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

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

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

Plugging this into our formula we attain

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

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

This integral can be evaluated using $\displaystyle u$-substitution.

$\displaystyle u = x^2$

$\displaystyle du=2x dx$

Then we can proceed as follows

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

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

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

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

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

Now we have

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

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

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

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

For this problem we evaluate

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

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

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

$\displaystyle =14$

In order to find the average value we must solve

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

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

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

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

For this problem we evaluate

$\displaystyle \frac{\int_0^{10}\,\frac{1}{5}x\,dx}{10-0}$

$\displaystyle =\frac{\frac{1}{10}x^2|_0^{10}}{10}$

$\displaystyle =\frac{\frac{1}{10}(10)^2-\frac{1}{10}(0)^2}{10}$

$\displaystyle =\frac{10}{10}$

$\displaystyle =1$

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

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

For this problem we evaluate

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

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

$\displaystyle =\frac{2(2)^2-2(0)^2}{2}$

$\displaystyle =\frac{8}{2}$

$\displaystyle =4$

Find the interval $\displaystyle \small [0,b]$ over which the average value of the function $\displaystyle f(x)$ defined below is equal to $\displaystyle \small 72$ (in other words find $\displaystyle b$). 

$\displaystyle \small f(x)=3x^2+2x$

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Definition of the average value of a function:

The average value of a function $\displaystyle \small f(x)$ over an interval $\displaystyle \small [a,b]$ is defined as, 

                      $\displaystyle \small Average_{\: \: over\: \: a,b}\small= \frac{1}{b-a}\int_a^bf(x)dx$

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We are given the left-hand end point $\displaystyle \small a$ of the interval, and must find the right-hand endpoint $\displaystyle \small b$ such that the average value works out to $\displaystyle \small 72$. 

$\displaystyle \small 72 =\frac{1}{b-0} \int_0^b(3x^2+2x)dx$

$\displaystyle \small 72 = \frac{1}{b}\left[x^3+x^2\right]_0^b$

$\displaystyle 72=\frac{1}{b}(b^3 +b^2)$

$\displaystyle 72 = b^2 + b$

$\displaystyle b^2 +b -72 = 0$

$\displaystyle (b+9)(b-8)=0$

$\displaystyle b = \left \{ -9, 8\right \}$

There are two real solutions for $\displaystyle b$, we want the positive solution, therefore $\displaystyle b = 8$. 

$\displaystyle \frac{1}{8} \int_0^8(3x^2+2x)dx = 72$

The formula for arc length for a function of x is

$\displaystyle S=\int_{a}^{b}\sqrt{1+ [f'(x)]^{2}} dx$

,where "S" is the arc length of f(x) from x=a to x=b.

Thus we first need to find $\displaystyle f'(x)$. Using the power rule on each term gives,

$\displaystyle f'(x)=2x+3$

Now substitute this into the arc length equation. Also put the bounds on the integral, x=0 and x=1, in for "a" and "b".

$\displaystyle S= \int_0^1 \sqrt{1+ (2x+3)^{2}}dx$

While it may be tempting to multiply out the $\displaystyle (2x+3)^{2}$, it is not necessary. The only way to integrate this arrangement is through a Trigonometric Substitution.

For this problem, we make the following substitution.

$\displaystyle \tan \theta =2x+3$

This will create a pythagorean trig identity inside the square root. But we need to find a few other parts using this substitution. Since we are changing the variable from $\displaystyle x$ to $\displaystyle \theta$, we have to change $\displaystyle dx$ into an equivalent expression in terms of $\displaystyle d\theta$.

Solve the substition for x.

$\displaystyle \tan \theta =2x+3$

$\displaystyle \tan \theta -3 = 2x$

$\displaystyle \frac{\tan \theta}{2} - \frac{3}{2}=x$

Now differentiate both sides.

$\displaystyle \frac { (\sec^{2}\theta)}{2}d\theta=dx$

This is what we will substitute in for $\displaystyle dx$.

Our bounds are x=0 and x=1, but now we are integrating with respect to the new variable, $\displaystyle \theta$. Thus we have find equivalent $\displaystyle \theta$ bounds. We do this by using the substitution we made and plugging in the x bounds, then solving for $\displaystyle \theta$.

Start with

$\displaystyle \tan \theta =2x+3$

For x=0, replace x with 0 and solve for $\displaystyle \theta$.

