What is #f(x) = int xe^(2-x) + 3x^2 dx# if #f(0 ) = 1 #?

1 Answer
Ken C.
Jul 7, 2016

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

Explanation:

Begin by using the sum rule for integrals and splitting these into two separate integrals:
#intxe^(2-x)dx+int3x^2dx#

The first of these mini-integrals is solved using integration by parts:
Let #u=x->(du)/dx=1->du=dx#
#dv=e^(2-x)dx->intdv=inte^(2-x)dx->v=-e^(2-x)#

Now using the integration by parts formula #intudv=uv-intvdu#, we have:
#intxe^(2-x)dx=(x)(-e^(2-x))-int(-e^(2-x))dx#
#=-xe^(2-x)+inte^(2-x)dx#
#=-xe^(2-x)-e^(2-x)#

The second of these is a case of the reverse power rule, which states:
#intx^ndx=(x^(n+1))/(n+1)#

So #int3x^2dx=3((x^(2+1))/(2+1))=3(x^3/3)=x^3#

Therefore, #intxe^(2-x)+3x^2dx=-xe^(2-x)-e^(2-x)+x^3+C# (remember to add the constant of integration!)

We are given the initial condition #f(0)=1#, so:
#1=-(0)e^(2-(0))-e^(2-(0))+(0)^3+C#
#1=-e^2+C#
#C=1+e^2#

Making this final substitution, we obtain our final solution:
#intxe^(2-x)+3x^2dx=-xe^(2-x)-e^(2-x)+x^3+1+e^2#

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