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AP Calculus BC : Application of Integrals

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Example Questions

Example Question #1 : Application Of Integrals

The temperature of an oven is increasing at a rate $r(t)=2e^{0.5t}$ degrees Fahrenheit per miniute for $0\le t\le10$ minutes. The initial temperature of the oven is T(0)=75 degrees Fahrenheit.

What is the temperture of the oven at t = 5 ? Round your answer to the nearest tenth.

Possible Answers:

323.9

119.7

212.2

239.4

Correct answer:

119.7

Explanation:

Integrating r(t) over an interval [a,b] will tell us the total accumulation, or change, in temperature over that interval. Therefore, we will need to evaluate the integral

$\int_{0}^{5}2e^{0.5t}dt$

to find the change in temperature that occurs during the first five minutes.

A substitution is useful in this case. Let $u=0.5t, du=0.5dt, \ and\ 2du=dt$ . We should also express the limits of integration in terms of . When t = 0, u=0 , and when t=5, u=2.5. Making these substitutions leads to the integral

$\int_{0}^{2.5}4e^udu$ .

To evaluate this, you must know the antiderivative of an exponential function.

In general,

$\int e^xdx=e^x+c$ .

Therefore,

$\int_{0}^{2.5}4e^udu=4e^u|_{0}^{2.5}=4e^{2.5}-4\approx 44.7$ .

This tells us that the temperature rose by approximately 44.7 degrees during the first five minutes. The last step is to add the initial temperature, which tells us that the temperature at t=5 minutes is

75+44.7=119.7 degrees.

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Example Question #2 : Application Of Integrals

Find the equation for the velocity of a particle if the acceleration of the particle is given by:

a(t)=16t^2+48

and the velocity at time t=0 of the particle is .

Possible Answers:

16t^2+78

$\frac{16}{3}t^3+96$

16t^2+48

$\frac{16}{3}t^3+48t+30$

Correct answer:

$\frac{16}{3}t^3+48t+30$

Explanation:

In order to find the velocity function, we must integrate the accleration function:

$\int (16t^2+48)dt=\frac{16}{3}t^3+48t+C$

We used the rule

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

to integrate.

Now, we use the initial condition for the velocity function to solve for C. We were told that

v(0)=30

so we plug in zero into the velocity function and solve for C:

$v(0)=30=\frac{16}{3}(0)^3+48(0)+C$

C is therefore 30.

Finally, we write out the velocity function, with the integer replacing C:

$v(t)=\frac{16}{3}t^3+48t+30$

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Example Question #1 : Application Of Integrals

Find the work done by gravity exerting an acceleration of $-10\frac{m}{s^2}$ for a 10kg block down from its original position with no initial velocity.

Remember that

$F_{grav}=mass*acceleration$

$W=\int_{a}^{b}F(x)dx$ , where F(x) is a force measured in Newtons , is work measured in Joules , and and are initial and final positions respectively.

Possible Answers:

250J

100J

-500J

1000J

Correct answer:

250J

Explanation:

The force of gravity is proportional to the mass of the object and acceleration of the object.

$F=ma= (10*-10)\frac{kg*m}{s^2}= -100N$

Since the block fell down 5 meters, its final position is and initial position is .

$\\ W=\int_{0}^{-5}-100dx\\ \\=\frac{-100x}{2}|_{0}^{-5} \\ \\=-50x|_{0}^{-5}\\ \\=-50(-5)--50(0)\\ \\=250J$

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Example Question #2 : Application Of Integrals

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

$f(x)=\int3x^2-e^x+2dx$

F(0)=3

Possible Answers:

F(x)=x^3-e^x+2x-4

F(x)=x^3-e^x+2x

F(x)=x^3+e^x-2x+4

F(x)=x^3-e^x+2x+4

Correct answer:

F(x)=x^3-e^x+2x+4

Explanation:

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

$f(x)=\int3x^2-e^x+2dx$

F(0)=3

To solve this problem,we need to evaluate the given integral, then solve for our constant of integration.

Let's begin by recalling the following integration rules:

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

$2) \int e^xdx=e^x+c$

Using these two, we can integrate f(x)

$f(x)=\int3x^2-e^x+2dx$

$\int3x^2-e^x+2dx=\frac{3x^3}{3}-e^x+2x+c=x^3-e^x+2x+c$

SO, we get:

F(x)=x^3-e^x+2x+c

We are almost there, but we need to find c. To do so, plug in our initial conditions and solve:

3=0^3-e^0+2(0)+c

3=-1+c

c=4

So, our answer is:

F(x)=x^3-e^x+2x+4

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

Determine the volume of the solid obtained by rotating the region with the following bounds about the x-axis:

y=x^2-4x+5

y=0

x=1

x=4

Possible Answers:

$\frac{36\pi }{7}$

$\frac{46\pi }{3}$

$32\pi$

$18\pi$

$\frac{78\pi }{5}$

Correct answer:

$\frac{78\pi }{5}$

Explanation:

From calculus, we know the volume of an irregular solid can be determined by evaluating the following integral:

$V=\int_{a}^{b}A(x)dx$

Where A(x) is an equation for the cross-sectional area of the solid at any point x. We know our bounds for the integral are x=1 and x=4, as given in the problem, so now all we need is to find the expression A(x) for the area of our solid.

