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)∫ u1/2 du = (1/4)(2/3) u3/2 = (1/6) u3/2
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.....