What is the integral from-infinity to 0 of 2^r ?

The integral from-infinity to 0 of 2^r is 1/(ln(2))

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integral from-infinity to 0 of 2^r

\int_{- \infty}^{0} 2^{r} d r

\frac{1}{ln ( 2 )}

Decimal

1.44269 \dots

Solution steps

\int_{- \infty}^{0} 2^{r} d r

Compute the indefinite integral:

\int 2^{r} d r = \frac{2 ^{r}}{ln ( 2 )} + C

Compute the boundaries:

\int_{- \infty}^{0} 2^{r} d r = \frac{1}{ln ( 2 )} - 0

= \frac{1}{ln ( 2 )} - 0

Simplify

= \frac{1}{ln ( 2 )}

What is the integral from-infinity to 0 of 2^r ?
The integral from-infinity to 0 of 2^r is 1/(ln(2))