Application of Integrals - AP Calculus BC

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The temperature of an oven is increasing at a rate degrees Fahrenheit per miniute for $0\le t\le10$ minutes. The initial temperature of the oven is degrees Fahrenheit.

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

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Integrating over an interval 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 , and when 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 degrees during the first five minutes. The last step is to add the initial temperature, which tells us that the temperature at minutes is

degrees.

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

and the velocity at time of the particle is .

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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

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

C is therefore 30.

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

Find the work done by gravity exerting an acceleration of for a block down from its original position with no initial velocity.

Remember that

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

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

$F=ma= $(10-10)$\frac{kgm}{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$

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

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

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)

SO, we get:

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

So, our answer is:

Give the arclength of the graph of the function $f(x) = $\frac{1}{2}$ \ln \left ( \cos 2 x \right )$ on the interval $\left [ 0, $\frac{\pi}{8}$ \right ]$ .

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The length of the curve of on the interval can be determined by evaluating the integral

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

$f(x) = $\frac{1}{2}$ \ln \left ( \cos 2 x \right )$ , so

$f'(x) = $\frac{1}{2}$ \cdot $\frac{\frac{\mathrm{d}$ }{\mathrm{d} x}\cos 2 x}{\cos 2 x} = $\frac{1}{2}$ \cdot $\frac{-2 \sin 2x}{\cos 2 x}$ = - \tan 2x$

The integral becomes

$\int_${0}^{ $\frac{\pi}${8}$} $\sqrt{1 + \left( - \tan 2x \right $)^{2}$$} ; \mathrm{d}x$

$= \int_${0}^{ $\frac{\pi}${8}$} $\sqrt{1 + \tan ^{2}$ 2x } ; \mathrm{d}x$

$= \int_${0}^{ $\frac{\pi}${8}$} $\sqrt{\sec ^{2}$ 2x } ; \mathrm{d}x$

$= \int_${0}^{ $\frac{\pi}${8}$} \sec 2x ; \mathrm{d}x$

Use substitution - set . Then $$\frac{\mathrm{d}$ u}{\mathrm{d} x} = 2$ , and $\mathrm{d} x = $\frac{1}{2}$\mathrm{d} u$ . The bounds of integration become $2 \cdot 0 = 0$ and $2 \cdot $\frac{\pi}{8}$ = $\frac{\pi}{4}$$ , and the integral becomes

$$\frac{1}{2}$ \int_${0}^{ $\frac{\pi}${4}$} \sec u ; \mathrm{d}u$

$=\left.\begin{matrix}$\frac{1}{2}$ \ln | \sec u + \tan u | \ \ \end{matrix}\right|\begin{matrix} $\frac{\pi}{4}$\ \ 0 \end{matrix}$

$=\frac{1}{2}$ \ln \left | \sec $\frac{\pi}{4}$ + \tan $\frac{\pi}{4}$ \right | - $\frac{1}{2}$ \ln | \sec 0 + \tan 0 |$

$=\frac{1}{2}$ \ln\left| $\sqrt{2}$ + 1 \right| -$\frac{1}{2}$ \ln | 1 + 0 |$

$=\frac{1}{2}$ \ln \left ( 1 + $\sqrt{2}$ \right ) -$\frac{1}{2}$ \ln 1$

$=\frac{1}{2}$ \ln\left( 1 + $\sqrt{2}$ \right) - 0$

$=\frac{1}{2}$ \ln \left (1 + $\sqrt{2}$ \right )$

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

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The length of the curve of on the interval 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}${2}$}$

so

$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 . 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}$$

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

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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:

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:

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

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

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.

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$\small \small W=\int (Fdx)$ , 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}$/2$ .

What is the length of the curve over the interval $0\le x\le4$ ?

