integral of xe^-x dx
Using integration by parts by letting u=x and dv/dx=e^-x. this implies that du/dx=1 and v=-e^-xThe By Parts formulae is u.v - integral(v.du/dx) = -xe^-x - integral(-e^-x).1 dx = -xe^-x + integral(e^-x) dx = -xe^-x -e^-x +c (where c is the constant of integration.)
BK
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