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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

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2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Volume Of Cross Sections And Area Of Region

Approximate the volume of a solid in the first quadrant revolved about the y-axis and bounded by the functions: y=9x and y=x^3 . Round the volume to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

Write the washer's method.

$V=\int_{a}^{b}\pi[R(y)^2-r(y)^2]\:dy$

Set the equations equal to each other to determine the bounds.

9x=x^3

9=x^2

x=3

The bounds are from 0 to 3.

Determine the big and small radius. Rewrite the equations so that they are in terms of y.

$R(y)= \sqrt[3]{y}$

$r(y)=\frac{y}{9}$

Set up the integral and solve for the volume.

$V=\int_{0}^{3}\pi[(\sqrt[3]{y})^2-(\frac{y}{9})^2]\:dy$

$V=\int_{0}^{3}\pi[y^{\frac{2}{3}}-\frac{y^2}{81}]\:dy=\pi[\frac{3}{5}y^\frac{5}{3}-\frac{y^3}{243}]_{0}^{3}=11.41353$

The volume to the nearest integer is:

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Example Question #4 : Volume Of A Solid

Determine the volume of a solid created by rotating the curve $y=\sqrt[3]{x}$ and the line x=1 by revolving around the -axis.

Possible Answers:

$\frac{3}{5}\pi$

$\frac{1}{5}\pi$

$\frac{12}{5}\pi$

$\frac{8}{5}\pi$

$\frac{2}{5}\pi$

Correct answer:

$\frac{3}{5}\pi$

Explanation:

Write the volume formula for cylindrical shells.

$V=\int_{a}^{b}2\pi (\textup{shell radius})(\textup{shell height}) dy$

The shell radius is .

The shell height is the function in terms of . Rewrite that equation.

$y=\sqrt[3]{x}$

x=y^3

$\textup{shell height}: 1-y^3$

The bounds lie on the y-axis since the thickness variable is . This is from 0 to 1, since the intersection of the line x=1 and $y=\sqrt[3]{x}$ is at (1,1) .

Substitute all the values and solve for the volume.

$V=\int_{0}^{1}2\pi (y)(1-y^3) dy=\int_{0}^{1}2\pi (y-y^4)dy$

$V=2\pi \int_{0}^{1}(y-y^4)dy= 2\pi \left[\frac{y^2}{2}-\frac{y^5}{5}\right]_{0}^{1}= 2\pi\left[\frac{1}{2}-\frac{1}{5}\right]= \frac{3}{5}\pi$

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Example Question #1 : Volume Of A Solid

What is the volume of the solid formed when the line $y=\sqrt{x}$ is rotated around the -axis from x=4 to x = 9 ?

Possible Answers:

$\frac{211\pi}{5}$

$\frac{21\pi}{2}$

$36\pi$

Correct answer:

$\frac{211\pi}{5}$

Explanation:

To rotate a curve around the y-axis, first convert the function so that y is the independent variable by solving $y=\sqrt{x}$ for x. This leads to the function x=y^2 .

We'll also need to convert the endpoints of the interval to y-values. Note that when $x=4, \ y=2$ , and when x = 9, y = 3. Therefore, the the interval being rotated is from $y=2\ to\ y=3$ .

The disk method is best in this case. The general formula for the disk method is

$V=\pi\int_{a}^{b}f^2(x)dx$ , where V is volume, $a\ and\ b$ are the endpoints of the interval, and f(x) the function being rotated.

Substuting the function and endpoints from the problem at hand leads to the integral

$V=\pi\int_{2}^{3}(y^2)^2dy=\pi\int_{2}^{3}y^4dy$ .

To evaluate this integral, you must know the power rule. Recall that the power rule is

$\int{y^ndy}=\frac{y^{n+1}}{n+1}+c$ .

$\pi\int_{2}^{3}y^4dy=\pi[\frac{y^5}{5}]_{2}^{3}$

$=\pi\left[\frac{3^5}{5}-\frac{2^5}{5}\right]=\pi\left[\frac{243}{5}-\frac{32}{5}\right]$

$=\frac{211\pi}{5}$ .

