\ln \left | x \right |+C$

Explanation:

The integral of $\frac{1}{x}$ is $\ln \left | x \right |+C$ . The constant 3 is simply multiplied by the integral.

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Example Question #2 : Finding Indefinite Integrals

$\int cos(x)\sin(x){\mathrm{d} x}=?$

Possible Answers:

$\sin(x)\cos(x)+c$

$-(\frac{1}{2}\cos^2(x)+c)$

$\frac{\cos(x)}{\sin(x)}+c$

$\cos^2(x)sin^2(x)+c$

$\frac{\sin(x)}{\cos(x)}+c$

Correct answer:

$-(\frac{1}{2}\cos^2(x)+c)$

Explanation:

To integrate $\cos(x)\sin(x)$ , we need to get the two equations in terms of each other. We are going to use "u-substitution" to create a new variable, , which will equal $\cos(x)$ .

Now, if $u=\cos(x)$ , then

$\frac{\mathrm{d}u }{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-sin(x)$

Multiply both sides by $\mathrm d {x}$ to get the more familiar:

${\mathrm{d} u}=-\sin(x)\mathrm{d}x$

Note that our $\mathrm d{u}=-\sin(x)$ , and our original equation was asking for a positive $\sin(x)$ .

That means if we want $\int\cos(x)\sin(x)$ in terms of , it looks like this:

$\int u\cdot (-\mathrm d{u}})$

Bring the negative sign to the outside:

$-\int u\mathrm d{u}$ .

We can use the power rule to find the integral of :

$-(\frac{1}{2}u^2+c)$

Since we said that $u=\cos(x)$ , we can plug that back into the equation to get our answer:

$-(\frac{1}{2}\cos^2(x)+c)$

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Example Question #3 : Finding Indefinite Integrals

Evaluate the integral below:

$f(x)=\int \frac{dx}{x^2-16}$

Possible Answers:

$\frac{1}{4} ln\left | \frac{x+2}{x-2} \right |+c$

$\frac{1}{4} ln\left | \frac{x-2}{x+2} \right |+c$

$\frac{1}{8} ln\left | \frac{x-4}{x+4} \right |+c$

$\frac{1}{8} ln\left | \frac{x+4}{x-4} \right |+c$

Correct answer:

$\frac{1}{8} ln\left | \frac{x-4}{x+4} \right |+c$

Explanation:

In this case we have a rational function as $\frac{N(x)}{D(x)}$ , where

N(x)=1

and

$D(x)=\frac{1}{x^2-16}$

D(x) can be written as a product of linear factors:

$\frac{1}{x^2-16}=\frac{1}{(x-4)(x+4)}\equiv \frac{A}{x-4}+\frac{B}{x+4}$

It is assumed that A and B are certain constants to be evaluated. Denominators can be cleared by multiplying both sides by (x - 4)(x + 4). So we get:

1=A(x+4)+B(x-4)

First we substitute x = -4 into the produced equation:

$1=-8B\Rightarrow B=-\frac{1}{8}$

Then we substitute x = 4 into the equation:

$1=8A\Rightarrow A=\frac{1}{8}$

Thus:

$\frac{1}{(x-4)(x+4)}=\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4}$

Hence:

$\int \frac{dx}{x^2-16}=\int(\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4})dx$

$=\frac{1}{8}ln\left | x-4 \right |-\frac{1}{8}ln\left |x+4 \right |+c$

$=\frac{1}{8}(ln\left | x-4 \right |-ln\left |x+4 \right |)+c$

$=\frac{1}{8}ln\left | \frac{x-4}{x+4}\right |+c$

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Example Question #2241 : High School Math

Determine the indefinite integral:

$\dpi{120}\int\frac{\sqrt{\log_{10} x}}{x}\; \; dx$

Possible Answers:

$\frac{ 2 \sqrt{ \log_{10} x} } {3 } + C$

$\frac{ 2 \sqrt{ \log_{10} x} \cdot \log_{10} x} {3 } + C$

$\frac{ 2 \log_{10} x} {3 } + C$

$\frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C$

$\frac{ 2 \sqrt{ \log_{10} x} } {3 \ln x} + C$

Correct answer:

