# Improper Integral

1. Mar 27, 2008

### RyanSchw

Improper Integral [Solved]

1. The problem statement, all variables and given/known data

$$\int_{0}^{\infty} (x-1)e^{-x}dx$$

2. Relevant equations

Integration by Parts
Improper Integrals

3. The attempt at a solution

$$\lim_{R\rightarrow \infty} \int_0^R~xe^{-x}-e^{-x}dx$$

Let u = x
du = dx

Let dv = e^-x
v = -e^-x

$$-xe^{-x} - \int -e^{-x}dx$$

$$-xe^{-x}-e^{-x} - \int e^{-x}dx$$

$$= -xe^{-x}$$

$$\lim_{R\rightarrow \infty}-xe^{-x} \mid_{0}^{R}=0$$

Now there is clearly area when I look at the graph, but the graph intersects the x-axis at x=1 not x=0 I’m wondering if this is some sort of trick with the bounds of integration that I don’t see or if I made a mistake with my integration.

Thanks

Last edited: Mar 27, 2008
2. Mar 27, 2008

### Dick

No integration mistake. But remember that the 'area' an integral measures is area above the x-axis minus area below the x-axis. In your case these areas are equal.

3. Mar 27, 2008

### RyanSchw

Thanks.

I guess since the area wasn’t symmetrical I just assumed it could not be zero, and that I had made a mistake.

4. Mar 27, 2008

### dynamicsolo

When you graph $$xe^{-x}-e^{-x}$$ , it does seem remarkable that that little sliver above the x-axis from x = 1 out to infinity is actually going to cancel out that curved wedge between x = 0 and x = 1. But you can satisfy yourself that you've made no mistake in your integration by evaluating the areas from x = 0 to x=1 and from x = 1 to x -> infinity separately. The areas are -1/e and +1/e , so you're OK...