What is #f(x) = int xe^(2-x) -2 x^2 dx# if #f(0 ) = 1 #?

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Answer 1 · 2018-03-22T13:33:11

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

#f(x)=int(xe^(2-x)-2x^2)dx#

and #f(0)=1#

#f(x)=int(xe^(2-x)-2x^x2)dx=color(red)(int(xe^(2-x))dx)-int2x^2dx#

the second integral is integrated using the power rule; the first integral is done by integrating by parts#

#I_1=int2x^2dx=2/3x^3#

#I_2=int(xe^(2-x))dx#

#IBP formula

#I_2=intu(dv)/(dx)dx=uv-intv(du)/(dx)dx#

#u=x=>du=dx#

#(dv)/(dx)=e^(2-x)=>v=-e^(2-x)#

#:. I_2=-xe^(2-x)-int(-e^(2-x))dx#

#I_2=-xe^(2-x)-e^(2-x)#

#:. f(x)=I_2-I_1#

#f(x)=-xe^(2-x)-e^(2-x)-2/3x^3+c#

#f(0)=1#

#=>1=0-e^2-0+c#

#:. c=1+e^2#

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

which can be simplified as required