So I need to integrate x(Sqrt[13-x^2]
I understand that the answer is -1/3 (13-x^2)^3/2
I just don't understand the steps involved to solve it.
I understand that the answer is -1/3 (13-x^2)^3/2
I just don't understand the steps involved to solve it.
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∫ [x][sqrt(13 - x^2)] dx
Let u = 13 - x^2
du = -2x dx
dx = -1/(2x) du
Sub that back in and that x's will cancel, leaving the -1/2 to be factored out:
(-1/2) ∫ sqrt(u) du
(-1/2) ∫ u^(1/2) du
(-1/2)(2/3)u^(3/2)
(-1/3)u^(3/2)
(-1/3)[13 - x^2]^(3/2) + C
Done!
Let u = 13 - x^2
du = -2x dx
dx = -1/(2x) du
Sub that back in and that x's will cancel, leaving the -1/2 to be factored out:
(-1/2) ∫ sqrt(u) du
(-1/2) ∫ u^(1/2) du
(-1/2)(2/3)u^(3/2)
(-1/3)u^(3/2)
(-1/3)[13 - x^2]^(3/2) + C
Done!