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AP Calculus AB : Functions, Graphs, and Limits

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 #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 #121 : Functions, Graphs, And Limits

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:

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

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

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

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 #121 : Functions, Graphs, And Limits

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

Possible Answers:

31.5

193.5

130.5

Undefined

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 #4 : Calculus 3

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

Possible Answers:

$1-e^{2}$

$e+1$

$e^{2}-e$

$e^{2}-1$

$undefined$

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 #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 #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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AP Calculus AB : Functions, Graphs, and Limits

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

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

Example Question #121 : Functions, Graphs, And Limits

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 #121 : Functions, Graphs, And Limits

Example Question #4 : Calculus 3

Example Question #1 : Finding Integrals By Substitution

Example Question #123 : Functions, Graphs, And Limits

Example Question #124 : Functions, Graphs, And Limits

Example Question #333 : Ap Calculus Ab

All AP Calculus AB Resources

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