# Improper Integral

## Homework Statement

$$\int_{-\infty}^{0} 2^{r}dr$$

## 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..

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$$\lim_{t \to -\infty} \frac{1}{ln2}-\frac{2^{t}}{ln2}=\lim_{t \to -\infty} \frac{1}{ln2}-\frac{1}{(ln2)(2^{t})}$$