What is the average value of y=x^3*sqrt(x^4+9) on the interval [0,2]?

So we have

∫x³ √(x⁴ + 9) dx

Let u = x⁴ + 9      du = 4x³ dx   ⇒   du/4 = x³ dx

I'll just continue in terms of "u" instead of "x," so that we don't have to "back-substitute."

So when x = 0, u = 9        And when x = 2, u = 25

So this gives us

(1/4)∫ u^1/2 du = (1/4)(2/3) u^3/2  = (1/6) u^3/2

And evaluating this from u = 9 to u = 25  gives us

(1/6)(25^(3/2)-9^(3/2))  =  49/3

And to get the average value, we divide this by the interval width, i.e., 2. 

This gives us......... 49/6

Note how the "average value" turns a "non-rectangular" area into a "rectangular area!!"

The "width" of the rectangle (2)  times the "height" (49/6) = 49/3 !!

And I think that's it.....