Write the formula for arc length.

$\displaystyle s=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$

Calculate the derivatives.

$\displaystyle \frac{dx}{dt}=-4sin(t)$

$\displaystyle \frac{dy}{dt}=4cos(t)$

Substitute the derivatives and the bounds into the integral.

$\displaystyle s=\int_{0}^{2\pi}\sqrt{(-4sin(t))^2+(4cos(t))^2}=\int_{0}^{2\pi}\sqrt{16sin^2(t)+16cos^2(t)}$

$\displaystyle s=\int_{0}^{2\pi} 4 dt= 4[t]_{0}^{2\pi} = 8\pi$

The average value of a function p(t) from t=a to t=b is found with the integral

$\displaystyle \frac{1}{b-a}\int_{a}^{b}p(t)dt$. 

In this case, we must compute the value of the integral

$\displaystyle \frac{1}{2-0}\int_{0}^{2}\frac{4t}{t^2+1}dt=\frac{1}{2}\int_{0}^{2}\frac{4t}{t^2+1}dt$.

A substitution makes this integral clearer. Let $\displaystyle u=t^2+1$. Then $\displaystyle du=2tdt$. We should also rewrite the limits of integration in terms of u. When t = 0, u=1, and when t = 2, u = 5. Making these substitutions results in the integral

$\displaystyle \frac{1}{2}\int_{1}^{5}\frac{2du}{u}=\frac{2}{2}\int_{1}^{5}\frac{du}{u}=\int_{1}^{5}\frac{du}{u}$

Evaluating this integral using the fact that

$\displaystyle \int \frac{du}{u}=ln(u)+c$ yields

$\displaystyle \int_{1}^{5}\frac{du}{u}=ln(5)-ln(1)=ln(5)$

In general, the average value of a function $\displaystyle f(x)$ over the interval $\displaystyle [a, b]$ is 

$\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)dx$

This means that the average value of $\displaystyle f(\theta)$ over $\displaystyle 0\le\theta\le4\pi$ is 

$\displaystyle \frac{1}{4\pi}\int_{0}^{4\pi}2cos(\theta)d\theta$.

Since the antiderivative of $\displaystyle cos(\theta)$ is $\displaystyle \int cos(\theta)d\theta=sin(\theta)+c$, the integral evaluates to 

$\displaystyle \frac{1}{4\pi}\int_{0}^{4\pi}2cos(\theta)d\theta=\frac{2}{4\pi}sin(\theta)|_{0}^{4\pi}$

$\displaystyle =\frac{1}{2\pi}(sin(4\pi)-sin(0))=0$.

The general formula for finding the length of a curve $\displaystyle f(x)$ over an interval $\displaystyle [a,b]$ is $\displaystyle \int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx$

In this example, the arc length can be found by computing the integral

$\displaystyle \int_{0}^{4}\sqrt{1+(\frac{d}{dx}x^{\frac{3}{2}})^2}\ dx$.

The derivative of $\displaystyle x^{\frac{3}{2}}$ can be found using the power rule, $\displaystyle \frac{d}{dx} x^n=nx^{n-1}$, which leads to 

$\displaystyle \int_{0}^{4}\sqrt{1+\left(\frac{3}{2}x^{\frac{1}{2}}\right)^2}\ dx=\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx$. 

At this point, a substitution is useful.

Let 

$\displaystyle u=1+\frac{9}{4}x,\ du=\frac{9}{4}dx,\frac{4}{9}du=dx$.

We can also express the limits of integration in terms of $\displaystyle u$ to simplify computation. When $\displaystyle x=0, u=1$, and when $\displaystyle x = 4, u=10$.

Making these substitutions leads to 

$\displaystyle \int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx=\int_{1}^{10}\frac{4}{9}\sqrt{u}du=\frac{4}{9}\int_{1}^{10}u^{\frac{1}{2}}du$.

Now use the power rule, which in general is $\displaystyle \int x^n=\frac{x^{n+1}}{n+1}+c$, to evaluate the integral. 

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

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

When f is integrable on [a,b], the average value of f(x) on [a,b] is defined to be:

$\displaystyle f(x)_{average}=\frac{1}{b-a}\int_{a}^{b}f(x)dx$

For the problem statement, we are given f(x) and the intervals [a,b]. All that needs to be done is solving the integral over this interval and dividing the result by the difference between the two intervals. 

So:

$\displaystyle f(x)_{average}=\frac{1}{(2)-(-1)}\int_{-1}^{2}(2x+3)^2dx=\frac{1}{3}\int_{-1}^{2}(2x+3)^2dx$

To solve this integral, we have two options. We can FOIL the terms out for $\displaystyle (2x+3)^2$ and solve the integral of the resulting polynomial, or we can use a simple u-substitution. Either way, the result will always be the same. We will try both ways, to prove that this holds true:

FOIL Method:

$\displaystyle \frac{1}{3}\int_{-1}^{2}(2x+3)^2dx=\frac{1}{3}\int_{-1}^{2}(4x^2+12x+9)dx$

$\displaystyle \frac{1}{3}[\frac{4}{3}x^3+6x^2+9x]_{-1}^{2}$

$\displaystyle \frac{1}{3}([\frac{4}{3}(2)^3+6(2)^2+9(2)]-[\frac{4}{3}(-1)^3+6(-1)^2+9(-1)])$

$\displaystyle \frac{1}{3}([\frac{32}{3}+24+18]-[-\frac{4}{3}+6-9])$

$\displaystyle \frac{1}{3}(57)=19$

This is one of the answer choices!

U-Substition Method:

$\displaystyle \frac{1}{3}\int_{-1}^{2}(2x+3)^2dx\rightarrow u=2x+3, du=2dx\rightarrow \frac{1}{2}du=dx$

Next, we must adjust the bounds of the new integral which will be in terms of u.

$\displaystyle u(-1)=2(-1)+3=1$

$\displaystyle u(2)=2(2)+3=7$

So the new integral becomes:

$\displaystyle \frac{1}{3}\int_{1}^{7}\frac{1}{2}u^2du=\frac{1}{6}\int_{1}^{7}u^2du$

$\displaystyle \frac{1}{6}[\frac{1}{3}u^3]_{1}^{7}=\frac{1}{18}[(7)^3-(1)^3]=19$

As you can see both methods results in the same answer!

When asked for the average value between intervals of form $\displaystyle x = a$ and $\displaystyle x = b$ then integral becomes 

$\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)$ so in this case we can rewrite our average value problem as 

$\displaystyle \frac{1}{2-0}\int_{0}^{2}e^{3x}$.

The integral comes $\displaystyle \frac{1}{3}e^{3x}$ evaluated at $\displaystyle x = 2$ and $\displaystyle x = 0$ which simplifies to 

$\displaystyle \frac{e^6 - 1}{3}$ (note that $\displaystyle e^0$ is 1).

However, since we are looking for the average value, we must divide the whole thing by b-a, which in this case is 2, so the final answer is 

$\displaystyle \frac{e^6 - 1}{6}$.

The average value of a function over an interval is defined to be the integral of the function divided 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 get:

$\displaystyle \frac{1}{2-0}\int_0^2 x^2dx=\frac{1}{2}\cdot \frac{8}{3}=\frac{4}{3}$

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:

$\displaystyle \frac{1}{4-0}\int_{0}^4 (x^3-1)dx=\frac{1}{4}(x^4/4-x)|_{0}^4=\frac{1}{4}(4^4/4-4)$

$\displaystyle =\frac{1}{4}(64-4)=\frac{60}{4}=15$

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:

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