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Example Question #1 : Volume Of A Solid

Find the volume of the solid generated when the function

f(x)=x

is revolved around the x-axis on the interval [0,8] .

Hint: Use the method of cylindrical disks.

Possible Answers:

$V=\frac{625\pi}{3}$ units cubed

$V=\frac{400\pi}{3}$ units cubed

$V=\frac{512\pi}{3}$ units cubed

$V=215\pi$ units cubed

Correct answer:

$V=\frac{512\pi}{3}$ units cubed

Explanation:

The formula for the volume is given as

$V=\pi \int_{a}^{b} r^2 \, dx$

where r=f(x) and the bounds on the integral come from the interval [a,b] .

As such,

$V=\pi \int_{0}^{8} x^2 \, dx$

When taking the integral, we will use the inverse power rule which states

$\int x^n=\frac{x^{n+1}}{n+1}$

Applying this rule we get

$\pi \int_{0}^{8} x^2 \, dx=\pi \left(\frac{x^3}{3}\right) \left.\begin{matrix} \\ \end{matrix}\right|_{0}^{8}$

And by the corollary of the First Fundamental Theorem of Calculus

$\pi \left(\frac{x^3}{3}\right) \left.\begin{matrix} \\ \end{matrix}\right|_{0}^{8}=\pi \left(\left[\frac{8^3}{3}\right]-\left[\frac{0^3}{3}\right]\right)$

$=\frac{512\pi}{3}$

As such,

$V=\frac{512\pi}{3}$ units cubed

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Example Question #1 : Volume Of Cross Sections And Area Of Region

Using the method of cylindrical disks, find the volume of the region revolved around the x-axis of the graph of

f(x)=-x^2+10x+3

on the interval

[0,10]

Possible Answers:

$\frac{13270}{3}\pi$ units cubed

$20000\pi$ units cubed

$15000\pi$ units cubed

$10000\pi$ units cubed

Correct answer:

$\frac{13270}{3}\pi$ units cubed

Explanation:

The formula for volume of the region revolved around the x-axis is given as

$V=\pi \int_{a}^{b} r^2\, dx$

where r=f(x)

As such

$V=\pi \int_{0}^{10} (-x^2+10x+3)^2\,dx = \pi \int_{0}^{10}x^4-20x^3+94x^2+60x+9\,dx$

When taking the integral, we use the inverse power rule which states

$\int x^n\,dx = \frac{x^{n+1}}{n+1}$

Applying this rule term by term we get

$= \pi \int_{0}^{10}x^4-20x^3+94x^2+60x+9\,dx = \pi\left ( \frac{x^5}{5} - \frac{20x^4}{4} + \frac{94x^3}{3} + \frac{60x^2}{2} + 9x\right |_{0}^{10}$

And by the corollary of the Fundamental Theorem of Calculus

$\pi\left ( \frac{x^5}{5} - \frac{20x^4}{4} + \frac{94x^3}{3} + \frac{60x^2}{2} + 9x\right |_{0}^{10}=\pi \left ( \left [ \frac{(10)^5}{5}-\frac{20(10)^4}{4}+\frac{94(10)^3}{3}+\frac{60(10)^2}{2}+9(10) \right ]-0 \right )$

$=\frac{13270}{3}\pi$

As such the volume is

$\frac{13270}{3}\pi$ units cubed

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Example Question #1 : Volume Of Cross Sections And Area Of Region

Find the volume of the solid generated by rotating about the y-axis the region under the curve $y =\sqrt{x}$ , from to .

Possible Answers:

$\frac{128}{5}\pi$

None of the other answers

$\frac{256}{7}\pi^2$

$\frac{64}{3}\pi$

$\frac{\pi}{2}$

Correct answer:

$\frac{128}{5}\pi$

Explanation:

Since we are revolving a function of around the y-axis, we will use the method of cylindrical shells to find the volume.

