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AP Calculus BC : Integrals

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Example Question #11 : Application Of Integrals

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:

$\frac{14}{3}$

$\frac{16}{3}$

$\frac{11}{3}$

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

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

Possible Answers:

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

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

$1+\frac{1 }{2} \ln 2$

$\frac{1 }{2}+\frac{1 }{2} \sqrt{2}$

$1 + \sqrt{2}$

Correct answer:

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

Explanation:

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

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

$f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ , 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 u = 2x . 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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Example Question #1 : Average Values And Lengths Of Functions

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

Possible Answers:

$\frac{4\pi}{3}$

$\frac{8}{27}(10\sqrt{10}-1)$

$4\sqrt2-1$

$\frac{56}{27}$

Correct answer:

$\frac{8}{27}(10\sqrt{10}-1)$

Explanation:

The general formula for finding the length of a curve f(x) over an interval [a,b] 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 x=0, u=1 , and when x = 4, u=10 .

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

Find the total distance traveled by a particle along the curve y=x^2 from x=0 to x=2 .

Possible Answers:

$\int_0^2\sqrt{1+2x^2}dx$

$\int_0^2\sqrt{1+2x}dx$

$\int_0^2\sqrt{1+4x^2}dx$

$\int_0^2\sqrt{1+x^2}dx$

Correct answer:

$\int_0^2\sqrt{1+4x^2}dx$

Explanation:

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 y'=f'(x)=2x . 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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Example Question #21 : Application Of Integrals

Solve the separable differential equation

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

with the condition y(0)=4 .

Possible Answers:

$y=4e^{\frac{x^3}{3}}$

$y=4e^{\frac{x^2}{3}}$

$y=Ce^{\frac{x^3}{3}}$

$y=Ce^{\frac{64}{3}}$

$y=Ce^{\frac{x^2}{2}}$

Correct answer:

$y=4e^{\frac{x^3}{3}}$

Explanation:

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

$\frac{dy}{y}=x^2dx$

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:

$y=e^{\frac{x^3}{3}}+e^C=e^{{\frac{x^3}{3}}\cdot C}=Ce^{\frac{x^3}{3}}$

To solve for C, we use our given condition:

4=Ce^0

4=C

Our final answer is

$y=4e^{\frac{x^3}{3}}$

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AP Calculus BC : Integrals

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #11 : Application Of Integrals

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

Example Question #1 : Average Values And Lengths Of Functions

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

Example Question #21 : Application Of Integrals

All AP Calculus BC Resources

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