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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 #291 : Ap Calculus Ab

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

Possible Answers:

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

4x^4+2x

3x^2

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

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 .

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

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 .

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

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

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Example Question #292 : Ap Calculus Ab

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

Possible Answers:

x^3+c

2x+3

x^2

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

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

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 #33 : Calculus Ii — Integrals

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

Possible Answers:

8.7

7.8

71.12

3.9

10.8

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

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

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 #294 : Ap Calculus Ab

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

Possible Answers:

$\frac{\pi}{2}$

$\pi$

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 $f(x)=\cos(x)$ , 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 #4 : Finding Indefinite Integrals

What is the indefinite integral of x+18 ?

Possible Answers:

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

Undefined

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

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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Example Question #295 : Ap Calculus Ab

What is the indefinite integral of $\sin(x)$ ?

Possible Answers:

$-\cos(x)$

$\cos(x)$

$-\cos(x)+c$

$-\sin(x)+c$

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

Correct answer:

$-\cos(x)+c$

Explanation:

$\sin(x)$ is a special function.

The indefinite integral is $\int \sin(x) {\mathrm{d} x}=-\cos(x)+c$ .

Even though it is a special function, we still need 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 .

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Example Question #2221 : High School Math

What is the indefinite integral of 6x^2+8 ?

Possible Answers:

2x^3-8x-c

12x

$\frac{2}{3}x^3+8x+c$

2x^3+8x+c

Correct answer:

2x^3+8x+c

Explanation:

To solve this problem, we can use the anti-power rule or reverse power rule. We raise the exponent on the variables by one and divide by the new exponent.

For this problem, we'll treat as 8x^0 since anything to the zero power is one.

$\int 6x^2+8{\mathrm{d} x}=\frac{6x^{2+1}}{2+1}+\frac{8x^{0+1}}{0+1}+c$

Since the derivative of any constant is , when we take the indefinite integral, we add a to compensate for any constant that might be there.

From here we can simplify.

$\int 6x^2+8{\mathrm{d} x}=\frac{6x^{3}}{3}+\frac{8x^{1}}{1}+c$

$\int 6x^2+8{\mathrm{d} x}=2x^3+8x+c$

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Example Question #2191 : High School Math

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

Possible Answers:

$\pi$

$2\pi$

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 $f(x)=\cos(2x)$ , we can't use the power rule. Instead we must use u-substituion. If $u = 2x, du = 2dx,and\ dx = \frac{du}{2}$ .

$\int f(x){\mathrm{d} x}=\int \cos(2x){\mathrm{d} x}= \int cos(u) \cdot \frac{1}{2}dx)= \frac{sin(u)}{2}+c=\frac{\sin(2x)}{2}+c$

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

Now we can plug that equation into our FToC equation:

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

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

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

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

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

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Example Question #33 : Calculus Ii — Integrals

$\int_5^{10}\2x^2{\mathrm{d} x}=?$

Possible Answers:

$58\tfrac{2}{3}$

750

$58\tfrac{1}{3}$

$583\tfrac{1}{3}$

$666\tfrac{2}{3}$

Correct answer:

$583\tfrac{1}{3}$

Explanation:

Remember the fundamental theorem of calculus!

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

Since our f(x)=2x^2 , we can use the reverse power rule to find that the antiderivative is:

$\int2x^2{\mathrm{d} x}=\frac{2}{3}x^3+c$

Remember to include a for any integral or antiderivative taken!

Now we can plug that equation into our FToC equation:

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

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

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

$\int_5^{10} 2x^2{\mathrm{d} x}=(\frac{2}{3}(10)^3)-(\frac{2}{3}(5)^3)$

$\int_5^{10} 2x^2{\mathrm{d} x}=(666\tfrac{2}{3})-(88\tfrac{1}{3})$

$\int_5^{10} 2x^2{\mathrm{d} x}=583\tfrac{1}{3}$

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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 #291 : Ap Calculus Ab

Example Question #292 : Ap Calculus Ab

Example Question #33 : Calculus Ii — Integrals

Example Question #51 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #294 : Ap Calculus Ab

Example Question #4 : Finding Indefinite Integrals

Example Question #295 : Ap Calculus Ab

Example Question #2221 : High School Math

Example Question #2191 : High School Math

Example Question #33 : Calculus Ii — Integrals

All AP Calculus AB Resources

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