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AP Calculus AB : Asymptotic and Unbounded Behavior

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Example Questions

Example Question #2 : Calculus 3

$\int_{-1}^{0}e^{1-t}dt =$

Possible Answers:

$undefined$

$e+1$

$e^{2}-e$

$1-e^{2}$

$e^{2}-1$

Correct answer:

$e^{2}-e$

Explanation:

We can use the substitution technique to evaluate this integral.

Let u=1-t .

We will differentiate with respect to .

$\frac{\mathrm{d}u }{\mathrm{d} t}=-1$ , which means that ${\mathrm{d} u}=-{\mathrm{d} t}$ .

We can solve for ${\mathrm{d}t }$ in terms of ${\mathrm{d} u}$ , which gives us ${\mathrm{d}t }=-{\mathrm{d} u}$ .

We will also need to change the bounds of the integral. When t=-1 , u=1-(-1) , and when t=0 , u=1-0=1 .

We will now substitute in for the 1-t , and we will substitute $-{\mathrm{d}u }$ for ${\mathrm{d} t}$ .

$\int_{2}^{1}-e^{u}du$

$\int_{2}^{1}-e^{u}du = -e^{u}|_{2}^{1}=-e^{1}-(-e^{2})=e^{2}-e^{1}$

The answer is $e^{2}-e$ .

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Example Question #3 : Finding Integrals By Substitution

Evaluate:

$\dpi{120} \int_{0}^{\frac{\pi }{ 2}}\sin x \sqrt{\cos x }\; \; dx$

Possible Answers:

$\frac{4}{3}$

$\frac{ 2} {3}$

Correct answer:

$\frac{ 2} {3}$

Explanation:

Set $u = \cos x$ .

Then $\frac{\mathrm{du} }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x} \cos x= -\sin x$ and $\sin x \; dx = - du$ .

Also, since $u = \cos x$ , the limits of integration change to $\cos \frac{\pi }{2} = 0$ and $\cos 0 = 1$ .

Substitute:

$\dpi{120} \int_{0}^{\frac{\pi }{ 2}}\sin x \sqrt{\cos x }\; \; dx$

$= \int_{1}^{0} \left (- \sqrt{u} \right ) \; du$

$= \int_{1}^{0} \left (-u ^{\frac{1}{2}} \right ) \; du$

$= \int_{0}^{1} u ^{\frac{1}{2}} \; du$

$\dpi{150} = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} & \\ & \end{matrix}\right|$ $\begin{matrix} 1\\ \\ 0 \end{matrix}$

$\dpi{150} = \left.\begin{matrix} \\ \frac{ 2u \sqrt{u}} {3} & \\ & \end{matrix}\right|$ $\begin{matrix} 1\\ \\ 0 \end{matrix}$

$=\frac{ 2\cdot 1\cdot \sqrt{1}} {3} - \frac{ 2\cdot 0\cdot \sqrt{0}} {3} = \frac{ 2} {3}$

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Example Question #123 : Functions, Graphs, And Limits

Evaluate the following integral:

$\int \frac{dx}{x^2+36}$

Possible Answers:

$\ln(x^2+36x)$

$\sin^{-1}(x+36)$

$\frac{1}{6}\tan^{-1}\bigg(\frac{x}{6}\bigg)+C$

$\frac{x^3}3{+ 36x}$

Correct answer:

$\frac{1}{6}\tan^{-1}\bigg(\frac{x}{6}\bigg)+C$

Explanation:

First you must know that:

$tan^{-1}(x)= \frac{1}{x^2+1}$ and $\int \frac{1}{a^2+x^2}=\bigg(\frac{1}{a}\bigg)tan^{-1}\bigg(\frac{x}{a}\bigg)+C$

Therefore we can rewrite our problem in this form:

$\int\frac{1}{x^2+6^2}dx$

where a=6 .

Thus the integral becomes,

$\int \frac{1}{6^2+x^2}=\bigg(\frac{1}{a}\bigg)tan^{-1}\bigg(\frac{x}{a}\bigg)+C=\bigg(\frac{1}{6}\bigg)tan^{-1}\bigg(\frac{x}{6}\bigg)+C$

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Example Question #124 : Functions, Graphs, And Limits

Evaluate:

$\int_{0}^{2} 2x+3$ .

Possible Answers:

Correct answer:

Explanation:

$\int2x+3 = x^2+3x$

Setting the limits from zero to two we can find that,

$\int2x+3 = x^2+3x$

${}=[(2)^2+3(2)]-[(0^2+3(0)]$

=4+6-0

=10

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Example Question #333 : Ap Calculus Ab

Evaluate:

$\int_{0}^{3} \left | x-1\right |dx$ .

Possible Answers:

2.5

Correct answer:

2.5

Explanation:

Seeing that the equation contains an absolute value you should know that the graph must always remain positive therefore resulting in a V-shaped graph.

