# Find the average value of $2x^2 + 5x + 2$ on the interval where $x \in [1,3]$.

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.

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?

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?

• I believe so, but just to be sure, do I add leave the 1/2 until I solve the definite integral? Feb 13, 2017 at 1:26
• @Michelle You leave the $\frac{1}{2}$. Feb 13, 2017 at 1:40

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

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