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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 #47 : Calculus Ii — Integrals

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

Possible Answers:

71.25

241.5

49.5

142.5

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 #48 : Calculus Ii — Integrals

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

Possible Answers:

$250\tfrac{2}{3}$

$358\tfrac{1}{3}$

$259\tfrac{1}{3}$

$457\tfrac{1}{3}$

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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Example Question #49 : Calculus Ii — Integrals

$\int_0^4-8x^2+15=?$

Possible Answers:

$230\tfrac{2}{3}$

$-110\tfrac{2}{3}$

569

$-230\tfrac{2}{3}$

Correct answer:

$-110\tfrac{2}{3}$

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.

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

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

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

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

Plug that back into FTOC:

$\int_a^b -8x^2+15{\mathrm{d} x}=(-\frac{8}{3}b^3+15b+c)-(-\frac{8}{3}a^3+15a+c)$

Notice that the 's cancel out.

Plug in our given numbers.

$\int_0^4 -8x^2+15{\mathrm{d} x}=(-\frac{8}{3}(4)^3+15(4))-(-\frac{8}{3}(0)^3+15(0))$ $\int_0^4 -8x^2+15{\mathrm{d} x}=(-\frac{8}{3}(64)+(60))-(0+0)$

$\int_0^4 -8x^2+15{\mathrm{d} x}=(-170\tfrac{2}{3}+(60))$

$\int_0^4 -8x^2+15{\mathrm{d} x}=-110\tfrac{2}{3}$

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Example Question #80 : Calculus Ii — Integrals

$\int 4x^2+5x-3\: dx=?$

Possible Answers:

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

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

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

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

8x+5

Correct answer:

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

Explanation:

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

$\int 4x^2+5x-3\: dx=\frac{4x^{2+1}}{2+1}+\frac{5x^{1+1}}{1+1}-\frac{3x^{0+1}}{0+1}+c$

Don't forget to include a to compensate for any constant!

$\int 4x^2+5x-3\: dx=\frac{4x^{3}}{3}+\frac{5x^{2}}{2}-\frac{3x^{1}}{1}+c$

$\int 4x^2+5x-3\: dx=\frac{4}{3}x^3+\frac{5}{2}x^2-3x+c$

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Example Question #81 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

What is the indefinite integral of 4x+3 with respect to ?

Possible Answers:

2x^3-c

2x^2+3x+c

2x^2+3x

4+c

Correct answer:

2x^2+3x+c

Explanation:

To find the indefinite integral, we're going to use the reverse power rule: raise the exponent of the variable by one and then divide by that new exponent.

$\int 4x+3 dx=(\frac{4x^{1+1}}{1+1})+(\frac{3x^{0+1}}{0+1})+c$

Be sure to include + c to compensate for any constant!

$\int 4x+3dx=(\frac{4x^{1+1}}{1+1})+(\frac{3x^{0+1}}{0+1})+c$

$\int 4x+3dx=(\frac{4x^{2}}{2})+(\frac{3x^{1}}{1})+c$

$\int 4x+3dx=2x^2+3x+c$

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Example Question #50 : Calculus Ii — Integrals

$\int_3^4x^3+2x^2+5=?$

Possible Answers:

$73\tfrac{5}{12}$

$179\tfrac{7}{12}$

$126\tfrac{2}{3}$

$53\tfrac{1}{4}$

Correct answer:

$73\tfrac{5}{12}$

Explanation:

The Fundamental Theorem of Calculus states that if g'(x)=f(x) , then $\int_a^bf(x)=g(b)-g(a)$ . Therefore, we need to find the indefinite integral of our given equation.

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.

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

Plug that into FTOC:

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

Notice that the 's cancel out.

Plug in our given numbers.

$\int_3^4 x^3+2x^2+5{\mathrm{d} x}=(\frac{1}{4}(4)^4+\frac{2}{3}(4)^3+5(4))-(\frac{1}{4}(3)^4+\frac{2}{3}(3)^3+5(3))$ $\int_3^4 x^3+2x^2+5{\mathrm{d} x}=(\frac{1}{4}(256)+\frac{2}{3}(64)+20)-(\frac{1}{4}(81)+\frac{2}{3}(27)+15)$

$\int_3^4 x^3+2x^2+5{\mathrm{d} x}=(64+42\tfrac{2}{3}+20)-(20\tfrac{1}{4}+18+15)$

$\int_3^4 x^3+2x^2+5{\mathrm{d} x}=(126\tfrac{2}{3})-(53\tfrac{1}{4})$

$\int_3^4 x^3+2x^2+5{\mathrm{d} x}=73\tfrac{5}{12}$

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Example Question #51 : Calculus Ii — Integrals

$\int_2^44x^2+\frac{2}{3}x{\mathrm{d} x}=?$

Possible Answers:

$78\tfrac{2}{3}$

308

708

$102\tfrac{2}{3}$

$78\tfrac{1}{3}$

Correct answer:

$78\tfrac{2}{3}$

Explanation:

To find the definite integral, we can use the Fundamental Theorem of Calculus that states that 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 start.

