What is #f(x) = int xe^(2-x) -2 x^2 +2 dx# if #f(0 ) = 2 #?

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Answer 1 · 2018-08-12T00:46:20

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

#=intxe^(2-x)dx-2intx^2dx+2int1dx#

#=intxe^(2-x)dx-2/3x^3+2x#

Let #X=2-x#

#x=2-X#

#dx=-1dX#

So:

#I=-int(2-X)e^Xdx-2/3x^3+2x#

Using integration by parts:
#f(X)=(2-X)#, #f'(X)=-1#, #g'(X)=e^X#, #g(X)=e^X#

So:

#I=-(2-X)e^X+int-e^XdX-2/3x^3+2x#

#I=(X-2)e^X-e^X-2/3x^3+2x#

#I=-xe^(2-x)-e^(2-x)-2/3x^3+2x+c#, #c in RR#

And knowing that #f(0)=2#,

#c-e^2=2#

So:

#I=-(x+1)e^(2-x)-2/3x^3+2x+2+e^2#