The average value of a function is found by taking the integral of the function over the interval and dividing by the length of the interval.

The general rule for this type of integration is as follows.

\(\displaystyle \frac{1}{b-a}\int_a^b \sin x dx=\frac{1}{b-a}\left[-\cos x\right]\left |_a^b\)

So we have:

\(\displaystyle \frac{1}{\pi/2-0}\int_{0}^{\pi/2} \sin x dx=\frac{2}{\pi}(-\cos x)|_{0}^{\pi/2}=\frac{2}{\pi}\)

The average value of a function is found by taking the integral of the function over the interval and dividing by the length of the interval.

The general rule for this type of integration is as follows.

\(\displaystyle \frac{1}{b-a}\int_a^b e^u dx=\frac{1}{b-a}\left[\frac{1}{\frac{du}{dx}}e^u\right]\left |_a^b\)

So we have:

\(\displaystyle \frac{1}{1-0}\int_{0}^1 e^{3x} dx=\left(\frac{1}{3}e^{3x}\right)|_{0}^1=\frac{1}{3}(e^{3}-1)\)

The average value of a function is found by taking the integral of the function over the interval and dividing by the length of the interval.

The general rule for this type of integration is as follows.

\(\displaystyle \frac{1}{b-a}\int_a^b e^u dx=\frac{1}{b-a}\left[\frac{1}{\frac{du}{dx}}e^u\right]\left |_a^b\)

So we have:

\(\displaystyle \frac{1}{5-0}\int_{0}^5 e^{2x} dx=\frac{1}{5}\left(\frac{1}{2}e^{2x}\right)|_{0}^5=\frac{1}{10}(e^{10}-1)\)

The average value of a function is found by taking the integral of the function over the interval and dividing by the length of the interval.

The general rule for this type of integration is as follows.

\(\displaystyle \frac{1}{b-a}\int_a^b x^n dx=\frac{1}{b-a}\left[\frac{x^{n+1}}{n+1}\right]\left |_a^b\)

So we have:

\(\displaystyle \frac{1}{2-0}\int_{0}^2 (3x^2+2x) dx=\frac{1}{2}(x^3+x^2)|_{0}^2=\frac{1}{2}(2^2+2^3)=\frac{12}{2}=6\)

To find the length of a line along a given function f(x) from a to b, utilize the equation

\(\displaystyle L=\int^{b}_{a}\sqrt{1+[f'(x)]^{2}} dx\)

Applying the equation for this problem, we first find the first derivative of the function:

\(\displaystyle f'(x)=2\sqrt{2}\)

Then, we use this expression in the above formula:

\(\displaystyle L=\int^{5}_{3}\sqrt{1+(2\sqrt{2})^{2}} dx=L=\int^{5}_{3}3dx=3x\left.\begin{matrix} \\ \end{matrix}\right|^{5}_{3}=6\)

To find the length of a line along a given function f(x) from a to b, utilize the equation

\(\displaystyle L=\int^{b}_{a}\sqrt{1+[f'(x)]^{2}} dx\)

Applying the equation for this problem and using the given expression for f'(x):

\(\displaystyle L=\int^{4}_{1}\sqrt{1+(\sqrt{x^{4}-1})^{2}} dx=\int^{4}_{1}x^{2}dx=\frac{x^{3}}{3}\left.\begin{matrix} \\ \end{matrix}\right|^{4}_{1}=\frac{63}{3}=21\)

The correct answer is \(\displaystyle \frac{2(\pi-1)}{\pi}\).

To find the average value of the function, we use

Average value \(\displaystyle = \frac{1}{b-a}\int_{a}^{b} x\sin(x) dx\),

with \(\displaystyle a = \frac{\pi}{2}, b=\pi\) in this case.

