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AP Calculus AB : Functions, Graphs, and Limits

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3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #12 : Finding Integrals

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

Possible Answers:

$2\pi$

$\pi$

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 #13 : Finding Integrals

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

Possible Answers:

$583\tfrac{1}{3}$

$58\tfrac{1}{3}$

750

$666\tfrac{2}{3}$

$58\tfrac{2}{3}$

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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Example Question #11 : Finding Integrals

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

Possible Answers:

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

$\frac{\pi }{n}$

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

$-2\pi$

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 #44 : Finding Integrals

What is the indefinite integral of x+1 ?

Possible Answers:

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

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

2x^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 #45 : Finding Integrals

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

Possible Answers:

144x+169

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

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

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

24x+13

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 #46 : Finding Integrals

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

Possible Answers:

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

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

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

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

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

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 #15 : Finding Integrals

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

Possible Answers:

102.5

76.5

128.5

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

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

Possible Answers:

106.25

206.25

53.5

306.25

100

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 #91 : Functions, Graphs, And Limits

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 #92 : Functions, Graphs, And Limits

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

Possible Answers:

5x+12+c

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

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

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

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

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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AP Calculus AB : Functions, Graphs, and Limits

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #12 : Finding Integrals

Example Question #13 : Finding Integrals

Example Question #11 : Finding Integrals

Example Question #44 : Finding Integrals

Example Question #45 : Finding Integrals

Example Question #46 : Finding Integrals

Example Question #15 : Finding Integrals

Example Question #61 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #91 : Functions, Graphs, And Limits

Example Question #92 : Functions, Graphs, And Limits

All AP Calculus AB Resources

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