=- ∫ (4-u)sqrt(u) du
=- ∫ (4-u)u^(1/2) du
***= -4∫u^.5du + ∫u^1.5 du
***= -4(u^1.5/1.5) + u^2.5/2.5 + c
***= -(8/3)u^(3/2) + (2/5)u^(5/2)+c
we disagree about sign of second term
=- ∫ 4u^(1/2) - u^(3/2) du
=-4∫ u^(1/2) - u^(3/2) du
=-4((2/3)u^(3/2) - (2/5)u^(5/2))
=-4((2/3)(4-x)^(3/2) - (2/5)(4-x)^(5/2)) + C is my answer.
I am given an integral to solve with given substitution values. I got an answer, but I'm not quite sure if it's correct as an online integral calculator gave a different answer.
∫ x sqrt(4-x) dx
Given that u = 4-x .
In this case, x = 4 - u
du = -dx
Now..
=- ∫ (4-u)sqrt(u) du
=- ∫ (4-u)u^(1/2) du
=- ∫ 4u^(1/2) - u^(3/2) du
=-4∫ u^(1/2) - u^(3/2) du
=-4((2/3)u^(3/2) - (2/5)u^(5/2))
=-4((2/3)(4-x)^(3/2) - (2/5)(4-x)^(5/2)) + C is my answer.
Is there anything wrong with my math during the process? Any help is greatly appreciated!
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