So we have

\(\displaystyle = \frac{1}{\pi - \frac{\pi}{2}}\int_{\pi/2}^{\pi} x\sin(x) dx\)

\(\displaystyle = \frac{2}{\pi}\int_{\pi/2}^{\pi} x\sin(x) dx\)

Using integration by parts with \(\displaystyle u = x, dv = \sin(x)dx\) and hence \(\displaystyle du=dx, v = -\cos(x)\), we have

\(\displaystyle =\frac{2}{\pi}\{[-x\cos(x)]_{\pi/2}^{\pi}-\int_{\pi/2}^{\pi}-\cos(x)dx\}\)

\(\displaystyle =\frac{2}{\pi}\{(\pi)- [-\sin(x)]_{\pi/2}^{\pi}\}\)

\(\displaystyle =\frac{2}{\pi}\{\pi- (0-(-1))\}\)

\(\displaystyle =\frac{2(\pi-1)}{\pi}\)

The arc length for a function \(\displaystyle f(x)\) on the interval \(\displaystyle [a,b]\) is given by the following integral:

\(\displaystyle L[a,b]=\int_a^b\sqrt{1+(f^{'}(x))^2}dx\).

Thus we must compute the value of \(\displaystyle f^{'}(x)\) for the function \(\displaystyle f(x)=ln(sec(x))\).

To do so, we use the chain rule: 

\(\displaystyle f^{'}(x)=\frac{1}{sec(x)}(\sec(x))^{'}=\frac{1}{\sec(x)}\sec(x)\tan(x)=\tan(x)\).

Now plugging \(\displaystyle f^{'}(x)\) to the arc length formula, and using the interval \(\displaystyle [-\frac{\pi}{4},\frac{\pi}{4}]\), we obtain, 

\(\displaystyle L[-\frac{\pi}{4},\frac{\pi}{4}]=\int_\frac{-\pi}{4}^\frac{\pi}{4}\sqrt{1+\tan^2(x)}dx\).

Using the trig identity \(\displaystyle 1+tan^{2}(x)=\sec^{2}(x)\), our integral becomes:

\(\displaystyle \int_\frac{-\pi}{4}^\frac{\pi}{4}\sqrt{\sec^2(x)}dx=\int_\frac{-\pi}{4}^\frac{\pi}{4}\sec(x)dx\)

\(\displaystyle =\ln\left | \sec(x)+tan(x)) \right |\biggr|_\frac{-\pi}{4}^\frac{\pi}{4}=\ln\left | \sec(\frac{\pi}{4})+tan(\frac{\pi}{4})) \right |-\ln\left | \sec(\frac{-\pi}{4})+tan(\frac{-\pi}{4})) \right |\)

\(\displaystyle =\ln\left | \sqrt2+1 \right |-\ln\left | \sqrt2-1\right |=\ln\left |\frac{\sqrt2+1}{\sqrt2-1} \right |\).

In the last step above, we invoked the property

\(\displaystyle \ln(a)-\ln(b)=\ln\frac{a}{b}\).

Since the expression in the absolute value is positive, we can get rid of the absolute value brackets.

Thus we obtain

\(\displaystyle L[-\frac{\pi}{4},\frac{\pi}{4}]=\ln\biggr(\frac{\sqrt2+1}{\sqrt2-1}\biggr)\).

To find the average value of a function, you must use the following equation:

\(\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)dx\).  Now, we simply substitute our values, and integrate.

\(\displaystyle \frac{1}{10}\int_{0}^{10}x^3+1dx=\frac{1}{10}[\frac{x^4}{4}+x]\).  Now, we plug in our bounds (upper - lower).

\(\displaystyle \frac{1}{10}[\frac{(10^4}{4}+10)-(\frac{0^4}{4}+0)]=\frac{1}{10}[2510]=251.\)

The average value equation is \(\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)dx=\frac{1}{e-1}\int_{1}^{e}\frac{3}{x}dx.\)

\(\displaystyle =\frac{1}{e-1}[3ln(x)].\)  Now, all we have to do is plug in our limits.

\(\displaystyle \frac{1}{e-1}[3ln(e)-3ln(1)]=\frac{3}{e-1}.\)

The second term drops out because the natural log of \(\displaystyle 1\) equals zero.  In the first term, natural log and \(\displaystyle e\) cancel each other out, just leaving the \(\displaystyle 3\) out front.