Question
Evaluate the definite integral from 0 to 1 of
(x^2)sqrt(5x+6)dx
(x^2)sqrt(5x+6)dx
Answers
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 = (u-6)/5:
∫[6,11] (u-6)^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
That's almost x^5/2, but for that pesky 5x+6. So, let u=5x+6 and we have x = (u-6)/5:
∫[6,11] (u-6)^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