# Thread: Average of a variable given the function?

1. ## Average of a variable given the function?

Can you help me with this: I chose 3/2 but I'm not sure if it's right. To get this I graphed the equation in the first quadrant and tried to figure out the average height, which I thought had to be about 1.5. Is this right? Thanks!

2. ## Re: Average of a variable given the function?

Let's find out. First, let's figure out where it crosses the $x$-axis:

$0 = 4x-x^3 = x(4-x^2) = x(2+x)(2-x)$

So, it crosses the axis at $x=-2, x=0, x=2$. We only care about positive $x$-values, and $y(1) >0, y(3)<0$. Hence it is in the 1st quadrant only when $0 < x < 2$. So, we use the average value formula:

\displaystyle \begin{align*}y_{\text{avg}} & = \dfrac{1}{b-a}\int_a^b ydx \\ & = \dfrac{1}{2-0}\int_0^2(4x-x^3)dx \\ & = \dfrac{1}{2}\left[2x^2-\dfrac{x^4}{4}\right]_0^2 \\ & = 2\end{align*}

So, the average height is 2, and 8/4 = 2, so that is the correct answer.

3. ## Re: Average of a variable given the function?

Originally Posted by SlipEternal
Let's find out. First, let's figure out where it crosses the $x$-axis:

$0 = 4x-x^3 = x(4-x^2) = x(2+x)(2-x)$

So, it crosses the axis at $x=-2, x=0, x=2$. We only care about positive $x$-values, and $y(1) >0, y(3)<0$. Hence it is in the 1st quadrant only when $0 < x < 2$. So, we use the average value formula:

\displaystyle \begin{align*}y_{\text{avg}} & = \dfrac{1}{b-a}\int_a^b ydx \\ & = \dfrac{1}{2-0}\int_0^2(4x-x^3)dx \\ & = \dfrac{1}{2}\left[2x^2-\dfrac{x^4}{4}\right]_0^2 \\ & = 2\end{align*}

So, the average height is 2, and 8/4 = 2, so that is the correct answer.
Thank you so much! This makes it so much more clear.