Using the formula for cylindrical shells, we have

$\int_{0}^{4}2\pi x (\sqrt{x})dx$

$=\int_{0}^{4}2\pi x^{3/2}dx$

$=[2\pi x^{5/2}]^{4}_{0}$

$=\frac{128}{5}\pi$ .

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Example Question #181 : Ap Calculus Bc

Find the area bound by the curve of g(t), the x and y axes, and the line $t=\frac{3 \pi}{4}$

g(t)=4sin(t)-6cos(t)

Possible Answers:

2.59

11.07

-1.41

-2.59

Correct answer:

2.59

Explanation:

Find the area bound by the curve of g(t), the x and y axes, and the line $t=\frac{3 \pi}{4}$

g(t)=4sin(t)-6cos(t)

We are asked to find the area under a curve. This sounds like a job for an integral.

We need to integrate our function from 0 to $t=\frac{3 \pi}{4}$ . These will be our limits of integration. With that in mind, our integral look like this.

$G(t)=\int_0^\frac{3\pi}{4}4sin(t)-6cos(t)dt$

Now, we need to recall the rules for integrating sine and cosine.

$1) \int sin(t)dt=-cos(t)+c$

$2) \int cos(t)dt=sin(t)+c$

With these rules in mind, we can integrate our original function to get:

$G(t)=\int_0^\frac{3\pi}{4}4sin(t)-6cos(t)dt=-4cos(t)-6sin(t)+c|_0^{\frac{3 \pi}{4}}$

$G(t)=-4cos(t)-6sin(t)+c|_0^{\frac{3 \pi}{4}}$

Now, we just need to evaluate our new function at the given limits. Let's start with 0

G(0)=-4cos(0)-6sin(0)+c=-4-0+c=-4+c

And our upper limit...

$G({\frac{3 \pi}{4}})=-4cos({\frac{3 \pi}{4}})-6sin({\frac{3 \pi}{4}})+c$

$-4cos({\frac{3 \pi}{4}})-6sin({\frac{3 \pi}{4}})+c=(-4)\frac{-\sqrt{2}}{2}-(6)\frac{-\sqrt{2}}{2}+c$

$(-4)\frac{-\sqrt{2}}{2}-(6)\frac{\sqrt{2}}{2}+c=2\sqrt{2}-3\sqrt{2}+c=-\sqrt{2}+c$

Now, we need to take the difference between our limits:

$G(\frac{3 \pi}{4})-G(0)=(-\sqrt{2}+c)-(-4+c)=-\sqrt{2}+4\approx2.585$

So, our answer is 2.59

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Example Question #171 : Integrals

Determine the length of the following function between $1\leq t\leq 4$

$v(t)=\frac{2}{3}(t-1)^{\frac{3}{2}}$

Possible Answers:

$\frac{16}{9}$

$\frac{14}{3}$

$-\frac{7}{3}$

$-\frac{14}{3}$

$\frac{7}{3}$

Correct answer:

$\frac{14}{3}$

Explanation:

In order to begin the problem, we must first remember the formula for finding the arc length of a function along any given interval:

$L=\int ds$

where ds is given by the equation below:

$ds=\sqrt{1+(\frac{dy}{dx})^{2}}dx$

We can see from our equation for ds that we must find the derivative of our function, which in our case is dv/dt instead of dy/dx, so we begin by differentiating our function v(t) with respect to t:

$\frac{dv}{dt}=(t-1)^{\frac{1}{2}}=\sqrt{t-1}$

Now we can plug this into the given equation to find ds:

$ds=\sqrt{1+(\frac{dv}{dt})^{2}}dt=\sqrt{1+(\sqrt{t-1})^{2}}=\sqrt{1+t-1}=\sqrt{t}$

Our last step is to plug our value for ds into the equation for arc length, which we can see only involves integrating ds. The interval along which the problem asks for the length of the function gives us our limits of integration, so we simply integrate ds from t=1 to t=4:

$L=\int ds=\int_{1}^{4}\sqrt{t}dt=[\frac{2}{3}t^{\frac{3}{2}}]_{1}^{4}$

$=\frac{2}{3}(4)^{\frac{3}{2}}-\frac{2}{3}(1)^{\frac{3}{2}}=\frac{16}{3}-\frac{2}{3}=\frac{14}{3}$

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Example Question #1 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

In physics, the work done on an object is equal to the integral of the force on that object dotted with its displacent.

