How do I solve this?

#int_0^1sqrt(xe)^sqrtx#

1 Answer
May 19, 2018

See below

Explanation:

Try change of variable #sqrtx=t# or equivalent #x=t^2#, then

#dx=2tdt# with this, our integral changes to

#I=int2t^2e^tdt=2intt^2e^tdt#

Lets do by parts #u=t^2# and #dv=e^tdt#, then #du=2tdt# and #v=e^t#

#I=2(t^2e^t-int2te^tdt)=2t^2e^t-4intte^tdt#. again by parts

#u=t# and #dv=e^tdt#. Then we have #du=dt# and #v=e^t#

#I=2t^2e^t-4(te^t-inte^tdt)=2t^2e^t-4te^t+4e^t=#

#=(2xe^(sqrtx)-4sqrtxe^(sqrtx)+4e^sqrtx)=#

#=(e^(sqrtx)(2x-4sqrtx+4))_0^1=-2e#