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Find the average value of the following function:

$p(x)  =  2x^2 + 5x + 2$

on the interval  $1  \le  x  \le  3$.

I know that I need to find $u$, $du$, $v$, and $dv$ and set it up into an definite integral but I don't know what to make them to sent up the equation and find the answer. How do you know what to make them? After I come up with the equation I believe I can solve it.

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As seen here, the average value is defined by $$\frac{1}{3-1}\int_{1}^{3} 2x^2+5x+2 \; \mathrm{d}x$$ And there is an nice explanaton for why we define it as such.

I think you can integrate this function. Can you continue from here?

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  • $\begingroup$ I believe so, but just to be sure, do I add leave the 1/2 until I solve the definite integral? $\endgroup$ – Michelle Feb 13 '17 at 1:26
  • $\begingroup$ @Michelle You leave the $\frac{1}{2}$. $\endgroup$ – S.C.B. Feb 13 '17 at 1:40
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The average value $A$ of a function $p$ over the interval $(a,b)$ is given by \begin{align*} A = \frac{1}{b-a}\int_a^b p(x)dx \end{align*} What are your $a$ and $b$ in this situation?

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You find the average of a function via applying the following formula:

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

$\frac {1}{2} \int_{1}^{3} 2x^2+5x+2 dx$

Wolfram says that the integral returns $124/3$, so

$\frac {1}{2}* \frac {124}{3} = 20.667$

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