Application of Integrals - AP Calculus BC

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Question

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 degrees Fahrenheit.

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

Answer

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.

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Question

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 .

Answer

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:

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

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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Question

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

Remember that

$F_{grav}=mass*acceleration$

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

Answer

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 .

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

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Question

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

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

Answer

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

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

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:

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

So, our answer is:

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Question

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 degrees Fahrenheit.

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

Answer

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.

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Question

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 .

Answer

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:

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

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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Question

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

Remember that

$F_{grav}=mass*acceleration$

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

Answer

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 .

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

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Question

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

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

Answer

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

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

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:

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

So, our answer is:

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Question

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

Answer

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

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

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

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

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

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

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

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

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Question

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

Answer

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

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

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

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

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

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

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Question

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

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

Answer

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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Question

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.

Answer

$\small \small W=\int (Fdx)$ , 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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Question

What is the length of the curve $g(x)=x^{\frac{3}{2}}$ over the interval $0\le x\le4$ ?

Answer

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 $x^{\frac{3}{2}}$ 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.

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

$=\frac{8}{27}(10^{\frac{3}{2}}-1^\frac{3}{2})=\frac{8}{27}(10\sqrt{10}-1)$

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Question

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

Answer

To find the required distance, we can use the arc length expression given by $\int_{x_0}^{x_1}\sqrt{1+f'(x)^2}dx$ .

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.

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Question

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

Answer

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

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

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

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

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

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

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

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

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Question

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

Answer

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

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

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

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

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

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

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Question

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

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

Answer

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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Question

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.

Answer

$\small \small W=\int (Fdx)$ , 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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Question

What is the length of the curve $g(x)=x^{\frac{3}{2}}$ over the interval $0\le x\le4$ ?

Answer

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 $x^{\frac{3}{2}}$ 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.

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

$=\frac{8}{27}(10^{\frac{3}{2}}-1^\frac{3}{2})=\frac{8}{27}(10\sqrt{10}-1)$

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Question

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

Answer

To find the required distance, we can use the arc length expression given by $\int_{x_0}^{x_1}\sqrt{1+f'(x)^2}dx$ .

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.

Compare your answer with the correct one above

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