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AP Calculus AB : Functions, Graphs, and Limits

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Example Questions

Example Question #101 : Functions, Graphs, And Limits

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

Possible Answers:

-16x

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

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

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

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 #41 : Calculus Ii — Integrals

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

Possible Answers:

160

1248

1088

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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Example Question #42 : Calculus Ii — Integrals

$\int_2^{12} x+1{\mathrm{d} x}=?$

Possible Answers:

Correct answer:

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 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.

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

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

According to FTOC:

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

Notice that the 's cancel out.

Plug in our given information and solve.

$\int_2^{12} x+1{\mathrm{d} x}=(\frac{1}{2}(12)^2+(12))-(\frac{1}{2}(2)^2+(2))$

$\int_2^{12} x+1{\mathrm{d} x}=(\frac{1}{2}*144+12)-(\frac{1}{2}*4+2)$

$\int_2^{12} x+1{\mathrm{d} x}=(72+12)-(2+2)$

$\int_2^{12} x+1{\mathrm{d} x}=(84)-(4)$

$\int_2^{12} x+1{\mathrm{d} x}=80$

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Example Question #43 : Calculus Ii — Integrals

$\int_3^512x^2+13x+4{\mathrm{d} x}=?$

Possible Answers:

682.5

861

178.5

504

Undefined

Correct answer:

504

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 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.

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

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

According to FTOC:

$\int_a^b 12x^2+13x+4{\mathrm{d} x}=(4b^3+\frac{13}{2}b^2+4b+c)-(4a^3+\frac{13}{2}a^2+4a+c)$

Notice that the 's cancel out.

Plug in our given numbers and solve.

$\int_3^5 12x^2+13x+4{\mathrm{d} x}=(4(5)^3+\frac{13}{2}(5)^2+4(5))-(4(3)^3+\frac{13}{2}(3)^2+4(3))$

$\int_3^5 12x^2+13x+4{\mathrm{d} x}=((4*125)+(\frac{13}{2}*25)+20)-((4*27)+(\frac{13}{2}*(9))+12)$

$\int_3^5 12x^2+13x+4{\mathrm{d} x}=(500+162.5+20)-(108+58.5+12)$

$\int_3^5 12x^2+13x+4{\mathrm{d} x}=(682.5)-(178.5)$

$\int_3^5 12x^2+13x+4{\mathrm{d} x}=504$

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Example Question #102 : Functions, Graphs, And Limits

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

Possible Answers:

6x+c

$\frac{1}{4}x^4+\frac{2}{3}x^3+5x+c$

3x^2+4x+c

4x^4+6x^3+5x

$\frac{1}{4}x^4+\frac{2}{3}x^3+5x$

Correct answer:

$\frac{1}{4}x^4+\frac{2}{3}x^3+5x+c$

Explanation:

To find the indefinite integral, we can use the reverse power rule. Raise the exponent of the variable by one and then divide by that new exponent.

We're going to treat as 5x^0 .

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

Remember to include the when taking the integral to compensate for any constant.

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

Simplify.

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

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Example Question #101 : Asymptotic And Unbounded Behavior

What is the indefinite integral of $4x^2+\frac{2}{3}x$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

Remember to include a to cover any potential constant that might be in our new equation.

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

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

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Example Question #103 : Functions, Graphs, And Limits

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

Possible Answers:

$\cos(x)+c$

$\tan(x)+c$

$\frac{1}{\sin(x)}+c$

$-\cos(x)+c$

$-\sin(x)+c$

Correct answer:

$-\cos(x)+c$

Explanation:

Just like with the derivatives, the indefinite integrals or anti-derivatives of trig functions must be memorized.

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

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Example Question #103 : Asymptotic And Unbounded Behavior

$\int_2^42x^4+\frac{1}{2}x{\mathrm{d} x}=?$

Possible Answers:

413.6

Undefined

427.4

399.8

Correct answer:

399.8

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.

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.

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

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

Apply the FTOC:

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

Notice that the 's cancel out.