This looks like $\small W=\int(F\cdot ds)$ ( $\small W$ is work, $\small F$ is force, and $\small ds$ is the infinitesimally small displacement vector). For a force whose direction is the line of motion, the equation becomes $\small \small W=\int (Fdx)$ .

If the force on an object as a function of displacement is $\small F(x)=3x^{2}+x$ , what is the work as a function of displacement $\small W(x)$ ? Assume $\small W(0)=0$ and the force is in the direction of the object's motion.

Possible Answers:

$\small \small W(x)=6x+1$

$\small \small W(x)=3x^{2}+x$

$\small W(x)=x^{3}+x^{2}/2$

$\small \small W(x)=3x^{3}/2+x^{2}$

Not enough information

Correct answer:

$\small W(x)=x^{3}+x^{2}/2$

Explanation:

$\small \small W=\int (Fdx)$ , so $\small W=\int(3x^{2}+x)dx$ .

Both the terms of the force are power terms in the form $\small ax^{n}$ , which have the integral $\small \frac{a}{n+1}x^{n+1}$ , so the integral of the force is $\small x^{3}+x^{2}/2+c$ .

We know

$\small W(0)=0\rightarrow 0^{3}+0^{2}/2+c=0 \therefore c=0$ .

This means

$\small W(x)=x^{3}+x^{2}/2$ .

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Example Question #2 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

Give the arclength of the graph of the function $f(x) = \frac{2}{3} x \sqrt{x}$ on the interval $\left [ 0,3 \right ]$ .

Possible Answers:

$\frac{14}{3}$

$\frac{11}{3}$

$\frac{16}{3}$

Correct answer:

$\frac{14}{3}$

Explanation:

The length of the curve of f(x) on the interval [a,b] can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x$ .

$f(x) = \frac{2}{3} x \sqrt{x} = \frac{2}{3} x^{\frac{3}{2}}$

$f(x) = \frac{2}{3} \cdot \frac{3}{2} x^{\frac{1}{2}} = \sqrt{x}$ .

The above integral becomes

$\int_{0}^{3} \sqrt{1 + (\sqrt{x}) ^{2}} \; \mathrm{d}x$

$\int_{0}^{3} \sqrt{x+1} \; \mathrm{d}x$

Substitute u = x + 1 . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 1$ , $\mathrm{d} u = \mathrm{d} x$ , and the integral becomes

$\int_{1}^{4} \sqrt{u} \; \mathrm{d}u = \int_{1}^{4} u^{\frac{1}{2}} \; \mathrm{d}u$

$= \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \\ \\ \end{matrix}\right| \begin{matrix} 4\\ \\ 1 \end{matrix}$

$= \left.\begin{matrix} \frac{2u \sqrt{u}}{3} \\ \\ \end{matrix}\right| \begin{matrix} 4\\ \\ 1 \end{matrix}$

$= \frac{2 \cdot 4 \sqrt{4}}{3} - \frac{2 \cdot 1 \sqrt{1}}{3}$

$= \frac{2 \cdot 4 \cdot 2 }{3} - \frac{2 \cdot 1 \cdot 1}{3} = \frac{16}{3} - \frac{2}{3} =\frac{14}{3}$

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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Volume Of Cross Sections And Area Of Region

Example Question #4 : Volume Of A Solid

Example Question #1 : Volume Of A Solid

Example Question #1 : Volume Of A Solid

Example Question #1 : Volume Of Cross Sections And Area Of Region

Example Question #1 : Volume Of Cross Sections And Area Of Region

Example Question #181 : Ap Calculus Bc

Example Question #171 : Integrals

Example Question #1 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

Example Question #2 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

All AP Calculus BC Resources

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