$\displaystyle \tan \theta =2(0)+3$

$\displaystyle \tan \theta =3$

$\displaystyle \theta = \tan^{-1}(3)$

Repeat for x=1

$\displaystyle \tan \theta =2(1)+3$

$\displaystyle \tan \theta =5$

$\displaystyle \theta =\tan^{-1}(5)$

Assembling all these pieces gives the following substitution:

$\displaystyle S={\color{Red} \int_0^1} \sqrt{1+ ({\color{Green} 2x+3})^{2}}{\color{Blue} dx}$

$\displaystyle S= {\color{Red} \int_{\tan^{-1}(3)}^{\tan^{-1}(5)}} \sqrt{1 + ({\color{Green} \tan \theta})^2} \cdot {\color{Blue} \frac{\sec^{2}\theta}{2}d\theta}$

Now apply the Pythagorean Trig. Identity, $\displaystyle 1 + \tan^{2}\theta=\sec^{2}\theta$.

$\displaystyle S=\int_{\tan^{-1}(3)}^{\tan^{-1}(5)} \sqrt{\sec^{2}\theta} \cdot \frac{\sec^{2}\theta}{2}d\theta$

Simplify by canceling the square and square root. Then combine the secants through exponent properties.

$\displaystyle S=\int_{\tan^{-1}(3)}^{\tan^{-1}(5)}\frac{\sec^{3}\theta}{2}d\theta$

The only way to integrate $\displaystyle \sec^{3}\theta$ is to use integration by parts, and solving for the integral. First we can pull the 2 in the denominator out of the integral as a constant. 

$\displaystyle S=\frac{1}{2}\int_{\tan^{-1}(3)}^{\tan^{-1}(5)}\sec^{3}\theta d\theta$

Then we separate the $\displaystyle \sec^{3} \theta d\theta$ into $\displaystyle \sec \theta$ and $\displaystyle \sec^{2} \theta d\theta$

$\displaystyle S=\frac{1}{2}\int_{\tan^{-1}(3)}^{\tan^{-1}(5)}\sec\theta \cdot \sec^{2}\theta d\theta$

For now we will focus on just the integral, and will multiply by 1/2 afterward. Now use integration by parts with the following "u" and "dv".

$\displaystyle u=\sec \theta$                                 $\displaystyle dv= \sec^{2} \theta d\theta$

then differentiate "u" to find "du" and integrate "dv" to find "v".

$\displaystyle du= \sec \theta \tan \theta d\theta$               $\displaystyle v=\tan \theta$

Now assemble the pieces. Remember that we are leaving the 1/2 off until the end.

$\displaystyle {\color{Red} \int_{a}^{b} u \cdot dv}= [{\color{Green} u} \cdot {\color{Blue} v}]|_{a}^{b} - \int_{a}^{b} {\color{Blue} v} \cdot {\color{Green} du}$

$\displaystyle {\color{Red} \int_{\tan^{-1}(3)}^{\tan^{-1}(5)}\sec\theta \cdot \sec^{2}\theta d\theta }= [{\color{Green} \sec\theta} \cdot {\color{Blue} \tan\theta}]|_{\tan^{-1}(3)}^{\tan^{-1}(5)}-\int_{\tan^{-1}(3)}^{\tan^{-1}(5)}{\color{Blue} \tan\theta} \cdot {\color{Green} \sec\theta \tan\theta d\theta}$

Now combine the two $\displaystyle \tan\theta$ in the integral into $\displaystyle \tan^{2}\theta$. Also, combine the $\displaystyle \sec\theta \cdot \sec^{2}\theta$ into $\displaystyle \sec^{3}\theta$.

$\displaystyle \int_{\tan^{-1}(3)}^{\tan^{-1}(5)}\sec^{3}\theta d\theta = [ \sec\theta \cdot \tan\theta]|_{\tan^{-1}(3)}^{\tan^{-1}(5)}-\int_{\tan^{-1}(3)}^{\tan^{-1}(5)}\tan^{2}\theta \cdot \sec\theta d\theta$

Now apply a Pythagorean Trig Identity to the $\displaystyle \tan^{2}\theta$ to make it $\displaystyle (\sec^{2}\theta-1)$.

$\displaystyle \int_{\tan^{-1}(3)}^{\tan^{-1}(5)}\sec^{3}\theta d\theta = [ \sec\theta \cdot \tan\theta]|_{\tan^{-1}(3)}^{\tan^{-1}(5)}-\int_{\tan^{-1}(3)}^{\tan^{-1}(5)}(\sec^{2}\theta -1) \sec\theta d\theta$

Now mulitiply the $\displaystyle (\sec^{2}\theta-1)\sec\theta$  in the integral. 

