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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 #7 : Finding Definite Integrals

$\int_2^6(5x^2+3x+2){\mathrm{d} x}=?$

Possible Answers:

$402\tfrac{2}{3}$

$204\tfrac{2}{3}$

$-12\tfrac{2}{3}$

$26\tfrac{2}{3}$

$401\tfrac{1}{3}$

Correct answer:

$402\tfrac{2}{3}$

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)=5x^2+3x+2 , we can use the power rule for all of the terms involved to find our anti-derivative:

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

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

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(\frac{5}{3}b^3+\frac{3}{2}b^2+2b+c)-(\frac{5}{3}a^3+\frac{3}{2}a^2+2a+c)$

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

$\int_{2}^{6}f(x){\mathrm{d} x}=(\frac{5}{3}(6)^3+\frac{3}{2}(6)^2+2(6))-(\frac{5}{3}(2)^3+\frac{3}{2}(2)^2+2(2))$

$\int_{2}^{6}f(x){\mathrm{d} x}=(360+54+12)-(\frac{40}{3}+6+4)$

$\int_{2}^{6}f(x){\mathrm{d} x}=(426)-(23\tfrac{1}{3})$

$\int_{2}^{6}f(x){\mathrm{d} x}=402\tfrac{2}{3}$

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Example Question #8 : Finding Definite Integrals

$\int_3^5\frac{x+3}{x}{\mathrm{d} x}=?$

Possible Answers:

1.33

6.12

3.53

3.21

5.67

Correct answer:

3.53

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 $f(x)=\frac{x+3}{x}$ , we can't use the power rule. We have to break up the quotient into separate parts:

$f(x)=\frac{x}{x}+\frac{3}{x}=1+\frac{3}{x}$ .

The integral of 1 should be no problem, but the other half is a bit more tricky:

$\int\frac{3}{x}{\mathrm{d} x}$ is really the same as $3\int\frac{1}{x}{\mathrm{d} x}$ . Since $\int\frac{1}{x}{\mathrm{d} x}=\ln{x}$ , $\int\frac{3}{x}{\mathrm{d} x}=3\ln{x}$ .

Therefore:

$\int\frac{x+3}{x}{\mathrm{d} x}=x+3\ln{x}+c$

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

Now we can plug that equation into our FToC equation:

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

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

$\int_{3}^{5}f(x){\mathrm{d} x}=(5+3\ln{5})-(3+3\ln{3})$

$\int_{3}^{5}f(x){\mathrm{d} x}=(9.83)-(6.30)$

$\int_{3}^{5}f(x){\mathrm{d} x}\approx 3.53$

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Example Question #9 : Finding Definite Integrals

$\int_2^5\sqrt{x}\text{ }{\mathrm{d} x}=?$

Possible Answers:

9.34

5.56

7.52

1.89

2.59

Correct answer:

5.56

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 $f(x)=\sqrt{x}$ , we can use the power rule, if we turn it into an exponent:

$f(x)=\sqrt{x}=x^{\frac{1}{2}}$

This means that:

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

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

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(\frac{2}{3}b^\frac{3}{2}+c)-(\frac{2}{3}a^\frac{3}{2}+c)$

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

$\int_{2}^{5}f(x){\mathrm{d} x}=(\frac{2}{3}(5)^\frac{3}{2})-(\frac{2}{3}(2)^\frac{3}{2})$

$\int_{2}^{5}f(x){\mathrm{d} x}\approx(7.45)-(1.89)$

$\int_{2}^{5}f(x){\mathrm{d} x}\approx5.56$

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Example Question #1 : Finding Indefinite Integrals

What is the anti-derivative of ?

Possible Answers:

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

x^2+c

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

2+c

Correct answer:

x^2+c

Explanation:

To find the indefinite integral of our expression, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the by one and then divide by that new exponent.

First we need to realize that 2x=2x^1 . From there we can solve:

$\int2x{\mathrm{d} x}=\frac{2x^{1+1}}{1+1}+c$

When taking an integral, be sure to include a at the end of everything. stands for "constant". Since taking the derivative of a constant whole number will always equal , we include the to anticipate the possiblity of the equation actually being x^2+5 or x^2-100 instead of just x^2 .

