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AP Calculus AB : AP Calculus AB

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Example Questions

Example Question #34 : Calculus Ii — Integrals

If n is a positive integer, find $\int_{0}^{2\pi}xsinxdx$ .

Possible Answers:

$-2\pi$

$\frac{\pi }{n}$

$\frac{2\pi }{n}$

$-\frac{\pi }{n}$

Correct answer:

$-2\pi$

Explanation:

We can find the integral using integration by parts, which is written as follows:

$\int udv=uv-\int vdu$

Let u=x and dv=sinxdx . We can get the box below:

$\begin{Bmatrix} u=x\\ du=dx\\ dv=sinxdx\\ v=-cosx+c \end{Bmatrix}$

Now we can write:

$\int_{0}^{2\pi}xsinxdx=\left |-xcosx+cx \right |_{0}^{2\pi }-\int_{0}^{2\pi} (-cosx+c)dx$

$=\left |-xcosx+cx+sinx-cx \right |_{0}^{2\pi }$

$=(-2\pi +0)-(0+0)$

$=-2\pi$

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Example Question #11 : Finding Indefinite Integrals

What is the indefinite integral of x+1 ?

Possible Answers:

2x^2+x+c

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

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

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

Correct answer:

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

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 1x^0 since anything to the zero power is one.

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

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.

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

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

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Example Question #12 : Finding Indefinite Integrals

What is the indefinite integral of 12x^2+13x+4 ?

Possible Answers:

24x+13

$4x^3-\frac{13}{2}x^2+4x-c$

$4x^3+\frac{13}{2}x^2+4x+c$

144x+169

$12x^3+\frac{13}{2}x^2+4x+c$

Correct answer:

$4x^3+\frac{13}{2}x^2+4x+c$

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 4x^0 since anything to the zero power is one.

$\int 12x^2+13x+4{\mathrm{d} x}=\frac{12x^{2+1}}{2+1}+\frac{13x^{1+1}}{1+1}+\frac{4x^{0+1}}{0+1}+c$

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.

$\int 12x^2+13x+4{\mathrm{d} x}=\frac{12x^{3}}{3}+\frac{13x^{2}}{2}+\frac{4x^{1}}{1}+c$

$\int 12x^2+13x+4{\mathrm{d} x}=4x^3+\frac{13}{2}x^2+4x+c$

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Example Question #91 : Asymptotic And Unbounded Behavior

What is the indefinite integral of $2x^4+\frac{1}{2}x$ ?

Possible Answers:

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

$8x^2+\frac{1}{2}+c$

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

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

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

Correct answer:

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

Explanation:

To find the indefinite integral, we can use the reverse power rule: we raise the exponent by one and then divide by our new exponent.

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

Remember when taking the indefinite integral to include a to cover any potential constants.

Simplify.

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

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

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Example Question #301 : Ap Calculus Ab

$\int^5_25x+8{\mathrm{d} x}$ ?

Possible Answers:

128.5

102.5

76.5

64.25

Correct answer:

76.5

Explanation:

Remember the fundamental theorem of calculus! If g'(x)=f(x) , then $\int^b_af(x){\mathrm{d} x}=g(b)-g(a)$ .

Since we're given f(x) , we need to find the indefinite integral of the equation to get g(x) .

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.

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

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

Now we can plug that back in:

$\int^b_a 5x+8{\mathrm{d} x}=(\frac{5}{2}b^2+8b +c)-(\frac{5}{2}a^2+8a +c)$

Notice that the 's cancel out.

Plug in our given numbers.

$\int^5_2 5x+8{\mathrm{d} x}=(\frac{5}{2}(5)^2+8*5)-(\frac{5}{2}(2)^2+8*2)$

$\int^5_2 5x+8{\mathrm{d} x}=(\frac{5}{2}*25+40)-(\frac{5}{2}*4+16)$

$\int^5_2 5x+8{\mathrm{d} x}=(62.5+40)-(10+16)$

$\int^5_2 5x+8{\mathrm{d} x}=(102.5)-(26)$

$\int^5_2 5x+8{\mathrm{d} x}=76.5$

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Example Question #302 : Ap Calculus Ab

$\int_4^5 x^3+2x+5$ ?

Possible Answers:

53.5

306.25

100

106.25

206.25

Correct answer:

106.25

Explanation:

Remember the fundamental theorem of calculus! If g'(x)=f(x) , then $\int^b_af(x){\mathrm{d} x}=g(b)-g(a)$ .

Since we're given f(x) , we need to find the indefinite integral of the equation to get g(x) .

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.

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

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

Plug that back into the FTOC:

$\int_a^b x^3+2x+5{\mathrm{d} x}=(\frac{1}{4}b^4+b^2+5b+c)-(\frac{1}{4}a^4+a^2+5a+c)$

Notice that the 's cancel out.

Plug in our given values from the problem.

