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Calculus 2 : Integral Applications

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Example Questions

Example Question #161 : Integral Applications

Determine the average value of the function
f(x)=x^2-3x+9
On the interval [0,5]

Possible Answers:

$\frac{295}{30}$

$\frac{11}{4}$

Correct answer:

$\frac{295}{30}$

Explanation:

The average value of a function on the interval [a,b] is defined as

$\frac{\int_{a}^{b}f(x)\,dx}{b-a}$

As such, we solve the expression

$\frac{\int_{0}^{5}x^2-3x+9\,dx}{5-0}$

$=\frac{(\frac{1}{3}x^3-\frac{3}{2}x^2+9x|_0^5}{5}$

$=\frac{[\frac{1}{3}(5)^3-\frac{3}{2}(5)^2+9(5)]-[\frac{1}{3}(0)^3-\frac{3}{2}(0)^2+9(0)]}{5}$

$=\frac{\frac{295}{6}}{5}$

As such, the average value is

$\frac{295}{30}$

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Example Question #20 : Average Values And Lengths Of Functions

Determine the average value of the function
f(x)=x^2
On the interval [2,6]

Possible Answers:

$\frac{217}{4}$

$\frac{52}{3}$

Correct answer:

$\frac{52}{3}$

Explanation:

The average value of a function on the interval [a,b] is defined as

$\frac{\int_{a}^{b}f(x)\,dx}{b-a}$

As such, we solve the expression

$\frac{\int_{2}^{6}x^2\,dx}{6-2}$

$=\frac{(\frac{1}{3}x^3|_2^6}{4}$

$=\frac{[\frac{1}{3}(6)^3]-[\frac{1}{3}(2)^3]}{4}$

$=\frac{\frac{208}{3}}{4}$

As such, the average value is

$\frac{52}{3}$

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Example Question #21 : Average Values And Lengths Of Functions

You will need a calculator for this question.

Find the surface area of revolution formed by the function, f(x)=x^2+1 , from x=0 to x=2, when revolved around the y axis. Round to the nearest hundredth.

Possible Answers:

$\approx 13.61$

$\approx 36.18$

$\approx 18.09$

$\approx 72.36$

Correct answer:

$\approx 36.18$

Explanation:

To find the surface area of a revolution, we must use the following formula.

$SA= 2\pi \int_{a}^{b} r \cdot ds$ , where "r" is the variable radius of the curve from the axis of revolution, and "ds" is the differential of the arc length of the curve at that radius. For this problem, "r" is simply the variable "x", because every point's radius from the axis of revolution (the y axis), is its x coordinate.

You can think of "ds" as the arc length formula after it is differentiated. Arc length is an integral, so the derivative of an integral is just what's inside that integral. Here that means $ds=\sqrt{1+(f'(x))^{2}} \cdot dx$ .

"a" and "b" will be the x=0 and x=2 part of the question.

From this information we can rewrite the surface area formula as

$SA = 2 \pi \int_{0}^{2} x \cdot \sqrt{1 +(f'(x))^2} \cdot dx$

Now we will find f'(x).

$f'(x)= \frac{\mathrm{d} }{\mathrm{d} x}(x^2 + 1)=2x$

Plug in to the formula.

$SA = 2 \pi \int_{0}^{2} x \cdot \sqrt{1 +(2x)^2} \cdot dx$

Now simplify (2x)^2 .

$SA = 2 \pi \int_{0}^{2} x \cdot \sqrt{1 +4x^2} \cdot dx$

Notice that the is part of the derivative of what is under the radical. This almost follows the basic integral form $\int{u^n \cdot du}$ ,with u= 1 + 4x^2 , and du = 8x . Let's force it to match the form exactly. First I will rewrite the square root as an exponent.

$SA = 2 \pi \int_{0}^{2} x \cdot (1 +4x^2)^{\frac{1}{2}} \cdot dx$

Now multiply by a form of 1 (highlighted in red) to force the "du" part of the integration form.

$SA = 2 \pi \cdot {\color{Red} \frac{1}{8}} \int_{0}^{2} {\color{Red} 8} \cdot x \cdot (1 +4x^2)^{\frac{1}{2}} \cdot dx$

It may help to rewrite the expression in terms of to make it easier to apply integration rules. We just need to remember to convert back to x after integrating, before we plug in the bounds.

$2\pi \cdot \frac{1}{8} \int u^\frac{1}{2} du$

Now integrate.

$\frac{\pi}{4} \cdot [\frac{2}{3}u^{\frac{3}{2}}]$

Now convert u back to x. Recall that u= 1 + 4x^2

$SA = \frac{\pi}{4} \cdot [\frac{2(1 + 4x^2)^{\frac{3}{2}}}{3}]|_{0}^{2}$

Now plug in the upper and lower bounds.

