Change the order of integration and solve. Best of luck.

Question

∫₀⁴∫₀^√y e^{12x-x^3}dxdy

If you can't read it, the inner integral is from 0->√y

Outermost integral is from 0->4

I believe I did the change of integration correctly, but actually evaluating that integral is tough.

∫_-2²∫⁴_x^2 e^{12x-x^3}dydx

Answer 1

I don't think you paid enough attention to what you got after the first step

Note that x is always greater than or equal to 0, so the range will need to be from 0 to 2 rather than -2 to 2.

Using I[a,b] for the integral from a to b and E[a, b] for the evaluation from a to b,

I[0, 2] I[x^2, 4] e^(12x-x^3) dy dx =

I[0, 2] e^(12x-x^3) y E[x^2, 4] dx =

I[0, 2] e^(12x-x^3) (4 - x^2) dx

Note that (12x - x^3) ' = 12 - 3x^2 = 3(4 - x^2)

Thus,

I[0, 2] e^(12x-x^3) (4 - x^2) dx = 1/3 e^(12x-x^3) E[0, 2] = (e^16 - 1)/3