Asked by Michael

Evaluate the integral from 0 to 24 of the square root of 9+3x dx.
Answered by Steve
if you let u = 9+3x, du = 3dx and you have

∫√(9+3x) dx = ∫√u (du/3) = 1/3 ∫u^(1/2) du

take it from there.
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