improper integrals: [integral sign] limits from −∞ to 6 ( 10xe^(2x) ) dx?

Answer 1

First, write the integral as a limit:

[a→-∞]lim [a, 6]∫10xe^(2x) dx

Now proceed by parts: let u=5x, du=5 dx, dv=2e^(2x) dx, v=e^(2x):

[a→-∞]lim 5xe^(2x) |[a, 6] - [a, 6]∫5e^(2x) dx
[a→-∞]lim 5xe^(2x) - 5/2 e^(2x) |[a, 6]

Now evaluate the limits of integration:

[a→-∞]lim 30e^12 - 5/2 e^12 - (5ae^(2a) - 5/2 e^(2a))
[a→-∞]lim 55/2 e^12 - (5ae^(2a) - 5/2 e^(2a))

Now simply evaluate the limit:

[a→-∞]lim 55/2 e^12 - (5ae^(2a) - 5/2 e^(2a))
55/2 e^12 - [a→-∞]lim 5ae^(2a) + [a→∞]lim 5/2 e^(2a)
55/2 e^12 - [a→-∞]lim 5a/e^(-2a) + 0

Since the [a→-∞]lim 5a/e^(-2a) is ∞/∞ form, we use L'hopital's rule:

55/2 e^12 - [a→-∞]lim 5/(-2e^(-2a))
55/2 e^12 - 0
55/2 e^12

And we are done.

Answer 2

First use integration-by-parts on 10x e^(2x) dx.

Let u = 5x. So du = 5 dx
Let dv = 2e^(2x)dx. Then v = e^(2x)

uv - ∫ v du =
5xe^(2x) - ∫ e^(2x) 5 dx =
5xe^(2x) - (5/2)e^(2x) + C

Now evaluate this from −∞ to 6
30e^(12) - (5/2)e^(12) - lim x→−∞ [5xe^(2x) - (5/2)e^(2x)]
(55/2)e^(12) - lim x→−∞ [5xe^(2x)] - 0
(55/2)e^(12) + lim x→−∞ [-5x / e^(-2x)]
You can use l'Hopital's to show the limit goes to 0, so you're left with just the first term.