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AP Calculus AB : Comparing relative magnitudes of functions and their rates of change

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Example Questions

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Example Question #61 : Asymptotic And Unbounded Behavior

Evaluate the integral below:

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

Possible Answers:

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

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

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

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

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 #1 : Finding Integrals By Substitution

Determine the indefinite integral:

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

Possible Answers:

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

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

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

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

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

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 #71 : Functions, Graphs, And Limits

Evaluate:

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

Possible Answers:

$\frac{2}{3}$

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

$2 \ln 3$

$2\sqrt{3}$

$2 \sqrt{\ln 3}$

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 #61 : Asymptotic And Unbounded Behavior

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

Possible Answers:

$5\tfrac{1}{2}$

$18\frac{2}{3}$

$22\tfrac{1}{3}$

$11\tfrac{3}{4}$

$28\tfrac{2}{3}$

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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Example Question #1 : Finding Definite Integrals

$\int_{1}^{2}\frac{1}{x}{\mathrm{d} x}=?$

Possible Answers:

0.81

1.2

0.75

0.30

0.69

Correct answer:

0.69

Explanation:

Remember the fundamental theorem of calculus!

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

As it turns out, since our $f(x)=\frac{1}{x}$ , the power rule really doesn't help us. $\frac{1}{x}$ has a special anti derivative: $\ln{x}$ .

$\int \frac{1}{x}{\mathrm{d} x}=\ln{x}+c$

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

Now we can plug that equation into our FToC equation:

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

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

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

$\int_{1}^{2}f(x){\mathrm{d} x}=(\ln(2))-(0)$

$\int_{1}^{2}f(x){\mathrm{d} x}\approx 0.69$

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Example Question #1 : Finding Definite Integrals

$\int_{3}^{40}e^x{\mathrm{d} x}=?$

Possible Answers:

$2.35\cdot 10^{9}$

$2.35\cdot 10^{19}$

$2.35\cdot 10^{15}$

$1.05\cdot 10^{17}$

$2.35\cdot 10^{17}$

Correct answer:

$2.35\cdot 10^{17}$

Explanation:

Remember the fundamental theorem of calculus!

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

As it turns out, since our f(x)=e^x , the power rule really doesn't help us. e^x is the only function that is it's OWN anti-derivative. That means we're still going to be working with e^x .

$\int e^x{\mathrm{d} x}=e^x+c$

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

Now we can plug that equation into our FToC equation:

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

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

$\int_{3}^{40}f(x){\mathrm{d} x}=(e^{40})-(e^3)$

$\int_{3}^{40}f(x){\mathrm{d} x}=(2.35\cdot 10^{17})-(20.09)$

Because e^3 is so small in comparison to the value we got for $e^{40}$ , our answer will end up being $2.35\cdot 10^{17}$

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Example Question #1 : Finding Indefinite Integrals

What is the indefinite integral of x^2+5x ?

Possible Answers:

$\frac{1}{3}x^3+\frac{5}{2}x^2+c$

x^4+5x^2+c

10x^2+c

3x+5

2x+5

Correct answer:

$\frac{1}{3}x^3+\frac{5}{2}x^2+c$

Explanation:

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent. For this problem, that would look like:

$\int x^2+5x{\mathrm{d} x}=\frac{x^{2+1}}{2+1}+\frac{5x^{1+1}}{1+1}+c$

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

$\int x^2+5x{\mathrm{d} x}=\frac{x^{3}}{3}+\frac{5x^{2}}{2}+c$

$\int x^2+5x{\mathrm{d} x}=\frac{1}{3}x^3+\frac{5}{2}x^2+c$

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Example Question #1 : Finding Indefinite Integrals

What is the indefinite integral of 5x+8 ?

Possible Answers:

16x^3+c

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

$\frac{5}{2}x^2+8x +c$

Correct answer:

$\frac{5}{2}x^2+8x +c$

Explanation:

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent. For this problem, that would look like:

$\int 5x+8{\mathrm{d} x}=\frac{5x^{1+1}}{1+1}+\frac{8x^{0+1}}{0+1}+c$

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

$\int 5x+8{\mathrm{d} x}=\frac{5x^{2}}{2}+\frac{8x^{1}}{1}+c$

$\int 5x+8{\mathrm{d} x}=\frac{5}{2}x^2+8x +c$

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Example Question #1 : Finding Indefinite Integrals

What is the indefinite integral of x^3+2x+5 ?

Possible Answers:

3x^2+2

$\frac{1}{4}x^4+2x^2+5x+c$

$\frac{1}{4}x^4+x^2+5x+c$

4x^4+x^2+5x+c

Correct answer:

$\frac{1}{4}x^4+x^2+5x+c$

Explanation:

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent.

We're going to treat as 5x^0 , as anything to the zero power is one.

For this problem, that would look like:

$\int x^3+2x+5{\mathrm{d} x}=\frac{x^{3+1}}{3+1}+\frac{2x^{1+1}}{1+1}+\frac{5x^{0+1}}{0+1}+c$

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

$\int x^3+2x+5{\mathrm{d} x}=\frac{x^{4}}{4}+\frac{2x^{2}}{2}+\frac{5x^{1}}{1}+c$

$\int x^3+2x+5{\mathrm{d} x}=\frac{1}{4}x^4+x^2+5x+c$

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Example Question #71 : Asymptotic And Unbounded Behavior

Determine the indefinite integral:

$\dpi{120} \int\cos x \sqrt{\sin x }\; \; dx$

Possible Answers:

$\dpi{120} \dpi{120} \frac{ 2 \sqrt{\cos x }} {3} +C$

$\dpi{120} \dpi{120} \frac{ 2 \cos x \cdot \sqrt{\cos x }} {3} +C$

$\dpi{120} \frac{ 2 \sin x \cdot \sqrt{\cos x }} {3} +C$

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

$\dpi{120} \frac{ 2 \sqrt{\sin x }} {3} +C$

Correct answer:

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

Explanation:

Set $u = \sin x$ . Then

$\frac{\mathrm{du} }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x} \sin x= \cos x$ .

and

$\cos x \; dx = du$

The integral becomes:

$\dpi{120} \int\cos x \sqrt{\sin x }\; \; dx$

$\dpi{120} =\int \sqrt{u }\; \; du$

$= \int u ^{\frac{1}{2}} \; du$

$\dpi{120} \dpi{150} = \frac{u^{\frac{1}{2}+1}}{{\frac{1}{2}+1}} + C$

$\dpi{150} = \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} + C$

$\dpi{120} =\frac{ 2u \sqrt{u}} {3} +C$

Substitute back:

$\dpi{120} =\frac{ 2 \sin x \cdot \sqrt{\sin x }} {3} +C$

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AP Calculus AB : Comparing relative magnitudes of functions and their rates of change

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #61 : Asymptotic And Unbounded Behavior

Example Question #1 : Finding Integrals By Substitution

Example Question #71 : Functions, Graphs, And Limits

Example Question #61 : Asymptotic And Unbounded Behavior

Example Question #1 : Finding Definite Integrals

Example Question #1 : Finding Definite Integrals

Example Question #1 : Finding Indefinite Integrals

Example Question #1 : Finding Indefinite Integrals

Example Question #1 : Finding Indefinite Integrals

Example Question #71 : Asymptotic And Unbounded Behavior

All AP Calculus AB Resources

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