AP Calculus AB : Functions, Graphs, and Limits

Study concepts, example questions & explanations for AP Calculus AB

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Example Questions

Example Question #61 : Functions, Graphs, And Limits

Evaluate the integral:

\displaystyle \int (e^x)dx

Possible Answers:

Cannot be evaluated 

\displaystyle e^{x^2}+C

\displaystyle e^0

\displaystyle e^x +C

\displaystyle \frac{e^{x^2}}{2}+C

Correct answer:

\displaystyle e^x +C

Explanation:

The derivative of \displaystyle e^x=e^x. Therefore, the antiderivative of \displaystyle e^x is equal to itself. 

Example Question #263 : Ap Calculus Ab

Evaluate:

\displaystyle \int_{-4}^{4} (x^{9} + tan(x) - x^{3}cos(x))dx

Possible Answers:

Can't be determined from the information given.

\displaystyle 25

\displaystyle 0

\displaystyle 5\pi

Correct answer:

\displaystyle 0

Explanation:

\displaystyle f(x) = x^{9} + tan(x) - x^{3}cos(x) and

\displaystyle f(-x) = (-x)^{9} + tan(-x) - (-x)^{3}cos(-x) =

\displaystyle -x^{9} - tan(x) + x^{3}(cos(x)) = 

\displaystyle -[x^{9} + tan(x) - x^{3}(cos(x))] = -f(x)

Recall that \displaystyle tan(x) is an odd function and \displaystyle cos(x) is an even function.

Thus, since \displaystyle f(x) is an odd function, the integral of this function from \displaystyle -4 to \displaystyle 4 will be zero.

 

 

Example Question #62 : Functions, Graphs, And Limits

Evaluate this indefinite integral:

\displaystyle \int sin^{4}(x)cos^{3}(x)dx

Possible Answers:

\displaystyle \frac{sin^{5}(x)}{5} - \frac{sin^{7}(x)}{7} + C

\displaystyle \frac{cos^{7}(x)}{7} - \frac{cos^{5}(x)}{5}

\displaystyle \frac{sin^{5}(x)}{5} + \frac{sin^{7}(x)}{7} + C

\displaystyle \frac{sin^{5}(x)}{5} - \frac{sin^{7}(x)}{7}

\displaystyle \frac{cos^{5}(x)}{5} - \frac{cos^{7}(x)}{7} + C

Correct answer:

\displaystyle \frac{sin^{5}(x)}{5} - \frac{sin^{7}(x)}{7} + C

Explanation:

To approach this problem, first rewrite the integral expression as shown below:

\displaystyle \int sin^{4}(x)cos^{2}(x)cos(x)dx.

Then, recognize that \displaystyle cos^2(x) = 1 - sin^2(x), and substitute this into the integral expression:

\displaystyle \int sin^{4}(x)(1-sin^{2}x)cos(x)dx

Use substitution, letting \displaystyle u = sin(x) and \displaystyle du = cos(x)dx.  The integral can then be rewritten as

\displaystyle \int u^{4}(1-u^{2})du = \int (u^{4} - u^{6})du.  

Evaluating this integral gives

\displaystyle \frac{u^{5}}{5} - \frac{u^{7}}{7} + C.

Finally, substituting \displaystyle u = sin(x) back into this expression gives the final answer:

\displaystyle \frac{sin^{5}(x)}{5} - \frac{sin^{7}(x)}{7} + C

(As this is an indefinite integral, \displaystyle C must be included).

Example Question #1 : Finding Integrals

Evaluate:

\displaystyle \int_{1}^{e^{100}} \frac{1}{x}dx

Possible Answers:

\displaystyle \frac{100}{e}

\displaystyle \frac{1}{100}

\displaystyle 100e

\displaystyle \ln 100

\displaystyle 100

Correct answer:

\displaystyle 100

Explanation:

\displaystyle \int_{1}^{e^{100}} \frac{dx}{x} = \ln e^{100} - \ln 1 = 100-0 = 100 

Example Question #1 : Finding Definite Integrals

Find  \displaystyle \int_{0}^{\pi} (\cos^{2}x+\sin^{2}x)dx

Possible Answers:

\displaystyle 2\pi

\displaystyle 1

\displaystyle \pi

\displaystyle 3

Correct answer:

\displaystyle \pi

Explanation:

This is most easily solved by recognizing that \displaystyle \sin^{2}x+\cos^{2}x=1.  

Example Question #2 : Finding Definite Integrals

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

Possible Answers:

\displaystyle 2

\displaystyle 0

\displaystyle \pi

\displaystyle 1

\displaystyle \frac{1}{2}

Correct answer:

\displaystyle 0

Explanation:

Remember the fundamental theorem of calculus!

\displaystyle \int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}

Since our \displaystyle f(x)=\sin(2x), we can't use the power rule. Instead we end up with: 

\displaystyle \int f(x){\mathrm{d} x}=\int \sin(2x){\mathrm{d} x}=\frac{-\cos(2x)}{2}+c

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

Now we can plug that equation into our FToC equation:

\displaystyle \int_{a}^{b}f(x){\mathrm{d} x}=(\frac{-\cos(2b)}{2}+c)-(\frac{-\cos(2a)}{2}+c)

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

\displaystyle \int_{\pi}^{2\pi}f(x){\mathrm{d} x}=(\frac{-\cos(4\pi)}{2})-(\frac{-\cos(2\pi)}{2})

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

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

Example Question #63 : Functions, Graphs, And Limits

\displaystyle \int \frac{3}{x}=

Possible Answers:

\displaystyle 3\ln \left | x \right |+C

\displaystyle 3x^{-2}+C

\displaystyle 3x^{-1}+C

\displaystyle -3e^{-x}

\displaystyle -3x^{-2}+C

Correct answer:

\displaystyle 3\ln \left | x \right |+C

Explanation:

The integral of \displaystyle \frac{1}{x} is \displaystyle \ln \left | x \right |+C.  The constant 3 is simply multiplied by the integral.  

