What is #f(x) = int xe^(x-1)-x^2e^-x dx# if #f(2) = 7 #?

1 Answer
Jun 27, 2018

#f(x)=e^(x-1)(x-1)+e^-x(x^2+2x+2)+7-e-10/e^2#

Explanation:

#intxe^(x-1)dx-intx^2e^-xdx#

Let us now deal with only the first integral

#intxe^(x-1)dx# Let #u=x-1# then #du=dx#

#int(u+1)e^udu#

Let #w=u+1#, #dw=du # , # dv=e^u du#, #v=e^u#

#(u+1)e^u-inte^udu#

#(u+1)e^u-e^u+c_1#

Substitute back in

#xe^(x-1)-e^(x-1)+c_1#

Now move onto the second integral

#-intx^2e^-xdx#

Let #u=-x# , #u^2=x^2# , #-du=dx#

#intu^2e^udu#

Let #w=u^2 , dw=2udu , dv=e^udu , v=e^u#

#u^2e^u-2intue^udu#

Let #w=u , dw=du , dv=e^udu , v=e^u#

#u^2e^u-2(ue^u-inte^udu)#

#u^2e^u-2ue^u+2e^u+c_2#

Substitute

#x^2e^-x+2xe^-x+2e^-x+c_2#

Combine both Integrals

#xe^(x-1)-e^(x-1)+c_1+ x^2e^-x+2xe^-x+2e^-x+c_2#

#f(x)=e^(x-1)(x-1)+e^-x(x^2+2x+2)+C#

Solve for C

#7=e+e^-2(2^2+2(2)+2)+C#

#7=e+10e^-2+C#

#C=7-e-10/e^2#

#f(x)=e^(x-1)(x-1)+e^-x(x^2+2x+2)+7-e-10/e^2#