What is the integral from-infinity to 0 of 6^r ?

The integral from-infinity to 0 of 6^r is 1/(ln(6))

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integral from-infinity to 0 of 6^r

\int_{- \infty}^{0} 6^{r} d r

\frac{1}{ln ( 6 )}

Decimal

0.55811 \dots

Solution steps

\int_{- \infty}^{0} 6^{r} d r

Compute the indefinite integral:

\int 6^{r} d r = \frac{6 ^{r}}{ln ( 6 )} + C

Compute the boundaries:

\int_{- \infty}^{0} 6^{r} d r = \frac{1}{ln ( 6 )} - 0

= \frac{1}{ln ( 6 )} - 0

Simplify

= \frac{1}{ln ( 6 )}

\frac{d y}{d t} + a y = b e^{- b t}

\frac{d}{d x} ((e^{x}) d x)

\frac{1}{s ^{4} - 9}

\frac{d}{d x} (\frac{( x - 4 ) ( x ^{2} + 4 x + 16 )}{x ^{3}})

\int \frac{1}{e ^{x} + 4} d x

What is the integral from-infinity to 0 of 6^r ?
The integral from-infinity to 0 of 6^r is 1/(ln(6))