What is the integral from-infinity to 0 of 4^r ?

The integral from-infinity to 0 of 4^r is 1/(2ln(2))

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integral from-infinity to 0 of 4^r

\int_{- \infty}^{0} 4^{r} d r

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

Decimal Notation

0.72134 \dots

Solution steps

\int_{- \infty}^{0} 4^{r} d r

Compute the indefinite integral:

\int 4^{r} d r = \frac{2 ^{2 r - 1}}{ln ( 2 )} + C

Compute the boundaries:

\int_{- \infty}^{0} 4^{r} d r = \frac{1}{2 ln ( 2 )} - 0

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

Simplify

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

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What is the integral from-infinity to 0 of 4^r ?
The integral from-infinity to 0 of 4^r is 1/(2ln(2))