Solving Improper Integral: $\int_{-\infty}^{0} 2^{r}dr$

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SUMMARY

The improper integral $\int_{-\infty}^{0} 2^{r}dr$ evaluates to approximately 1.4427, contrary to the initial assumption of infinity. The solution involves applying the limit process: $\lim_{t \to -\infty} \int_t^0 2^{r}dr = \lim_{t \to -\infty} \frac{2^{r}}{\ln 2}|_{t}^{0}$. The error in the initial calculation was clarified by recognizing that as $t$ approaches negative infinity, the term $\frac{2^{t}}{\ln 2}$ approaches zero, leading to the correct result of $\frac{1}{\ln 2}$.

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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:
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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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