Improper Integral

  • Thread starter RyanSchw
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  • #1
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Improper Integral [Solved]

Homework Statement



[tex]
\int_{0}^{\infty} (x-1)e^{-x}dx
[/tex]

Homework Equations



Integration by Parts
Improper Integrals


The Attempt at a Solution



[tex]
\lim_{R\rightarrow \infty} \int_0^R~xe^{-x}-e^{-x}dx
[/tex]

Let u = x
du = dx

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

[tex]
-xe^{-x} - \int -e^{-x}dx
[/tex]

[tex]
-xe^{-x}-e^{-x} - \int e^{-x}dx
[/tex]

[tex]
= -xe^{-x}
[/tex]

[tex]
\lim_{R\rightarrow \infty}-xe^{-x} \mid_{0}^{R}=0
[/tex]

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:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
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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
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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
dynamicsolo
Homework Helper
1,648
4
Thanks.

I guess since the area wasn’t symmetrical I just assumed it could not be zero, and that I had made a mistake.
When you graph [tex]xe^{-x}-e^{-x}[/tex] , 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...
 

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