From the given bounds, we know our unrotated region is bounded by the x-axis (y=0) at the bottom, and by the line y=x^2-4x+5 at the top. Because we are rotating about the x-axis, we know that the radius of our solid at any point x is just the distance y=x^2-4x+5. Now that we have a function that describes the radius of the solid at any point x, we can plug the function into the formula for the area of a circle to give us an expression for the cross-sectional area of our solid at any point:

$A(x)=\pi r^2=\pi (x^2-4x+5)^2=\pi (x^4-8x^3+26x^2-40x+25)$

We now have our equation for the cross-sectional area of the solid, which we can integrate from x=1 to x=4 to find its volume:

$V=\int_{1}^{4}\pi (x^4-8x^3+26x^2-40x+25)dx$

$=\pi [\frac{1}{5}x^5-2x^4+\frac{26}{3}x^3-20x^2+25x]_{1}^{4}=\frac{78\pi }{5}$

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Example Question #2 : Volume Of Cross Sections And Area Of Region

Suppose the functions f(x) = 0 , x=y^2 , and x=4 form a closed region. Rotate this region across the x-axis. What is the volume?

Possible Answers:

$\frac{32\pi}{5}$

$4\pi$

$8\pi$

$\frac{64}{5}\pi$

$\frac{128\pi}{13}$

Correct answer:

$8\pi$

Explanation:

Write the formula for cylindrical shells, where is the shell radius and is the shell height.

$V=\int_{a}^{b} 2\pi rh \:dy$

Determine the shell radius.

r=y

Determine the shell height. This is done by subtracting the right curve, x=4 , with the left curve, x=y^2 .

h=4-x=4-y^2

Find the intersection of x=y^2 and x=4 to determine the y-bounds of the integral.

4=y^2

2=y

The bounds will be from 0 to 2. Substitute all the givens into the formula and evaluate the integral.

$V=\int_{a}^{b} 2\pi rh \:dy = \int_{0}^{2} 2\pi \cdot y\cdot \left ( 4-y^2)dy$

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

$V=2\pi \left[ 2y^2 - \frac{1}{4}y^4\right]_{0}^{2} = 2\pi \left[2(2)^2 -\frac{1}{4}(2)^4\right] = 2\pi[8-4]= 8\pi$

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

Find the volume of the solid generated by revolving the region bounded by {}y = sin(x) and the -axis in the first quadrant about the -axis.

Possible Answers:

$\frac{1}{2}\pi$

$\frac{1}{2}\pi^2$

$\pi$

$\pi^2$

$\frac{1}{4}\pi$

Correct answer:

$\frac{1}{2}\pi^2$

Explanation:

Since we are revolving a region of interest around a horizontal line {}y = 0 , we need to express the inner and outer radii in terms of x.

Recall the formula:

$Volume = \pi \int_{b}^{a} [(\textup{outer radius})^{2} - (\textup{inner radius})^{2}]dx$

The outer radius is y = sin(x) and the inner radius is y = 0 . The x-limits of the region are between x=0 and $x=\pi$ . So the volume set-up is:

$V = \pi \int_{0}^{\pi} sin^{2}x$

Using trigonometric identities, we know that:

$sin^{2}x = \frac{1}{2} - \frac{1}{2}cos 2x$

Hence:

$V = \pi \int_{0}^{\pi} sin^2x dx =\pi \int_{0}^{\pi} \left(\frac{1}{2} - \frac{1}{2}cos2x\right) dx$

$= \pi \left[\frac{1}{2}x - \frac{1}{4}sin2x\right]_{0}^{\pi} = \frac{1}{2}\pi^2$

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Example Question #4 : Volume Of Cross Sections And Area Of Region

A man fills up a cup of water by leaving it outside during a rainstorm. The rate at which the height of the cup changes is equal to $\small 5t^{1/2}$ . What is the height of water at $\small t=9$ ? Assume the cup is empty at ${}\small t=0$ .

Possible Answers:

$\small 15$

$\small 90$

$\small 27$

$\small 270$

$\small \frac{5}{6}$

Correct answer:

$\small 90$

Explanation:

The rate at which the height changes is $\small \frac{\mathrm{d} h}{\mathrm{d} t}$ , which means $\small \small \frac{\mathrm{d} h}{\mathrm{d} t}=5t^{1/2}$ .

To find the height after nine seconds, we need to integrate to get ${}\small h(t)$ .

We can multiply both sides by $\small dt$ to get $\small dh=5t^{1/2}dt$ and then integrate both sides.

This gives us

$\small h=\frac{10t^{3/2}}{3}+c$ .

Since the cup is empty at $\small t=0, h(0)=0$ , so $\small c=0$ .

This means $\small \small h(9)=10(9)^{3/2}/3=90$ . No units were given in the problem, so leaving the answer unitless is acceptable.

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Example Question #5 : 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 #6 : Volume Of Cross Sections And Area Of Region

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:

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

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

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

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

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

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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AP Calculus BC : Application of Integrals

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Application Of Integrals

Example Question #2 : Application Of Integrals

Example Question #1 : Application Of Integrals

Example Question #2 : Application Of Integrals

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

Example Question #2 : Volume Of Cross Sections And Area Of Region

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

Example Question #4 : Volume Of Cross Sections And Area Of Region

Example Question #5 : Volume Of Cross Sections And Area Of Region

Example Question #6 : Volume Of Cross Sections And Area Of Region

All AP Calculus BC Resources

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