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

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

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integral from 1 to infinity of 1/(2^x)

\int_{1}^{\infty} \frac{1}{2 ^{x}} d x

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

Decimal

0.72134 \dots

Solution steps

\int_{1}^{\infty} \frac{1}{2 ^{x}} d x

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

(2 x^{2} + 1)^{3} (x^{2} - 4)^{2}

\int_{0}^{l n (3)} e^{4 x} d x

\int (4 x y) d x

\int 5 x cos (5 x) d x

f (x) = x^{2} - 5 x + 3

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