Asked by Anon
Evaluate
integral of x times square root of (2x-1)dx
integral of x times square root of (2x-1)dx
Answers
Answered by
Damon
x (2x-1)^.5 dx
u = x
dv = (2x-1)^.5 dx
du = dx
v = (1/3)(2x-1)^1.5
so
(x/3)(2x-1)^(1.5) -(1/3) (2x-1)^1.5 dx
(x/3)(2x-1)^1.5-(1/3)(1/5)(2x-1)^2.5
(1/3)[ x(2x-1)^1.5 - (1/5)(2x-1)^2.5 ]
u = x
dv = (2x-1)^.5 dx
du = dx
v = (1/3)(2x-1)^1.5
so
(x/3)(2x-1)^(1.5) -(1/3) (2x-1)^1.5 dx
(x/3)(2x-1)^1.5-(1/3)(1/5)(2x-1)^2.5
(1/3)[ x(2x-1)^1.5 - (1/5)(2x-1)^2.5 ]
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