What is #f(x) = int xe^(x+2)-x dx# if #f(-1) = 2 #?

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Answer 1 · 2016-08-22T16:31:53

#f(x)=xe^(x+2)-e^(x+2)-x^2/2+5/2+2e#.

#f(x)=int(xe^(x+2)-x)dx=intxe^(x+2)dx-intxdx=I-x^2/2#, where,

#I=intxe^(x+2)dx#.

To evaluate #I#, we use the Rule of Integration by Parts :

#intuvdx=uintvdx-int[(du)/dxintvdx]dx#.

We take, #u=x rArr (du)/dx=1#, and,

#v=e^(x+2) rArr intvdx=e^(x+2)#. Hence,

#I=xe^(x+2)-inte^(x+2)dx=xe^(x+2)-e^(x+2)#

#:. f(x)=xe^(x+2)-e^(x+2)-x^2/2+C...........(1)#.

To determine #C#, we use the cond. #: f(-1)=2# in #(1)#.

#:. -e-e-1/2+C=2 rArr C=5/2+2e#. Therefore, #(1)# gives,

#f(x)=xe^(x+2)-e^(x+2)-x^2/2+5/2+2e#.

Enjoy Maths.!