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AP Calculus AB : Asymptotic and Unbounded Behavior

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Example Questions

Example Question #53 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

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

Possible Answers:

$2\pi$

$\frac{\pi}{2}$

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 $f(x)=\sin(x)$ , we can't use the power rule, as it has a special antiderivative:

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

Remember to include the + c for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(-\cos(b)+c)-(-\cos(a)+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

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

$\int_{\pi}^{2\pi}f(x){\mathrm{d} x}=(-1)-(-(-1))$

$\int_{\pi}^{2\pi}f(x){\mathrm{d} x}=-2$

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Example Question #54 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

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

Possible Answers:

$\pi$

$\frac{\pi}{2}$

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 $f(x)=\cos(x)$ , we can't use the power rule, as it has a special antiderivative:

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

Remember to include the + c for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(\sin(b)+c)-(\sin(a)+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{\pi}^{2\pi}f(x){\mathrm{d} x}=(\sin({2\pi}))-(\sin(\pi))$

$\int_{\pi}^{2\pi}f(x){\mathrm{d} x}=(0)-(0)$

$\int_{\pi}^{2\pi}f(x){\mathrm{d} x}=0$

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Example Question #81 : Functions, Graphs, And Limits

What is the indefinite integral of x+18 ?

Possible Answers:

$\frac{1}{2}x^2+18x+c$

2x^2+18x+c

$\frac{1}{2}x^3+18x^2+cx$

Undefined

Correct answer:

$\frac{1}{2}x^2+18x+c$

Explanation:

To find the indefinite integral of our equation, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the by one and then divide by that new exponent.

Remember that, when taking the integral, we treat constants as that number times $x^{0}$ , since anything to the zero power is . Treat as 18x^0 .

$\int x+18{\mathrm{d} x}=\frac{x^{1+1}}{1+1}+\frac{18x^{0+1}}{0+1}+c$

When taking an integral, be sure to include a . stands for "constant". Since taking the derivative of a constant whole number will always equal , we include the to anticipate the possiblity of the equation actually being x^2+5 or x^2-100 instead of just x^2 .

$\int x+18{\mathrm{d} x}=\frac{x^{2}}{2}+\frac{18x^{1}}{1}+c$

$\int x+18{\mathrm{d} x}=\frac{1}{2}x^2+18x+c$

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Example Question #62 : Calculus Ii — Integrals

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

Possible Answers:

$\cos(x)$

$-\cos(x)$

$2\sin(x^2)+c$

$-\sin(x)+c$

$-\cos(x)+c$

Correct answer:

$-\cos(x)+c$

Explanation:

$\sin(x)$ is a special function.

The indefinite integral is $\int \sin(x) {\mathrm{d} x}=-\cos(x)+c$ .

Even though it is a special function, we still need to include a . stands for "constant". Since taking the derivative of a constant whole number will always equal , we include the to anticipate the possiblity of the equation actually being x^2+5 or x^2-100 instead of just x^2 .

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Example Question #82 : Functions, Graphs, And Limits

What is the indefinite integral of 6x^2+8 ?

Possible Answers:

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

12x

2x^3-8x-c

2x^3+8x+c

Correct answer:

2x^3+8x+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 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.

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

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

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Example Question #11 : Finding Definite 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 #12 : Finding Definite Integrals

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

Possible Answers:

$666\tfrac{2}{3}$

$58\tfrac{2}{3}$

750

$583\tfrac{1}{3}$

$58\tfrac{1}{3}$

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 #13 : Finding Definite Integrals

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

Possible Answers:

$\frac{\pi }{n}$

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

$-2\pi$

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

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 #301 : Ap Calculus Ab

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 #302 : Ap Calculus Ab

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

Possible Answers:

24x+13

144x+169

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

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

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

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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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 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #54 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #81 : Functions, Graphs, And Limits

Example Question #62 : Calculus Ii — Integrals

Example Question #82 : Functions, Graphs, And Limits

Example Question #11 : Finding Definite Integrals

Example Question #12 : Finding Definite Integrals

Example Question #13 : Finding Definite Integrals

Example Question #301 : Ap Calculus Ab

Example Question #302 : Ap Calculus Ab

All AP Calculus AB Resources

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