# Improper Integral

1. Feb 11, 2013

### iRaid

1. The problem statement, all variables and given/known data
$$\int_{-\infty}^{0} 2^{r}dr$$

2. Relevant equations

3. The attempt at a solution
$$\int_{-\infty}^{0} 2^{r}dr = \lim_{t \to -\infty} \int_t^0 2^{r}dr=\lim_{t \to -\infty} \frac{2^{r}}{ln2}|_{t}^{0} = \lim_{t \to -\infty} \frac{1}{ln2}-\frac{2^{t}}{ln2}$$

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

Last edited: Feb 11, 2013
2. Feb 11, 2013

### iRaid

I think I found my error...
$$\lim_{t \to -\infty} \frac{1}{ln2}-\frac{2^{t}}{ln2}=\lim_{t \to -\infty} \frac{1}{ln2}-\frac{1}{(ln2)(2^{t})}$$

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