Need some help finding the definite integral of xe^(-x) on [1,infinity]?

Answer 1

f(x) = -xe^(-x) - e^(-x)

f'(x) = -e^(-x) +xe^(-x) +e^(-x) = xe^(-x)

So value f(x) over the interval [1,infinity)
f(1) = -2e(-1) = -2/e
f(infinity) = 0?
answer is 2/e

my head hurts

Answer 2

∞
∫ (x e^(-x) dx )
1

First off, since this is an improper integral, we *have* to express it as a limit.

lim ∫ (1 to t, x e^(-x) dx )
t → ∞

Now, we use integration by parts.

Let u = x. dv = e^(-x) dx
du = dx. v = -e^(-x)

lim [ -xe^(-x) - ∫(-e^(-x) dx ) ]
t → ∞

lim [ -xe^(-x) + ∫(e^(-x) dx ) ]
t → ∞

lim [ -xe^(-x) - e^(-x) ]{evaluated from 1 to t}
t → ∞

Evaluate the limit at the bounds.

lim { [ -te^(-t) - e^(-t) ] - [ -(1)e^(-1) - e^(-1) ] }
t → ∞

The next few steps will be simplifications (grouping like terms, cancellations, etc..)

lim { [ -te^(-t) - e^(-t) ] - [ -e^(-1) - e^(-1) ] }
t → ∞

lim [ -te^(-t) - e^(-t) ] + e^(-1) + e^(-1) ]
t → ∞

lim [ -te^(-t) - e^(-t) + 2e^(-1)]
t → ∞

The only troublesome term is the first one, because evaluating it at infinity yields the form [∞ * 0]. Let's change that by moving e^(-t) to the denominator.

lim [ (-t/e^t) - e^(-t) + 2e^(-1)]
t → ∞

Let's evaluate the other two terms, to get them out of the way. As t approaches infinity, e^(-t) approaches 0. Also,
As t approaches a constant, the constant is just itself. Therefore, we have

lim [ (-t/e^t) ] - 0 + 2e^(-1)
t → ∞


lim [ (-t/e^t) ] + 2e^(-1)
t → ∞

Applying L'Hospital's rule, we get

lim [ (-1/e^t ] + 2e^(-1)
t → ∞

Evaluating the limit gives us the form [1/∞], which is 0, so our final answer is

0 + 2e^(-1)

2e^(-1)

2/e