What is #int_(0)^(4) xe^-xdx #?

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Answer 1 · 2016-05-11T00:36:36

#int_0^4xe^(-x)dx=1 -5/e^4#

We will proceed using integration by parts .

Let #u = x# and #dv = e^(-x)dx#
Then #du = dx# and #v = -e^(-x)#

Applying the formula #int_a^budv = (uv)_a^b - int_a^bvdu#

#int_0^4xe^(-x)dx = (-xe^(-x))_0^4 - int_0^4-e^-xdx#

#=(-4e^(-4) +0e^(0)) - (e^(-x))_0^4#

#=-4e^-4 - (e^-4 - e^0)#

#=-4e^-4 - e^-4 + 1#

#=1 -5/e^4#