What is #f(x) = int x^2e^x-x^2 dx# if #f(-1) = 5 #?

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Answer 1 · 2017-10-29T03:09:33

#f(x)=(x^2-2x+2)*e^x-(x^3)/3+14/3-5e^(-1)#

#f(x)=int x^2*e^x*dx#-#int x^2*dx#

=#(x^2-2x+2)*e^x-(x^3)/3+C#

After imposing #f(-1)=5# condition,

#5e^(-1)+1/3+C=5#

Hence #C=14/3-5e^(-1)#

Thus, #f(x)=(x^2-2x+2)*e^x-(x^3)/3+14/3-5e^(-1)#

Note: I used tabular integration for first integral.