Question

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

Answer 1

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`

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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)`