# How do you find the average value of the function for f(x)=2xsqrt(1+x^2), -3<=x<=3?

Nov 30, 2017

0

#### Explanation:

To find the average, we take the integral and divide through the length of the interval.

F(x) = $\int f \left(x\right) \mathrm{dx}$ = int sqrt(1+x²) d(1+x²)
= (2/3) (1+x²)^(3/2)

Now evaluate between -3 and 3 :

F(3) - F(-3) = 0

So 0/6 = 0.

Nov 30, 2017

$0$

#### Explanation:

The average value of a function over an interval is equal to the definite integral of that interval divided by the length of the interval. If we wanted to find the average value of $f \left(x\right)$ on the interval $\left[a , b\right]$, we can express it in general terms like this:
$\frac{{\int}_{a}^{b} f \left(x\right) \setminus \mathrm{dx}}{b - a}$

If we plug in our function, we get:
$\frac{{\int}_{-} {3}^{3} 2 x \sqrt{1 + {x}^{2}} \setminus \mathrm{dx}}{3 - \left(- 3\right)}$

Let's first start by computing the anti-derivative of the function. We can quite quickly see that we have the derivative of $1 + {x}^{2}$, $2 x$, on the outside of the square root. This is a tell-tale sign that we can use u-substitution with $u = 1 + {x}^{2}$. We knew that the derivative was $2 x$, so we can just divide through by $2 x$:
$\int \setminus 2 x \sqrt{1 + {x}^{2}} \setminus \mathrm{dx} = \int \setminus \frac{\cancel{2 x} \sqrt{1 + {x}^{2}}}{\cancel{2 x}} \setminus \mathrm{du} = \int \setminus \sqrt{u} \setminus \mathrm{du}$

By knowing $\sqrt{u} = {u}^{\frac{1}{2}}$, we can use the power rule:
$\int \setminus \sqrt{u} \setminus \mathrm{du} = \int \setminus {u}^{\frac{1}{2}} \setminus \mathrm{du} = {u}^{\frac{3}{2}} / \left(\frac{3}{2}\right) = \frac{2}{3} {u}^{\frac{3}{2}} = \frac{2}{3} {\left(1 + {x}^{2}\right)}^{\frac{3}{2}}$

Now we can evaluate the definite integral:
${\int}_{-} {3}^{3} 2 x \sqrt{1 + {x}^{2}} \setminus \mathrm{dx} = {\left[\frac{2}{3} {\left(1 + {x}^{2}\right)}^{\frac{3}{2}}\right]}_{-} {3}^{3}$

$= \frac{2}{3} {\left(1 + {3}^{2}\right)}^{\frac{3}{2}} - \frac{2}{3} {\left(1 + {\left(- 3\right)}^{2}\right)}^{\frac{3}{2}} = 0$

Now we divide by $3 - \left(- 3\right) = 6$, which gives:
$\frac{0}{6} = 0$

So, the average value of the function is $0$ on the interval $\left[- 3 , 3\right]$.

In fact, since the function is mirrored upside down on the other side of the x-axis you can say that for any real number $a$, the average value on the interval $\left[- a , a\right]$ will be equal to $0$.