Verified answer

let u = 1 - x^2

=> du/dx = -2x

=> dx = - du / 2x

so the INT becomes :

∫ x e^u (-du / 2x)

= -1/2 ∫ e^u du = -1/2 e^u + C

= -1/2 e^(1-x^2) [from 2 to -1]

= -1/2 [e^0 - e^-3]

= -1/2 [ 1 - 0.05]

= - 0.475

x is the derivative of x^2 (times a constant), so an obvious substitution is u = (1 - x^2), du = -2x dx

Then x dx = (-1/2)du and the integral is (-1/2) e^u du which integrates to (-1/2)e^u.

2

∫ x e^(1 - x ²) dx

-1

Let :-

u = 1 - x ²

du = - 2x dx

- du/2 = x dx

When x = - 1 , u = 0

When x = 2 , u = - 3

-3

-1/2 ∫ e^u du

0

0

1/2 ∫ e^u du

-3

(1/2) [ e^u ]-------------limits - 3 to 0

(1/2) [ 1 - 1/e³ ]

let u = 1-x^2

du/dx = -2x

dx = du/-2x

∫ (x e^(1-x^2) dx = ∫ (x e^(u) du/-2x = ∫ -e^(u)/2du = -1/2∫ e^(u)du

= -(1/2)*e^(1-x^2) + C

You can plug in the limits of integration.