What is #f(x) = int x^2e^(2x-1)-x^3e^x dx# if #f(2) = 7 #?

I got:

#(x^2/2 - x/2 + 1/4)e^(2x - 1) - (x^3 - 3x^2 + 6x - 6)e^x - 5/4e^3 + 2e^2 + 7#

This has several integration by parts. But we do have access to an integral table.

#int x^me^(ax)dx = 1/a x^me^(ax) - m/aint x^(m-1)e^(ax)dx# for #m >= 2#
#int xe^(ax)dx = 1/a^2e^(ax)(ax - 1) + C#

So the first integral should give:

#int x^2e^(2x-1)dx#

#= 1/2 x^2e^(2x-1) - int xe^(2x-1)dx#

#= x^2/2e^(2x-1) - 1/4 e^(2x-1)(2x-1)#

#= x^2/2e^(2x-1) - x/2 e^(2x-1) + 1/4e^(2x - 1)#

#= (x^2/2 - x/2 + 1/4)e^(2x - 1)#

and the second integral should give:

#int x^3e^xdx#

#= x^3e^x - 3intx^2e^xdx#

#= x^3e^x - 3(x^2e^x - 2int xe^xdx)#

#= x^3e^x - 3[x^2e^x - 2(e^x(x-1))]#

#= x^3e^x - 3x^2e^x + 6xe^x - 6e^x#

#= (x^3 - 3x^2 + 6x - 6)e^x#

These two integral solutions combine to give:

#color(green)((x^2/2 - x/2 + 1/4)e^(2x - 1) - (x^3 - 3x^2 + 6x - 6)e^x + C)#

Further, you know that #f(2) = 7#, so set this equation equal to #7# at #x = 2#:

#((2)^2/2 - (2)/2 + 1/4)e^(2(2)-1) - ((2)^3 - 3(2)^2 + 6(2) - 6)e^(2) + C = 7#

#(2 - 1 + 1/4)e^3 - (8 - 12 + 12 - 6)e^2 + C = 7#

#5/4e^3 - 2e^2 + C = 7#

Thus, your integration constant is #color(green)(C = -5/4e^3 + 2e^2 + 7)#, and your answer overall becomes:

#color(blue)(int x^2e^(2x-1) - x^3e^xdx)#

#= color(blue)((x^2/2 - x/2 + 1/4)e^(2x - 1) - (x^3 - 3x^2 + 6x - 6)e^x - 5/4e^3 + 2e^2 + 7)#