`int_1^4 ((4 + 6u)/sqrt(u)) du` Evaluate the integral

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Chapter 5, 5.4 - Problem 29 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`int_1^4(4+6u)/sqrt(u)du`

`=int_1^4(4/sqrt(u)+(6u)/sqrt(u))du`

`=int_1^4(4u^(-1/2)+6u^(1/2))du`

`=[4(u^(-1/2+1)/(-1/2+1))+6(u^(1/2+1)/(1/2+1))]_1^4`

`=[4(u^(1/2)/(1/2))+6(u^(3/2)/(3/2))]_1^4`

`=[8u^(1/2)+6*2/3u^(3/2)]_1^4`

`=[8(4)^(1/2)+4(4)^(3/2)]-[8(1)^(1/2)+4(1)^(3/2)]`

`=(8(2^2)^(1/2)+4(2^2)^(3/2))-(8+4)`

`=(16+4(2^3))-12`

`=(16+32)-12`

=36

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