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

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

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integral from-infinity to 0 of 2^x

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

\frac{1}{ln ( 2 )}

Decimal Notation

1.44269 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

a \to 0 lim (\frac{cos ( a )}{a})

\frac{\partial}{\partial x} (6 x^{9 y})

n = 0 \sum \infty \frac{e}{π}

\int 10 cos (4 x) d x

x \to 1 lim (x^{3} + 2 x^{2} - x - 10)

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