Question

How do you find the average value of `f(x)=x/sqrt(x^2+1), 0<=x<=4`?

Answer 1

The average value of `f(x)` on `0 ≤ x ≤ 4` is `1/4(sqrt(17) - 1)`, which is approximately `0.78`

Explanation:

The average value of a continuous function `f(x)` on `[a, b]`, is given by the formula `A = 1/(b - a) int_a^b f(x) dx`.

`A = 1/(4 - 0)int_0^4 x/sqrt(x^2 + 1)dx`

`A = 1/4int_0^4 x/sqrt(x^2 +1)dx`

This now becomes a trig substitution problem. Let `x = tantheta`. Then `dx = sec^2theta d theta`. Adjust the bounds of integration accordingly.

`A = 1/4int_0^(tan4) tantheta/sqrt(tan^2theta + 1) * sec^2theta d theta`

Use the pythagorean identity `tan^2alpha + 1 = sec^2alpha`.

`A = 1/4int_0^(tan4) tantheta/sqrt(sec^2theta) * sec^2theta d theta`

`A = 1/4int_0^(tan4) tantheta/sectheta * sec^2theta`

`A = 1/4int_0^(tan4) tanthetasectheta`

This is a known integral.

`A = 1/4[sectheta]_0^(tan4)`

We know from our initial substitution that `x/1 = tantheta`. This means that `sqrt(x^2 + 1)` is the hypotenuse. That's to say `sectheta = sqrt(x^2 + 1)`.

`A = 1/4[sqrt(x^2 +1)]_0^4`

Evaluate this using the 2nd fundamental theorem of calculus.

`A = 1/4(sqrt((4)^2 + 1) - sqrt(0^2 + 1))`

`A = 1/4sqrt(17) - 1/4`

`A = 1/4(sqrt(17) - 1)`

Hopefully this helps!

Answer 2

Alternative to get `1/4(sqrt17-1)`:

Explanation:

Like HSBC244 said, The average value of a continuous function `f(x)` on `[a, b]`, is given by the formula

`A = 1/(b - a) int_a^b f(x) dx`.

`A = 1/(4 - 0)int_0^4 x/sqrt(x^2 + 1)dx`

`A = 1/4int_0^4 x/sqrt(x^2 +1)dx`

Instead of doing a trig substitution like HSBC244, you can do a u substitution instead. Set `u=x^2+1` which makes `du=2x`. Multiply the integral by `2/2` to get the 2 sufficient for the u substitution and put the 2 in the denominator on the outside.

`A = 1/8int_1^17 1/sqrt(u)du`

Note that the limits of integration changed because we were doing a u substitution. 0 became 1 because:

`0^2+1=1`

Similarly plugging in 4 became 17 because:

`4^2+1=17`

Now rewrite it so it's easier to integrate by treating the square root in the denominator:

`A = 1/8int_1^17 u^(1/2)`

This integral becomes

`A = 1/8[2sqrt(u)]_1^17`

Take the 2 out and plug in 17 and 1. As you can see, this will give you the same answer as HSBC244's answer but does not require a trig substitution.