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Improper Integral

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex]\int_{-\infty}^{0} 2^{r}dr[/tex]

    2. Relevant equations



    3. 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: Feb 11, 2013
  2. jcsd
  3. Feb 11, 2013 #2
    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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