$SA = \frac{\pi}{4} \cdot [(\frac{2(1 +4(2)^2)^{\frac{3}{2}}}{3})-(\frac{2(1 + 4(0)^2)^{\frac{3}{2}}}{3})]$

Now use a calculator to find the approximate answer of

$SA \approx 36.18$

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Example Question #21 : Average Values And Lengths Of Functions

Evaluate the average value of the function on the interval [0,10] .

f(x)=2x

Possible Answers:

Correct answer:

Explanation:

To solve for the average value on the interval [a,b] we follow the following formula

$\frac{\int_{a}^{b}f(x)\,dx}{b-a}$

For this problem we evaluate

$\frac{\int_{0}^{10}2x\,dx}{10-0}$

$=\frac{x^2|_0^{10}}{10}$

$=\frac{10^2-0^2}{10}$

$=\frac{100}{10}$

=10

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Example Question #22 : Average Values And Lengths Of Functions

Evaluate the average value of the function on the interval [0,3] .

f(x)=3x^2

Possible Answers:

Correct answer:

Explanation:

To solve for the average value on the interval [a,b] we follow the following formula

$\frac{\int_{a}^{b}f(x)\,dx}{b-a}$

For this problem we evaluate

$\frac{\int_{0}^{3}3x^2\,dx}{3-0}$

$=\frac{x^3|_0^{3}}{3}$

$=\frac{3^3-0^3}{3}$

$=\frac{27}{3}$

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Example Question #162 : Integral Applications

Find the average value of f(x)=x^3-2x^2+4 on [0,3]

Possible Answers:

$\frac{54}{7}$

$\frac{19}{4}$

$\frac{57}{4}$

$\frac{19}{2}$

$\frac{57}{2}$

Correct answer:

$\frac{19}{4}$

Explanation:

Recall the formula for average value of a function, f(x), on a closed interval, [a,b], is: $f(avg)=\frac{1}{b-a}\int_{a}^{b}f(x)dx$

For this we first need to integrate our function on the defined interval:

$\int_{0}^{3}(x^3-2x^2+4)dx=(\frac{x^4}{4}-\frac{2x^3}{3}+4x)_{x=0}^{x=3}=\frac{57}{4}$

Plugging this into our formula we attain

$\frac{1}{b-a}\int_{a}^{b}f(x)dx=\frac{1}{3}\cdot \frac{57}{4}=\frac{19}{4}$

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Example Question #163 : Integral Applications

Find the average value of the function $f(x)=\frac{3x}{1+x^4}$ over the interval $[0,\frac{\pi}{2}]$

Possible Answers:

$\frac{3}{\pi}\tan^{-1}\frac{\pi^2}{4}$

$\frac{3}{\pi}\sin^{-1}\frac{\pi^2}{4}$

None of the other answers

$\frac{3}{2}\tan^{-1}\frac{\pi^2}{2}$

Correct answer:

$\frac{3}{\pi}\tan^{-1}\frac{\pi^2}{4}$

Explanation:

To find the average value, we need to evaluate $\frac{1}{b-a}\int_a^bf(x)dx =\frac{1}{\pi/2-0}\int_0^{\pi/2}\frac{3x}{1+x^4}dx$ .

This integral can be evaluated using -substitution.

u = x^2

du=2x dx

Then we can proceed as follows

$\int \frac{3x}{1+x^4}dx$ (Start)

$= \frac{3}{2}\int \frac{2x}{1+x^4}dx$ (Factor out a 3/2 , leaving a in the numerator)

$=\frac{3}{2}\int \frac{1}{1+u^2}du$ (Substitute the equations for u, du )

$=\frac{3}{2}\tan^{-1}(u)+C$ (Integrate, recall that $\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ )

$=\frac{3}{2}\tan^{-1}(x^2)+C$ (Substitute x^2 back in)

Now we have

$\frac{1}{b-a}\int_a^bf(x)dx =\frac{1}{\pi/2-0}\int_0^{\pi/2}\frac{3x}{1+x^4}dx = \frac{2}{\pi}[\frac{3}{2}\tan^{-1}(x^2)]_0^{\pi/2}$

$\frac{3}{\pi}\tan^{-1}\frac{\pi^2}{4}$ .

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Example Question #164 : Integral Applications

Find the average value of the function

f(x)=2x+4

on the interval

[0,10]

Possible Answers:

140

Correct answer:

Explanation:

To solve for the average value on the interval [a,b] we follow the following formula

$\frac{\int_a^bf(x)\,dx}{b-a}$

For this problem we evaluate

$\frac{\int_0^{10}2x+4\,dx}{10-0}$

$=\frac{x^2+4x|_0^{10}}{10}$

$=\frac{10^2+4(10)-[0^2+4(0)]}{10}$

=14

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Example Question #165 : Integral Applications

Find the average value of the function

f(x)=3x^2+2

on the interval

[3,5]

Possible Answers:

Correct answer:

Explanation:

To solve for the average value on the interval [a,b] we follow the following formula

$\frac{\int_a^bf(x)\,dx}{b-a}$

For this problem we evaluate

$\frac{\int_3^{5}3x^2+2\,dx}{5-3}$

$=\frac{x^3+2x|_3^{5}}{2}$

$=\frac{5^3+2(5)-[3^3+2(3)]}{2}$

=51

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Example Question #161 : Integral Applications

Find the average value of $f(x)=\frac{1}{x}$ on $1\leq x\leq e$ .

Possible Answers:

$\frac{1}{e-1}$

1-e

$\frac{1}{1-e}$

e-1

$\frac{1}{e}$

Correct answer:

$\frac{1}{e-1}$

Explanation:

In order to find the average value we must solve

$\frac{1}{e-1}\int_{1}^{e} \frac{1}{x}dx=\frac{1}{e-1}ln(x)$ evaluated between 1 and e.

$=\frac{1}{e-1}(1-0)=\frac{1}{e-1}$

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Calculus 2 : Integral Applications

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #161 : Integral Applications

Example Question #20 : Average Values And Lengths Of Functions

Example Question #21 : Average Values And Lengths Of Functions

Example Question #21 : Average Values And Lengths Of Functions

Example Question #22 : Average Values And Lengths Of Functions

Example Question #162 : Integral Applications

Example Question #163 : Integral Applications

Example Question #164 : Integral Applications

Example Question #165 : Integral Applications

Example Question #161 : Integral Applications

All Calculus 2 Resources

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