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AP Calculus AB : Comparing relative magnitudes of functions and their rates of change

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Example Questions

Example Question #91 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Find the indefinite ingtegral for $\int\frac{x+1}{\sqrt{x}} dx$ .

Possible Answers:

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

$(x-2)^\frac{3}{2}$

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

Correct answer:

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

Explanation:

First, bring up the radical into the numerator and distribute to the (x+1) term.

$(x+1)x^{\frac{-1}{2}}$

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

Then integrate.

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

Since it's indefinite, don't forget to add the C:

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

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Example Question #92 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Integrate this function: $\int(x+6)dx$ .

Possible Answers:

$\frac{x^2}{2}+6x+C$

x^2+6x

x^2 + 6x+C

Correct answer:

$\frac{x^2}{2}+6x+C$

Explanation:

First, divide up into two different integral expressions:

$\int xdx + \int6dx$

Then, integrate each:

$\frac{x^2}{2} + 6x$

Don't forget "C" because it is an indefinite integral:

$\frac{x^2}{2} + 6x+C$

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

Integrate the following expression: $\int(2x^4-6x^3+9x)dx$ .

Possible Answers:

x^5-x^4+x^2+C

2x^5-3x^4+9x^2

$\frac{2x^5}{5}-\frac{3x^4}{2}+\frac{9x^2}{2}+C$

Correct answer:

$\frac{2x^5}{5}-\frac{3x^4}{2}+\frac{9x^2}{2}+C$

Explanation:

First, divide up into three different expressions so you can integrate each x term separately:

$\int2x^{4}dx-\int6x^{3}dx+\int9xdx$

Then, integrate and simplify:

$2(\frac{x^{5}}{5})-6(\frac{x^4}{4})+9(\frac{x^2}{2})$

$\frac{2x^5}{5}-\frac{3x^4}{2}+\frac{9x^2}{2}$

Don't forget "C" because it's an indefinite integral:

$\frac{2x^5}{5}-\frac{3x^4}{2}+\frac{9x^2}{2}+C$

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Example Question #94 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Find the general solution of $\int\frac{1}{x^3}dx$ to find the particular solution that satisfies the intitial condition F(1)=0

Possible Answers:

$F(x)=\frac{1}{2x^2}+\frac{1}{2}$

$F(x)=-\frac{1}{2x^2}+\frac{1}{2}$

$F(x)=-2x^2+\frac{1}{2}$

Correct answer:

$F(x)=-\frac{1}{2x^2}+\frac{1}{2}$

Explanation:

To start the problem, it's easier if you bring up the denominator and make it a negative exponent:

$\int x^{-3}dx$

Then, integrate:

$\frac{x^{-2}}{-2}$

Simplify and add the "C" for an indefinite integral:

$-\frac{1}{2x^2}+ C$

Plug in the initial conditions [F(1)=0] to find C and generate the particular solution:

$-\frac{1}{2x^2}+C=0$

$-\frac{1}{2(1^2)}+C=0$

$C=\frac{1}{2}$

Thus, your final equation is:

$F(x)=-\frac{1}{2x^2}+\frac{1}{2}$

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Example Question #95 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Integrate:

$\int(2x-3x^2)dx$

Possible Answers:

x^3-x^2+C

-x^3+x^2 +C

$\frac{4}{3}x^3+\frac{9}{2}x^2+C$

$4x^2-\frac{9}{2}x^3+C$

Correct answer:

-x^3+x^2 +C

Explanation:

First, split up into 2 integrals:

$\int2xdx-\int3x^2dx$

Then integrate and simplify:

$2(\frac{x^2}{2})-3(\frac{x^3}{3})$

x^2-x^3

Don't forget to add C because it's an indefinite integral:

-x^3+x^2 +C

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Example Question #96 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Integrate:

$\int(x^3+2)^2dx$

Possible Answers:

7x^7+x^4+4x+C

$\frac{x^7}{7}-x^4+4x+C$

$\frac{x^7}{7}+x^4+4x+C$

x^7+x^4+4x+C

Correct answer:

$\frac{x^7}{7}+x^4+4x+C$

Explanation:

First, FOIL the binomial:

(x^3+2)(x^3+2)= x^6+4x^3+4

Once that's expanded, integrate each piece separately:

$\int x^6dx+\int4x^3dx+\int4dx$

$\frac{x^7}{7}+4(\frac{x^4}{4})+4x$

Then simplify and add C because it's an indefinite integral:

$\frac{x^7}{7}+x^4+4x+C$

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Example Question #31 : Finding Definite Integrals

$\int^6_3 x^2+5x\ dx =$

Possible Answers:

193.5

Undefined

31.5

130.5

Correct answer:

130.5

Explanation:

Remember the Rundamental Theorem of Calculus: If g'(x)=f(x) , then $\int^b_af(x){\mathrm{d} x}=g(b)-g(a)$ .

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent. For this problem, that would look like:

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

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

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

Now we can plug that back into the problem.

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

Notice that the 's cancel out. Plug in the values given in the problem:

$\int^6_3 x^2+5x{\mathrm{d} x}=(\frac{1}{3}(6)^3+\frac{5}{2}(6)^2)-(\frac{1}{3}(3)^3+\frac{5}{2}(3)^2)$ $\int^6_3 x^2+5x{\mathrm{d} x}=(\frac{1}{3}*216+\frac{5}{2}*36)-(\frac{1}{3}*27+\frac{5}{2}*9)$

$\int^6_3 x^2+5x{\mathrm{d} x}=(72+90)-(9+22.5)$

$\int^6_3 x^2+5x{\mathrm{d} x}=(162)-(31.5)$

$\int^6_3 x^2+5x{\mathrm{d} x}=130.5$

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Example Question #1 : Finding Integrals By Substitution

Evaluate:

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

Possible Answers:

$\frac{ 2} {3}$

$\frac{4}{3}$

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

Evaluate:

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

Possible Answers:

$124-5\ln 5$

$125-5\ln 5$

104

$124-10\ln 5$

124

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

Example Question #92 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

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

Example Question #94 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #95 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #96 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #31 : Finding Definite Integrals

Example Question #1 : Finding Integrals By Substitution

Example Question #97 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

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

All AP Calculus AB Resources

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