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. For this problem that would look like this:

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

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

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

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

Plug that into FTOC:

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

Notice that the 's cancel out.

Plug in our given values.

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}(4)^3+\frac{1}{3}(4)^2)-(\frac{4}{3}(2)^3+\frac{1}{3}(2)^2)$

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}(64)+\frac{1}{3}(16))-(\frac{4}{3}(8)+\frac{1}{3}(4))$

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{256}{3}+\frac{16}{3})-(\frac{32}{3}+\frac{4}{3})$

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{272}{3})-(\frac{36}{3})$

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=\frac{236}{3}$

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=78\tfrac{2}{3}$

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Example Question #52 : Calculus Ii — Integrals

$\int_3^64x^2+\frac{2}{3}x{\mathrm{d} x}=?$

Possible Answers:

300

261

7.69

339

Correct answer:

261

Explanation:

To find the definite integral, we can use the Fundamental Theorem of Calculus which states that 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 start.

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.

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

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

Plug that into FTOC:

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

Notice that the 's cancel out.

Plug in our given values.

$\int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}(6)^3+\frac{1}{3}(6)^2)-(\frac{4}{3}(3)^3+\frac{1}{3}(3)^2)$

$\int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}(216)+\frac{1}{3}(36))-(\frac{4}{3}(27)+\frac{1}{3}(9))$

$\int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(288+12)-(36+3)$

$\int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(300)-(39)$

$\int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=261$

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Example Question #111 : Asymptotic And Unbounded Behavior

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

Possible Answers:

$2\pi$

Correct answer:

Explanation:

The fundamental theorem of calculus states that if f'(x)=g(x) , then $\int_a^bg(x){\mathrm{d} x}=f(b)-f(a)$ .

First, we need to find the indefinite integral of our given equation. Just like with the derivatives, the indefinite integrals or anti-derivatives of trig functions must be memorized.

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

Don't forget the to compensate for any potential constant!

Plug this in to our FTOC:

$\int_b^a \sin(x){\mathrm{d} x}=(-\cos(b)+c)-(-\cos(a)+c)$ .

Notice that the 's cancel out.

$\int_b^a \sin(x){\mathrm{d} x}=-\cos(b)+\cos(a)$ .

Now plug in the given values.

$\int_{-\pi}^{\pi} \sin(x){\mathrm{d} x}=-\cos(\pi)+\cos(-\pi)$

$\int_{-\pi}^{\pi} \sin(x){\mathrm{d} x}=0+0$

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

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Example Question #21 : Finding Integrals

$\int_2^8 4x^2+5x-3\: dx=?$

Possible Answers:

804

216

$818\tfrac{2}{3}$

270

810

Correct answer:

804

Explanation:

To solve for the definite integral, use the fundamental theorem of calculus. If g'(x)=f(x) , then $\int_a^bf(x)=g(b)-g(a)$ .

First we need to find the indefinite integral.

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

$\int 4x^2+5x-3\: dx=\frac{4x^{2+1}}{2+1}+\frac{5x^{1+1}}{1+1}-\frac{3x^{0+1}}{0+1}+c$

Don't forget to include a to compensate for any constant!

$\int 4x^2+5x-3\: dx=\frac{4x^{3}}{3}+\frac{5x^{2}}{2}-\frac{3x^{1}}{1}+c$

$\int 4x^2+5x-3\: dx=\frac{4}{3}x^3+\frac{5}{2}x^2-3x+c$

Plug this into our first FTOC equation:

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

Notice that the 's cancel out.

$\int_a^b 4x^2+5x-3\: dx=(\frac{4}{3}b^3+\frac{5}{2}b^2-3b)-(\frac{4}{3}a^3+\frac{5}{2}a^2-3a)$

Plug in our given values.

$\int_2^8 4x^2+5x-3\: dx=(\frac{4}{3}(8)^3+\frac{5}{2}(8)^2-3(8))-(\frac{4}{3}(2)^3+\frac{5}{2}(2)^2-3(2))$

$\int_2^8 4x^2+5x-3\: dx=(682\tfrac{2}{3}+160-24)-(10\tfrac{2}{3}+10-6)$

$\int_2^8 4x^2+5x-3\: dx=(818\tfrac{2}{3})-(14\tfrac{2}{3})$

$\int_2^8 4x^2+5x-3\: dx=804$

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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 #47 : Calculus Ii — Integrals

Example Question #48 : Calculus Ii — Integrals

Example Question #49 : Calculus Ii — Integrals

Example Question #80 : Calculus Ii — Integrals

Example Question #81 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #50 : Calculus Ii — Integrals

Example Question #51 : Calculus Ii — Integrals

Example Question #52 : Calculus Ii — Integrals

Example Question #111 : Asymptotic And Unbounded Behavior

Example Question #21 : Finding Integrals

All AP Calculus AB Resources

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