#f(x) = int xe^(x^2-1)dx - intx^2e^(x)dx#
For the first integral:
#int xe^(x^2-1)dx = intxe^(x^2)*1/edx = 1/eintxe^(x^2)dx#
Let #u = e^(x^2)#. The exponential function rule states that
#[e^g(x)]' = g'(x)e^g(x) => [e^(x^2)]' = 2xe^(x^2)#
In other words, #dx = 1/(2xe^(x^2)) du = 1/(2x) * 1/u du#.
#1/eint xe^(x^2) dx = 1/eint xu * 1/(2xu)du=1/(2e)intdu = 1/(2e)u +C_1 = 1/2 e^(x^2-1)+C_1#
For the second one:
#intx^2e^xdx#
We can solve this by integration by parts.
#intalphabeta' = alphabeta - intalpha'beta#
In our case, #alpha = x^2 => alpha' = 2x# and #beta' =e^x => beta =e^x#. Plug in these functions.
#intx^2e^xdx = x^2e^x - 2color(blue)(intxe^xdx#
In order to solve the #color(blue)("blue")# integral, you can use integration by parts, again.
#alpha_2 =x => alpha_2' = 1#
#beta_2' = e^x => beta_2 = e^x#
#intxe^xdx = xe^x - inte^xdx = xe^x-e^x + C_2 = e^x(x-1) + C_2#
Then
#intx^2e^xdx = x^2e^x -2e^x(x-1) + C_2 = e^x(x^2-2x+2) + C_2#
Therefore, we have:
#f(x) = 1/2 e^(x^2-1) + e^x(2x-2-x^2) + C_1 + C_2#
However, a constant plus a constant is another constant. Let this be #c#.
#C_1 + C_2 = c#
#f(x) = 1/2 e^(x^2-1) + e^x(2x-2-x^2) + c#
We know that #f(color(red)2) = color(blue)4#.
#f(2) = 1/2 e^(color(red)2^2 - 1) + e^color(red)2(2*color(red)2-2-color(red)2^2) + c = 1/2 e^3 - 2e^2 + c = color(blue)4#
Therefore, #c = 4- e^2(e/2 + 2)#.
Finally,
#color(red)(f(x)) = 1/2 e^(x^2-1) + e^x(2x-2-x^2) + 4 - 1/2e^3 -2e^2#