$\frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C$

Explanation:

$\log_{10} x = \frac{\ln x}{\ln 10}$ , so this can be rewritten as

$\dpi{120}\int\frac{\sqrt{\log_{10} x}}{x}\; \; dx$

$\dpi{120}=\int\frac{\sqrt{\ln x}}{x \sqrt{\ln 10} }\; \; dx$

$\dpi{120}= \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }\; \; dx$

Set $u = \ln x$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$

and

$du = \frac{dx}{x}$

Substitute:

$\dpi{120} \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }\; \; dx$

$\dpi{120} =\frac{1}{\sqrt{\ln 10}} \int\sqrt{u} \; du$

$\dpi{120} =\frac{1}{\sqrt{\ln 10}} \int u ^{\frac{1}{2}} \; du$

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} + C \right )$

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2u \sqrt{u}} {3} + C \right )$

The outer factor can be absorbed into the constant, and we can substitute back:

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2 \ln x \cdot \sqrt{ \ln x}} {3} \right )+ C$

$= \frac{ 2 \sqrt{ \ln x} \cdot \ln x} {3 \cdot \sqrt{\ln 10}} + C$

$= \frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C$

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Example Question #2242 : High School Math

Evaluate:

$\dpi{120}\int_{1}^{e^{ 3}}\frac{\sqrt{\ln x}}{x}\; \; dx$

Possible Answers:

$\frac{2}{3}$

$2 \sqrt{\ln 3}$

$2\sqrt{3}$

$\frac{2\sqrt{3}}{3}$

$2 \ln 3$

Correct answer:

$2\sqrt{3}$

Explanation:

Set $u = \ln x$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$

and

$du = \frac{dx}{x}$

Also, since $u = \ln x$ , the limits of integration change to $\ln e^{3} = 3$ and $\ln 1= 0$ .

Substitute:

$\dpi{120}\int_{1}^{e^{ 3}}\frac{\sqrt{\ln x}}{x}\; \; dx$

$\dpi{120} = \int_{0}^{3}\sqrt{u} \; du$

$= \int_{0}^{3} u ^{\frac{1}{2}} \; du$

$\dpi{150} = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} & \\ & \end{matrix}\right|$ $\begin{matrix} 3\\ \\ 0 \end{matrix}$

$\dpi{150} = \left.\begin{matrix} \\ \frac{ 2u \sqrt{u}} {3} & \\ & \end{matrix}\right|$ $\begin{matrix} 3\\ \\ 0 \end{matrix}$

$=\frac{ 2\cdot 3\cdot \sqrt{3}} {3} - \frac{ 2\cdot 0\cdot \sqrt{0}} {3} =2\sqrt{3}$

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Example Question #2 : Finding Integrals

$\int_{2}^{4}x^2{\mathrm{d} x}=?$

Possible Answers:

$18\frac{2}{3}$

$28\tfrac{2}{3}$

$22\tfrac{1}{3}$

$11\tfrac{3}{4}$

$5\tfrac{1}{2}$

Correct answer:

$18\frac{2}{3}$

Explanation:

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our f(x)=x^2 , we can use the reverse power rule to find the indefinite integral or anti-derivative of our function:

$\int (x^2){\mathrm{d} x}=\frac{1}{3}x^3+c$

Remember to include the + c for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(\frac{1}{3}b^3+c)-(\frac{1}{3}a^3+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{2}^{4}f(x){\mathrm{d} x}=(\frac{1}{3}\cdot 4^3)-(\frac{1}{3}\cdot 8^3)$

$\int_{2}^{4}f(x){\mathrm{d} x}=(\frac{64}{3})-(\frac{8}{3})$

$\int_{2}^{4}f(x){\mathrm{d} x}=\frac{56}{3}=18\frac{2}{3}$

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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 #33 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #1 : Integrals

Example Question #1 : Integrals

Example Question #1 : Finding Integrals

Example Question #1 : Finding Indefinite Integrals

Example Question #2 : Finding Indefinite Integrals

Example Question #3 : Finding Indefinite Integrals

Example Question #2241 : High School Math

Example Question #2242 : High School Math

Example Question #2 : Finding Integrals

All AP Calculus AB Resources

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