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

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

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integral from 0 to infinity of 2^{-x}

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

\frac{1}{ln ( 2 )}

Decimal

1.44269 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

5 x + 6 x^{- 1}

f (x) = x (x - 1) (x - 2)

f (x) = 5^{c o s (2 \frac{π}{2})}

\frac{\partial}{\partial x} (\frac{2 y}{x + y ^{2}})

\int_{- 5}^{5} (e^{x} - e^{- x})^{2} d x

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