Question

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

Answer 1

0

Explanation:

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

F(x) = `int f(x) 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.

Answer 2

`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(x)` on the interval `[a,b]`, we can express it in general terms like this:
`(int_a^b f(x)\ dx)/(b-a)`

If we plug in our function, we get:
`(int_-3^3 2xsqrt(1+x^2)\ dx)/(3-(-3))`

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`, `2x`, 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 `2x`, so we can just divide through by `2x`:
`int\ 2xsqrt(1+x^2)\ dx=int\ (cancel(2x)sqrt(1+x^2))/cancel(2x)\ du=int\ sqrtu\ du`

By knowing `sqrt(u)=u^(1/2)`, we can use the power rule:
`int\ sqrtu\ du=int\ u^(1/2)\ du=u^(3/2)/(3/2)=2/3u^(3/2)=2/3(1+x^2)^(3/2)`

Now we can evaluate the definite integral:
`int_-3^3 2xsqrt(1+x^2)\ dx=[2/3(1+x^2)^(3/2)]_-3^3`

`=2/3 (1+3^2)^(3/2)-2/3(1+(-3)^2)^(3/2)=0`

Now we divide by `3-(-3)=6`, which gives:
`0/6=0`

So, the average value of the function is `0` on the interval `[-3,3]`.

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 `[-a,a]` will be equal to `0`.