Find the definite integral from 0 to 1 ==> ( x ^ 2 ) * ( e ^ x ) dx?

Answer 1

Use integration by parts, twice. The first time, let u = x^2 and dv = e^x dx; the second time, let u = 2x and let dv = e^x dx.

First time:
∫x^2e^x dx = x^2e^x - ∫2xe^x dx

Second time:
x^2e^x - ∫2xe^x dx = x^2e^x - (2xe^x - ∫2e^x)

Simplify/take antiderivative:
x^2e^x - (2xe^x - ∫2e^x) = x^2e^x - 2xe^x +∫2e^x = x^2e^x - 2xe^x + 2e^x = (x^2 - 2x + 2)e^x.

Now evaluate from 0 to 1.

(1 - 2 + 2)e^1 - (0 - 0 + 2)e^0 = e - 2.

Answer 2

[0, 1]∫x²e^x dx

Proceed with integration by parts: u=x², du=2x dx, v=e^x, dv = e^x dx.

x²e^x | [0, 1] - [0, 1]∫2xe^x dx
e - [0, 1]∫2xe^x

Integrate by parts again: u=2x, du=2 dx, v=e^x, dv=e^x dx.

e - (2xe^x | [0, 1] - [0, 1]∫2e^x dx)
e - 2e + [0, 1]∫2e^x dx

Integrating directly:

e - 2e + 2e^x | [0, 1]
e - 2e + 2e - 2

Simplifying:

e-2

Answer 3

Are you making people do your homework for you?

Answer 4

no idea. google it.