$\int_4^5 x^3+2x+5{\mathrm{d} x}=(\frac{1}{4}(5)^4+(5)^2+5(5))-(\frac{1}{4}(4)^4+(4)^2+5(4))$ $\int_4^5 x^3+2x+5{\mathrm{d} x}=(\frac{1}{4}*625+25+25)-(\frac{1}{4}*256+16+20)$

$\int_4^5 x^3+2x+5{\mathrm{d} x}=(156.25+25+25)-(64+16+20)$

$\int_4^5 x^3+2x+5{\mathrm{d} x}=(206.25)-(100)$

$\int_4^5 x^3+2x+5{\mathrm{d} x}=106.25$

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Example Question #11 : Finding Indefinite Integrals

What is the indefinite integral of 3x+12 ?

Possible Answers:

$\frac{3}{2}x^2+12x+c$

$\frac{2}{3}x^2-12+c$

$\frac{3}{2}x^2+12x$

Correct answer:

$\frac{3}{2}x^2+12x+c$

Explanation:

To find the indefinite integral, we can use the reverse power rule: we raise the exponent by one and then divide by our new exponent.

We are going to treat as 12x^0 since anything to the zero power is one.

$\int 3x+12{\mathrm{d} x}=\frac{3x^{1+1}}{1+1}+\frac{12x^{0+1}}{0+1}+c$

Remember when taking the indefinite integral to include a to cover any potential constants.

Simplify.

$\int 3x+12{\mathrm{d} x}=\frac{3x^{2}}{2}+\frac{12x^{1}}{1}+c$

$\int 3x+12{\mathrm{d} x}=\frac{3}{2}x^2+12x+c$

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Example Question #301 : Ap Calculus Ab

What is the indefinite integral of 5x^2+12x ?

Possible Answers:

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

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

$\frac{3}{5}x^3+\frac{1}{6}x^2+c$

5x+12+c

$\frac{5}{3}x^3+6x^2+c$

Correct answer:

$\frac{5}{3}x^3+6x^2+c$

Explanation:

To find the indefinite integral, we use the reverse power rule. That means we raise the exponent on the variables by one and then divide by the new exponent.

$\int 5x^2+12x{\mathrm{d} x}=\frac{5x^{2+1}}{2+1}+\frac{12x^{1+1}}{1+1}+c$

Remember to include a when computing integrals. This is a place holder for any constant that might be in the new expression.

$\int 5x^2+12x{\mathrm{d} x}=\frac{5x^{3}}{3}+\frac{12x^{2}}{2}+c$

$\int 5x^2+12x{\mathrm{d} x}=\frac{5}{3}x^3+\frac{12}{2}x^2+c$

$\int 5x^2+12x{\mathrm{d} x}=\frac{5}{3}x^3+6x^2+c$

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Example Question #302 : Ap Calculus Ab

What is the indefinite integral of -8x^2+15 ?

Possible Answers:

-16x

$-\frac{8}{3}x^3+15x$

$\frac{8}{3}x^3-15x+c$

$-\frac{8}{3}x^3+15x+c$

Correct answer:

$-\frac{8}{3}x^3+15x+c$

Explanation:

To find the indefinite integral, we use the reverse power rule. That means we raise the exponent on the variables by one and then divide by the new exponent.

$\int -8x^2+15{\mathrm{d} x}=\frac{-8x^{2+1}}{2+1}+\frac{15x^{0+1}}{0+1}+c$

Remember to include a when doing integrals. This is a placeholder for any constant that might be in the new expression.

$\int -8x^2+15{\mathrm{d} x}=\frac{-8x^{3}}{3}+\frac{15x^{1}}{1}+c$

$\int -8x^2+15{\mathrm{d} x}=-\frac{8}{3}x^3+15x+c$

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Example Question #42 : Calculus Ii — Integrals

$\int_4^8 6x^2+8{\mathrm{d} x}=?$

Possible Answers:

1248

1088

160

928

1488

Correct answer:

928

Explanation:

Use the Fundamental Theorem of Calculus: If g'(x)=f(x) , then $\int_a^bf(x)=g(b)-g(a)$ .

Therefore, we need to find the indefinite integral of our equation first.

To do that, 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.

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

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

That means that $\int_a^b6x^2+8{\mathrm{d} x}=(2b^3+8b+c)-(2a^3+8a+c)$ .

Notice that the 's cancel out.

From here, plug in our numbers.

$\int_4^86x^2+8{\mathrm{d} x}=(2(8)^3+8(8))-(2(4)^3+8(4))$

$\int_4^86x^2+8{\mathrm{d} x}=(2*512+64)-(2*64+32)$

$\int_4^86x^2+8{\mathrm{d} x}=(1024+64)-(128+32)$

$\int_4^86x^2+8{\mathrm{d} x}=(1088)-(160)$

$\int_4^86x^2+8{\mathrm{d} x}=928$

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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 #34 : Calculus Ii — Integrals

Example Question #11 : Finding Indefinite Integrals

Example Question #12 : Finding Indefinite Integrals

Example Question #91 : Asymptotic And Unbounded Behavior

Example Question #301 : Ap Calculus Ab

Example Question #302 : Ap Calculus Ab

Example Question #11 : Finding Indefinite Integrals

Example Question #301 : Ap Calculus Ab

Example Question #302 : Ap Calculus Ab

Example Question #42 : Calculus Ii — Integrals

All AP Calculus AB Resources

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