Improper Integral

  • Thread starter iRaid
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  • #1
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Homework Statement


[tex]\int_{-\infty}^{0} 2^{r}dr[/tex]

Homework Equations





The Attempt at a Solution


[tex]\int_{-\infty}^{0} 2^{r}dr = \lim_{t \to -\infty} \int_t^0 2^{r}dr=\lim_{t \to -\infty} \frac{2^{r}}{ln2}|_{t}^{0} = \lim_{t \to -\infty} \frac{1}{ln2}-\frac{2^{t}}{ln2}[/tex]

Which I thought = ∞, but I guess not. It's supposed to be 1.4427 according to wolfram..
 
Last edited:

Answers and Replies

  • #2
559
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I think I found my error...
[tex]\lim_{t \to -\infty} \frac{1}{ln2}-\frac{2^{t}}{ln2}=\lim_{t \to -\infty} \frac{1}{ln2}-\frac{1}{(ln2)(2^{t})}[/tex]

So then the 1/(ln2)(2^t) becomes 1/0 and then 1/ln2 = what wolfram gets..
 

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