# Find the average value of the function on the given interval. `g(t) = t/(sqrt(3 + t^2))` [1,3]

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We will apply the formula:

`g_(avg) = 1/(b-a) int_a^b g(t)dt`

Therefore,` a = 1` and b` = 3` .

We will have:

`g_avg = 1/(3-1) int_1^3 t/sqrt(3 + t^2)dt`

Let us set `u = sqrt(3 + t^2)` or `(3 + t^2)^(1/2)`

Differentiating both sides,

`du = 1/2(3 + t^2)^(1/2 - 1)*(2t)`

`= 1/2(3 + t^2)^(-1/2)*(2t)`

`= (2t)/(2sqrt(3 + t^2))`

`du = (t)/(sqrt(3 + t^2))dt`

Therefore, we can replace the `(t)/(sqrt(3 + t^2))dt` by `du` .

`g_(avg) = 1/(3-1) int_1^3 t/sqrt(3 + t^2)dt`

`= 1/(2)int_1^3 du `

Integrating that we will have:

`1/2int_1^3du= 1/2 u|_1^3`

`=1/2sqrt(3 + t^2)`

Plug-in the limits,

`1/2((sqrt(3 + 3^2) - sqrt(3 + 1^2))= 1/2((sqrt(3 + 9) - sqrt(3 + 1))`

`= 1/2(sqrt(12) - sqrt(4))`

Extracting the square roots.

`g_(avg) = 1/2(sqrt(12) - sqrt(4))`

`= 1/2(2sqrt(3) - 2)`

`= (2sqrt(3) - 2)/(2)`

Factoring out the `2` from the top,

`g_(avg) = (2(sqrt(3) - 1))/(2)`

Cancel out common factor on top and bottom.

`g_(avg) =` `sqrt(3) - 1`

That is it! :)

What does this result actually mean from a geometric standpoint?

Remember that the mean value theorem for integrals was used to create this result. The area under the curve g(t) from t = 1 to 3 is equal to the area of the rectangle whose base is 2 and height is g(avg). If you graph all of this on graph paper, you will be able to really get a good understanding of what I am referring to.

I have provided a link to a video tutorial which shows a relevant example to further illuminate this concept.

This is really great math!