Example Question #64 : Functions, Graphs, And Limits

\displaystyle \int cos(x)\sin(x){\mathrm{d} x}=?

Possible Answers:

\displaystyle \sin(x)\cos(x)+c

\displaystyle -(\frac{1}{2}\cos^2(x)+c)

\displaystyle \frac{\cos(x)}{\sin(x)}+c

\displaystyle \cos^2(x)sin^2(x)+c

\displaystyle \frac{\sin(x)}{\cos(x)}+c

Correct answer:

\displaystyle -(\frac{1}{2}\cos^2(x)+c)

Explanation:

To integrate \displaystyle \cos(x)\sin(x), we need to get the two equations in terms of each other. We are going to use "u-substitution" to create a new variable, \displaystyle u, which will equal \displaystyle \cos(x).

Now, if \displaystyle u=\cos(x), then 

\displaystyle \frac{\mathrm{d}u }{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-sin(x) 

Multiply both sides by \displaystyle \mathrm d {x} to get the more familiar: 

\displaystyle {\mathrm{d} u}=-\sin(x)\mathrm{d}x

Note that our \displaystyle \mathrm d{u}=-\sin(x), and our original equation was asking for a positive \displaystyle \sin(x).

That means if we want \displaystyle \int\cos(x)\sin(x) in terms of \displaystyle u, it looks like this:

Bring the negative sign to the outside:

\displaystyle -\int u\mathrm d{u}.

We can use the power rule to find the integral of \displaystyle u:

\displaystyle -(\frac{1}{2}u^2+c)

Since we said that \displaystyle u=\cos(x), we can plug that back into the equation to get our answer:

\displaystyle -(\frac{1}{2}\cos^2(x)+c)

Example Question #3 : Finding Indefinite Integrals

Evaluate the integral below:

 

\displaystyle f(x)=\int \frac{dx}{x^2-16}

Possible Answers:

1

\displaystyle \frac{1}{4} ln\left | \frac{x+2}{x-2} \right |+c

\displaystyle \frac{1}{4} ln\left | \frac{x-2}{x+2} \right |+c

\displaystyle \frac{1}{8} ln\left | \frac{x-4}{x+4} \right |+c

\displaystyle \frac{1}{8} ln\left | \frac{x+4}{x-4} \right |+c

Correct answer:

\displaystyle \frac{1}{8} ln\left | \frac{x-4}{x+4} \right |+c

Explanation:

In this case we have a rational function as \displaystyle \frac{N(x)}{D(x)}, where

\displaystyle N(x)=1

and

 \displaystyle D(x)=\frac{1}{x^2-16}

\displaystyle D(x) can be written as a product of linear factors:

 

\displaystyle \frac{1}{x^2-16}=\frac{1}{(x-4)(x+4)}\equiv \frac{A}{x-4}+\frac{B}{x+4}

 

It is assumed that A and B are certain constants to be evaluated. Denominators can be cleared by multiplying both sides by (x - 4)(x + 4). So we get:

 

\displaystyle 1=A(x+4)+B(x-4)

First we substitute x = -4 into the produced equation:

\displaystyle 1=-8B\Rightarrow B=-\frac{1}{8}

Then we substitute x = 4 into the equation:

\displaystyle 1=8A\Rightarrow A=\frac{1}{8}

Thus:

\displaystyle \frac{1}{(x-4)(x+4)}=\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4}

 

Hence:

\displaystyle \int \frac{dx}{x^2-16}=\int(\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4})dx

\displaystyle =\frac{1}{8}ln\left | x-4 \right |-\frac{1}{8}ln\left |x+4 \right |+c

\displaystyle =\frac{1}{8}(ln\left | x-4 \right |-ln\left |x+4 \right |)+c

\displaystyle =\frac{1}{8}ln\left | \frac{x-4}{x+4}\right |+c

 

 

Example Question #65 : Functions, Graphs, And Limits

Determine the indefinite integral:

Possible Answers:

\displaystyle \frac{ 2 \log_{10} x} {3 } + C

\displaystyle \frac{ 2 \sqrt{ \log_{10} x} } {3 } + C

\displaystyle \frac{ 2 \sqrt{ \log_{10} x} } {3 \ln x} + C

\displaystyle \frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C

\displaystyle \frac{ 2 \sqrt{ \log_{10} x} \cdot \log_{10} x} {3 } + C

Correct answer:

\displaystyle \frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C

Explanation:

\displaystyle \log_{10} x = \frac{\ln x}{\ln 10}, so this can be rewritten as 

 

Set \displaystyle u = \ln x. Then

\displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}

and 

\displaystyle du = \frac{dx}{x}

Substitute:

The outer factor can be absorbed into the constant, and we can substitute back:

\displaystyle = \frac{ 2 \sqrt{ \ln x} \cdot \ln x} {3 \cdot \sqrt{\ln 10}} + C

\displaystyle = \frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C

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