Calc 2
posted by Anonymous .
Evaluate the definite integral from 0 to 1 of
(x^2)sqrt(5x+6)dx

Calc 2 
Steve
∫[0,1] x^2 √(5x+6) dx
That's almost x^5/2, but for that pesky 5x+6. So, let u=5x+6 and we have x = (u6)/5:
∫[6,11] (u6)^2/25 √u du/5
= 1/125 ∫[6,11] u^(5/2)  12u^(3/2) + 36u^(1/2) du
= 1/125 (2/7 u^7/2  24/5 u^5/2 + 24u^3/2)[6,11]
= 2/4375 (576√6  1111√11)
= 1.03948
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