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The general formula for finding the length of a curve over an interval is $\int_${a}^{b}$$$\sqrt{1+\left(\frac{dy}{dx}$\right)^2$}\ dx$

In this example, the arc length can be found by computing the integral

$\int_${0}^{4}$$$\sqrt{1+(\frac{d}{dx}$x^{$\frac{3}$${2}$})^2$}\ dx$ .

The derivative of can be found using the power rule, $\frac{d}{dx}$ $x^n$$=nx^{n-1}$ , which leads to

$\int_${0}^{4}$$$\sqrt{1+\left(\frac{3}{2}$x^{$\frac{1}$${2}$}\right)^2$}\ dx=\int_${0}^{4}$$\sqrt{1+\frac{9}{4}$x}\ dx$ .

At this point, a substitution is useful.

Let

$u=1+\frac{9}{4}$x,\ du=\frac{9}{4}$dx,$\frac{4}{9}$du=dx$ .

We can also express the limits of integration in terms of to simplify computation. When , and when .

Making these substitutions leads to

$\int_${0}^{4}$$\sqrt{1+\frac{9}{4}$x}\ dx=\int_${1}^{10}$$\frac{4}{9}$$\sqrt{u}$du=\frac{4}{9}$\int_${1}^{10}$$u^{$\frac{1}${2}$}du$ .

Now use the power rule, which in general is $\int $x^n$$=\frac{x^{n+1}$$}{n+1}+c$ , to evaluate the integral.

$\frac{4}{9}$\int_${1}^{10}$$u^{$\frac{1}$${2}$}du=\frac{4}{9}$$\frac{u^{\frac{3}${2}$}}{$\frac{3}{2}$}|_${1}^{10}$

Find the total distance traveled by a particle along the curve from to .

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To find the required distance, we can use the arc length expression given by .

Taking the derivative of our function, we have . Plugging in our values for our integral bounds, we have

$\int_$0^2$$$\sqrt{1+(2x)^2$}$dx = \int_$0^2$ $$\sqrt{1+4x^2$}$dx$ .

As with most arc length integrals, this integral is too difficult (if not, outright impossible) to evaluate explicitly by hand. So we will just leave it this form, or evaluate it with some computer software.

Solve the separable differential equation

$\frac{\mathrm{d}$ y}{\mathrm{d} x}= $yx^2$

with the condition .

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To solve the separable differential equation, we must separate x and y, dx and dy respectively to opposite sides:

Integrating both sides, we get

$\ln\left | y\right |=\frac{ $x^3$}{3}$+C$

The rules of integration used were

$\int $\frac{dx}{x}$=\ln\left| x\right|+C$ , $\int $x^ndx$= $$\frac{x^{n+1}$$}{n+1}+C$

The constants of integration merged into one.

Now, we exponentiate both sides of the equation to solve for y, and use the properties of exponents to simplify:

To solve for C, we use our given condition:

Our final answer is

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

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

Find the volume of the solid generated by revolving the region bounded by and the -axis in the first quadrant about the -axis.

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Since we are revolving a region of interest around a horizontal line , 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 and the inner radius is . The x-limits of the region are between and $x=\pi$ . So the volume set-up is:

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

Using trigonometric identities, we know that:

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]_${0}^{\pi}$ = $$\frac{1}{2}$\pi^2$$

Suppose the functions , , and form a closed region. Rotate this region across the x-axis. What is the volume?

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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.

Determine the shell height. This is done by subtracting the right curve, , with the left curve, .

Find the intersection of and to determine the y-bounds of the integral.

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$

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 ${}\small t=0$ .

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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.

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

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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.

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

$\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 and $y=\sqrt[3]{x}$ is at .

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$

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

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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.

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:

Find the volume of the solid generated when the function

is revolved around the x-axis on the interval .

Hint: Use the method of cylindrical disks.

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The formula for the volume is given as

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

where and the bounds on the integral come from the interval .

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

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

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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 .

We'll also need to convert the endpoints of the interval to y-values. Note that when $x=4, \ y=2$ , and when 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 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

.

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

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

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

on the interval

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

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

where

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