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Math Help - Average of a variable given the function?

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

    Can you help me with this: Average of a variable given the function?-capture3.jpgI 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!
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  2. #2
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    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.
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  3. #3
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    Re: Average of a variable given the function?

    Quote Originally Posted by SlipEternal View Post
    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.
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