$\displaystyle \int_{\tan^{-1}(3)}^{\tan^{-1}(5)}\sec^{3}\theta d\theta = [ \sec\theta \cdot \tan\theta]|_{\tan^{-1}(3)}^{\tan^{-1}(5)}-\int_{\tan^{-1}(3)}^{\tan^{-1}(5)} (\sec^{3}\theta - \sec\theta) d\theta$

Now split the integral into two integrals. Remember to distribute the integral's negative sign to both new integrals.

$\displaystyle \int_{\tan^{-1}(3)}^{\tan^{-1}(5)}\sec^{3}\theta d\theta = [ \sec\theta \cdot \tan\theta]|_{\tan^{-1}(3)}^{\tan^{-1}(5)}-\int_{\tan^{-1}(3)}^{\tan^{-1}(5)} \sec^{3}\theta d\theta + \int_{\tan^{-1}(3)}^{\tan^{-1}(5)}\sec\theta d\theta$

Notice that we have $\displaystyle \int_{\tan^{-1}(3)}^{\tan^{-1}(5)}\sec^{3}\theta d\theta$on both sides of the equation now. Thus we can solve for it algebraicly. Add it to both sides, so it cancels off the right side and then combine it on the left.

$\displaystyle 2\int_{\tan^{-1}(3)}^{\tan^{-1}(5)} \sec^{3}\theta d\theta= [ \sec\theta \cdot \tan\theta]|_{\tan^{-1}(3)}^{\tan^{-1}(5)} + \int_{\tan^{-1}(3)}^{\tan^{-1}(5)}\sec\theta d\theta$

Before we divide both sides by 2, we should go ahead an evalute the integral on the right, and plug in the bounds for both terms. This will get a constant that will be easier to work with. (The following work only shows the right side of the equation.)

First, evaluate the $\displaystyle \int_{\tan^{-1}(3)}^{\tan^{-1}(5)}\sec\theta d\theta$

$\displaystyle [ \sec\theta \cdot \tan\theta]|_{\tan^{-1}(3)}^{\tan^{-1}(5)} + [\ln\left | \sec\theta + \tan\theta \right |]|_\tan^{-1}(3)}^{\tan^{-1}(5)$

Then we will plug in the upper and lower bounds for each part. This is where a scientific calculator is required.

$\displaystyle [[\sec(\tan^{-1}(5)) \cdot \tan(\tan^{-1}(5))]-[\sec(\tan^{-1}(3)) \cdot \tan(\tan^{-1}(3))]]+[[\ln\left | \sec(\tan^{-1}(5)) +\tan(\tan^{-1}(5)) \right |] - [\ln\left | \sec(\tan^{-1}(3)) +\tan(\tan^{-1}(3)) \right |]]$

Plugging all this into a calculator gives $\displaystyle \approx 16.50$

Now let's introduce this number in the integration by parts equation.

$\displaystyle 2\int_{\tan^{-1}(3)}^{\tan^{-1}(5)} \sec^{3}\theta d\theta={\color{Red} [ \sec\theta \cdot \tan\theta]|_{\tan^{-1}(3)}^{\tan^{-1}(5)} + \int_{\tan^{-1}(3)}^{\tan^{-1}(5)}\sec\theta d\theta}$

$\displaystyle 2\int_{\tan^{-1}(3)}^{\tan^{-1}(5)} \sec^{3}\theta d\theta\approx {\color{Red} 16.50}$

Now divide by the 2.

$\displaystyle \int_{\tan^{-1}(3)}^{\tan^{-1}(5)} \sec^{3}\theta d\theta\approx 8.25$

Now we plug this back into the Arc Length formula.

$\displaystyle S=\frac{1}{2}{\color{Red} \int_{\tan^{-1}(3)}^{\tan^{-1}(5)}\sec^{3}\theta d\theta}$

$\displaystyle S\approx \frac{1}{2}\cdot {\color{Red} (8.25)}$

Evaluating this gives us our answer, rounded to the hundredths digit.

$\displaystyle S\approx 4.13$

$\displaystyle \int x ^{n} \;\mathrm{d} x = \frac{ x ^{n+ 1 }}{n+1}$,

so

$\displaystyle \int_{-1}^{1} x ^{888}\; \mathrm{d} x$

$\displaystyle =\frac{ x ^{888+ 1 } }{888+ 1} \left| \begin{matrix} 1 \\ -1 \end{matrix}\right.$

$\displaystyle =\frac{ x ^{889} }{889} \left| \begin{matrix} 1 \\ -1 \end{matrix}\right.$

$\displaystyle =\frac{ 1 ^{889} }{889} - \frac{ (-1) ^{889} }{889} = \frac{ 1 }{889} - \frac{ -1 }{889} = \frac{ 1 }{889} +\frac{ 1 }{889} = \frac{ 2}{889}$