$\int2x{\mathrm{d} x}=\frac{2x^{2}}{2}+c$

$\int2x{\mathrm{d} x}=x^2+c$

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Example Question #8 : Finding Indefinite Integrals

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

Possible Answers:

4x^4+2x

3x^2

$\frac{1}{3}x^4+2+c$

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

6x+6+c

Correct answer:

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

Explanation:

To find the indefinite integral of our equation, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the by one and then divide by that new exponent.

Remember that, when taking the integral, we treat constants as that number times $x^{0}$ since anything to the zero power is . For example, treat as 2x^0 .

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

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

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

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Example Question #9 : Finding Indefinite Integrals

What is the indefinite integral of $\frac{1}{3}x^2$ ?

Possible Answers:

x^3+c

x^2

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

$\frac{1}{9}x^3+c$

2x+3

Correct answer:

$\frac{1}{9}x^3+c$

Explanation:

To find the indefinite integral of our equation, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the by one and then divide by that new exponent.

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

When taking an integral, be sure to include a . stands for "constant". Since taking the derivative of a constant whole number will always equal , we include the to anticipate the possiblity of the equation actually being x^2+5 or x^2-100 instead of just x^2 .

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

$\int \frac{1}{3}x^2{\mathrm{d} x}=\frac{1}{9}x^3+c$

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Example Question #10 : Finding Definite Integrals

$\int_{12}^{15}\ln(x){\mathrm{d} x}=?$

Possible Answers:

7.8

8.7

71.12

10.8

3.9

Correct answer:

7.8

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 $f(x)=\ln(x)$ , we can't use the power rule, as it has a special antiderivative:

$\int f(x){\mathrm{d} x}=\int \ln(x){\mathrm{d} x}=x\ln(x)-x+c$

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

Now we can plug that equation into our FToC equation:

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

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

$\int_{12}^{15}f(x){\mathrm{d} x}=(15\ln(15)-15)-(12\ln(12)-12)$

$\int_{12}^{15}f(x){\mathrm{d} x}\approx(25.62)-(17.82)$

$\int_{12}^{15}f(x){\mathrm{d} x}\approx7.8$

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Example Question #31 : Calculus Ii — Integrals

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

Possible Answers:

$\frac{\pi}{2}$

$2\pi$

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 $f(x)=\sin(x)$ , we can't use the power rule, as it has a special antiderivative:

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

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

Now we can plug that equation into our FToC equation:

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

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

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

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

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

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Example Question #11 : Finding Definite Integrals

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

Possible Answers:

$\pi$

$\frac{\pi}{2}$

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 $f(x)=\cos(x)$ , we can't use the power rule, as it has a special antiderivative:

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

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

Now we can plug that equation into our FToC equation:

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

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

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

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

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

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Example Question #10 : Finding Indefinite Integrals

What is the indefinite integral of x+18 ?

Possible Answers:

Undefined

$\frac{1}{2}x^3+18x^2+cx$

$\frac{1}{2}x^2+18x+c$

2x^2+18x+c

Correct answer:

$\frac{1}{2}x^2+18x+c$

Explanation:

To find the indefinite integral of our equation, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the by one and then divide by that new exponent.

Remember that, when taking the integral, we treat constants as that number times $x^{0}$ , since anything to the zero power is . Treat as 18x^0 .

$\int x+18{\mathrm{d} x}=\frac{x^{1+1}}{1+1}+\frac{18x^{0+1}}{0+1}+c$

$\int x+18{\mathrm{d} x}=\frac{x^{2}}{2}+\frac{18x^{1}}{1}+c$

$\int x+18{\mathrm{d} x}=\frac{1}{2}x^2+18x+c$

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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 #7 : Finding Definite Integrals

Example Question #8 : Finding Definite Integrals

Example Question #9 : Finding Definite Integrals

Example Question #1 : Finding Indefinite Integrals

Example Question #8 : Finding Indefinite Integrals

Example Question #9 : Finding Indefinite Integrals

Example Question #10 : Finding Definite Integrals

Example Question #31 : Calculus Ii — Integrals

Example Question #11 : Finding Definite Integrals

Example Question #10 : Finding Indefinite Integrals

All AP Calculus AB Resources

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