Since the equation is y=x-1 , when x=1, y=0 then the vertex of the graph is at (1,0) .

The graph contains a triangle ranging from 0 to 1 and a triangle from 1 to 3. Remebering that taking the interal of a function is the same as finding the area under the curve we can use these triangles to solve our problem.

The area of the triangle from 0 to 1 is,

$\frac{1}{2}bh=\frac{1}{2}(1)(1)=0.5$ .

The area of the triangle from 1 to 3 is,

$\frac{1}{2}bh=\frac{1}{2}(2)(2)=\frac{4}{2}=2$ .

Thus the evaluated integral must be these areas added together,

0.5+2=2.5 .

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Example Question #121 : Asymptotic And Unbounded Behavior

Evaluate:

$\int_{0}^{2}\sqrt{4+3x} dx$ .

Possible Answers:

$\frac{2\sqrt{1000}-16}{9}$

$\frac{17}{9}$

$\frac{20}{9}-\frac{16}{9}$

Correct answer:

$\frac{2\sqrt{1000}-16}{9}$

Explanation:

For this problem we need to use the U Substitution Method.

Using the U-du Rule you can set u=4+3x and du=3dx .

Because we only have a dx in our problem we need to solve for dx, thus

$\frac{1}{3}du=dx$ .

When x=2, u=10 and when x=0, u=4 .

Therefore your new equation will be:

$\int_{4}^{10}\sqrt{u}\frac{1}{3}du= \frac{1}{3}\int_{4}^{10}\sqrt{u}du$

$=\frac{1}{3}[\frac{2}{3}u^{\frac{3}{2}}]$ .

Plugging in our interval we get,

$\frac{1}{3}[\frac{2}{3}10^{\frac{3}{2}}-\frac{2}{3}4^{\frac{3}{2}}]$

$=\frac{1}{3}[\frac{2\sqrt{1000}-2\sqrt{64}}{3}]$

$=\frac{1}{3}[\frac{2\sqrt{1000}-16}{3}]$

$=\frac{2\sqrt{1000}-16}{9}$

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Example Question #93 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Evaluate:

$\int_{1}^{5} (3x^{2} - \frac{5}{x}) dx$

Possible Answers:

104

$124-10\ln 5$

$124-5\ln 5$

124

$125-5\ln 5$

Correct answer:

$124-5\ln 5$

Explanation:

The first step is to find the antiderivative, recalling that:

$\int \frac{1}{x} dx= ln \left | x\right |$ .

For this integral:

$\int_{1}^{5} (3x^{2} - \frac{5}{x}) dx = x^{3} - 5 ln(x)$ ,

where the intergral would be evaluated from x=5 to x=1 (the absolute value bar is not necessary, since both limits of integration are greater than zero):

$5^{3} - 5 ln(5) - (1^3 - 5 ln (1) =$

125 - 5 ln(5) - (1 - 0) =

124 - 5 ln(5)

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Example Question #98 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Evaluate the following indefinite integral:

$\int tan^{5}(x)sec^{2}(x) dx$

Possible Answers:

$\frac{tan^{6}(x)sec^{2}x}{6} + C$

$\frac{tan^{6}(x)}{6} + C$

$\frac{cot^{6}(x)}{6} + C$

${tan^{6}(x)} + C$

$\frac{tan^{6}(x)}{6}$

Correct answer:

$\frac{tan^{6}(x)}{6} + C$

Explanation:

Use substitution, where $u=\tan (x)$ and $du=\sec ^{2}(x)dx$ . Thus, the integral can be rewritten as:

$\int tan^{5}(x)sec^{2}(x) dx= \int u^{5} du = \frac{u^{6}}{6} + C$ .

Substitution of $u=\tan (x)$ back into this expression gives the final answer:

$\frac{tan^{6}(x)}{6} + C$

Note that since this is an indefinite integral, the addition of a constant term (C) is required.

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Example Question #121 : Asymptotic And Unbounded Behavior

Evaluate the limit:

$\lim_{x\rightarrow \infty} \frac{e^x}{x^3}$

Possible Answers:

$\infty$

$-\infty$

The limit does not exist

Correct answer:

$\infty$

Explanation:

When evaluating the limit as x approaches infinity, we must compare the magnitude of the functions. The exponential function in the numerator grows faster than the polynomial function (and any polynomial, for that matter) in the denominator, so the numerator dominates and the limit equals $\infty$ .

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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 #2 : Calculus 3

Example Question #3 : Finding Integrals By Substitution

Example Question #123 : Functions, Graphs, And Limits

Example Question #124 : Functions, Graphs, And Limits

Example Question #333 : Ap Calculus Ab

Example Question #121 : Asymptotic And Unbounded Behavior

Example Question #93 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #98 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #121 : Asymptotic And Unbounded Behavior

All AP Calculus AB Resources

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