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AP Calculus AB : Asymptotic and Unbounded Behavior

Study concepts, example questions & explanations for 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 #53 : Finding Integrals

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

Possible Answers:

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

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

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

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

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

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 #54 : Finding Integrals

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

Possible Answers:

$-\cos(x)+c$

$\cos(x)+c$

$-\sin(x)+c$

$\tan(x)+c$

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

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 #21 : Integrals

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

Possible Answers:

413.6

427.4

Undefined

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 #22 : Integrals

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

Possible Answers:

241.5

71.25

142.5

192

49.5

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 #23 : Integrals

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

Possible Answers:

$259\tfrac{1}{3}$

$457\tfrac{1}{3}$

$358\tfrac{1}{3}$

$250\tfrac{2}{3}$

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 #24 : Integrals

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

Possible Answers:

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

569

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

$230\tfrac{2}{3}$

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 #71 : Calculus Ii — Integrals

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

Possible Answers:

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

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

8x+5

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

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

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 #104 : Asymptotic And Unbounded Behavior

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

Possible Answers:

2x^2+3x+c

4+c

2x^3-c

2x^2+3x

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

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

Possible Answers:

$179\tfrac{7}{12}$

$73\tfrac{5}{12}$

$126\tfrac{2}{3}$

$53\tfrac{1}{4}$

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

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

Possible Answers:

708

308

$102\tfrac{2}{3}$

$78\tfrac{1}{3}$

$78\tfrac{2}{3}$

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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AP Calculus AB : Asymptotic and Unbounded Behavior

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #53 : Finding Integrals

Example Question #54 : Finding Integrals

Example Question #21 : Integrals

Example Question #22 : Integrals

Example Question #23 : Integrals

Example Question #24 : Integrals

Example Question #71 : Calculus Ii — Integrals

Example Question #104 : Asymptotic And Unbounded Behavior

Example Question #83 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #84 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

All AP Calculus AB Resources

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