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

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Answer 1 · 2016-06-04T02:21:25

#f(x)=xe^x-e^x-5/2x^2+2e^(-1)+15/2#

The given

#int (xe^x-5x) dx# and #f(-1)=5#

Integration by parts for the #int (xe^x) dx#

Let #u=x# and #dv=e^x*dx# and #v=e^x# and #du=dx#

Using the integration by parts formula

#int u*dv=uv-int v*du#

#int x*e^x*dx=xe^x-int e^x*dx#

#int x*e^x*dx=xe^x- e^x#

Therefore

#f(x)=int (xe^x-5x) dx=xe^x- e^x-5/2x^2+C#

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

At #f(-1)=5#

#f(-1)=5=-1*e^(-1)- e^(-1)-5/2(-1)^2+C#

#5=-1*e^(-1)- e^(-1)-5/2(-1)^2+C#

#5=-2*e^(-1)-5/2+C#

#C=15/2+2e^(-1)#

finally

#f(x)=xe^x- e^x-5/2x^2+15/2+2e^(-1)#

God bless....I hope the explanation is useful.