Asked by Amy
The integral from 1 to 2 of x*the square root of (x-1) dx.
Using the substitution method, I said that u = x-1, du = 1 dx, and dx = du. I'm not sure how to plug u back into the equation since I still have that first x. Suggestions are greatly appreciated.
Using the substitution method, I said that u = x-1, du = 1 dx, and dx = du. I'm not sure how to plug u back into the equation since I still have that first x. Suggestions are greatly appreciated.
Answers
Answered by
bobpursley
Let u^2=x-1
2udu=dx
INT (u^2+1)u 2udu=
INT (2u^4+2U^2) du
revise your limits.
2udu=dx
INT (u^2+1)u 2udu=
INT (2u^4+2U^2) du
revise your limits.
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