Plug in our given numbers and solve.

$\int_2^4 2x^4+\frac{1}{2}x{\mathrm{d} x}=(\frac{2}{5}(4)^5+\frac{1}{4}(4)^{2})-(\frac{2}{5}(2)^5+\frac{1}{4}(2)^{2})$

$\int_2^4 2x^4+\frac{1}{2}x{\mathrm{d} x}=(\frac{2}{5}*1024+\frac{1}{4}*16)-(\frac{2}{5}*32+\frac{1}{4}*4)$

$\int_2^4 2x^4+\frac{1}{2}x{\mathrm{d} x}=(409.6+4)-(12.8+1)$

$\int_2^4 2x^4+\frac{1}{2}x{\mathrm{d} x}=(413.6)-(13.8)$

$\int_2^4 2x^4+\frac{1}{2}x{\mathrm{d} x}=399.8$

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Example Question #2201 : High School Math

$\int_3^83x+12{\mathrm{d} x}=?$

Possible Answers:

71.25

49.5

142.5

241.5

192

Correct answer:

142.5

Explanation:

Use the Fundamental Theorem ofCcalculus. 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.

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.

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

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

Plug that into our Fundamental Theorem of Calculus:

$\int_a^b 3x+12{\mathrm{d} x}=(\frac{3}{2}b^2+12b+c)-(\frac{3}{2}a^2+12a+c)$

Notice that the 's cancel out.

Plug in our given numbers and solve.

$\int_3^8 3x+12{\mathrm{d} x}=(\frac{3}{2}(8)^2+12(8))-(\frac{3}{2}(3)^2+12(3))$

$\int_3^8 3x+12{\mathrm{d} x}=(\frac{3}{2}*64+12(8))-(\frac{3}{2}*9+12(3))$

$\int_3^8 3x+12{\mathrm{d} x}=(96+96)-(13.5+36)$

$\int_3^8 3x+12{\mathrm{d} x}=(192)-(49.5)$

$\int_3^8 3x+12{\mathrm{d} x}=142.5$

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Example Question #2202 : High School Math

$\int_3^55x^2+12x{\mathrm{d} x}=?$

Possible Answers:

$259\tfrac{1}{3}$

$358\tfrac{1}{3}$

$250\tfrac{2}{3}$

$457\tfrac{1}{3}$

Correct answer:

$259\tfrac{1}{3}$

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.

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.

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

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

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

Now plug that back into the FTOC:

$\int_a^b 5x^2+12x{\mathrm{d} x}=(\frac{5}{3}b^3+6b^2+c)-(\frac{5}{3}a^3+6a^2+c)$

Notice that the 's cancel out.

Plug in our given numbers.

$\int_3^5 5x^2+12x{\mathrm{d} x}=(\frac{5}{3}(5)^3+6(5)^2)-(\frac{5}{3}(3)^3+6(3)^2)$

$\int_3^5 5x^2+12x{\mathrm{d} x}=(\frac{5}{3}(125)+6(25))-(\frac{5}{3}(27)+6(9))$

$\int_3^5 5x^2+12x{\mathrm{d} x}=(208\tfrac{1}{3}+150)-(45+54)$

$\int_3^5 5x^2+12x{\mathrm{d} x}=(358\tfrac{1}{3})-(99)$

$\int_3^5 5x^2+12x{\mathrm{d} x}=259\tfrac{1}{3}$

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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 #101 : Functions, Graphs, And Limits

Example Question #41 : Calculus Ii — Integrals

Example Question #42 : Calculus Ii — Integrals

Example Question #43 : Calculus Ii — Integrals

Example Question #102 : Functions, Graphs, And Limits

Example Question #101 : Asymptotic And Unbounded Behavior

Example Question #103 : Functions, Graphs, And Limits

Example Question #103 : Asymptotic And Unbounded Behavior

Example Question #2201 : High School Math

Example Question #2202 : High School Math

All AP Calculus AB Resources

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