How do you evaluate #int y = x(e^(-x))# from 0 to 2?

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Answer 1 · 2017-02-22T03:02:42

#int_0^2xe^-xdx=(e^2-3)/e^2#

I believe you're asking for the integral:

#int_0^2xe^-xdx#

To do this, first find the antiderivative without the bounds. Do this through integration by parts, which takes the form #intudv=uv-intvdu#.

For #intxe^-xdx#, let:

#{(u=x,=>,du=dx),(dv=e^-xdx,=>,v=-e^-x):}#

Note that going from #dv# to #v# means finding #inte^-xdx#, which takes the substitution #t=-x#.

Through the IBP formula, we see that:

#intxe^-xdx=uv-intvdu=-xe^-x+inte^-xdx#

Which is an integral we previously found:

#intxe^-xdx=-xe^-x-e^-x=-e^-x(x+1)#

Then, with bounds:

#int_0^2xe^-xdx=[-e^-x(x+1)]_0^2#

#=-e^-2(2+1)-(-e^-0(0+1))#

#=-3e^-2+1#

#=(e^2-3)/e^2#