What is f(x) = int xe^(x^2-1)-x^2e^x dx if f(2) = 4 ?

1 Answer
Apr 5, 2018

f(x)= 1/2 e^(x^2-1) + e^x(2x-2-x^2) + 4 - 1/2e^3 -2e^2

Explanation:

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