What is the integral from 1 to 2 of 4^{-x} ?

The integral from 1 to 2 of 4^{-x} is 3/(32ln(2))

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integral from 1 to 2 of 4^{-x}

\int_{1}^{2} 4^{- x} d x

\frac{3}{32 ln ( 2 )}

Decimal Notation

0.13525 \dots

Solution steps

\int_{1}^{2} 4^{- x} d x

Compute the indefinite integral:

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

Compute the boundaries:

\int_{1}^{2} 4^{- x} d x = - \frac{1}{32 ln ( 2 )} - (- \frac{1}{8 ln ( 2 )})

= - \frac{1}{32 ln ( 2 )} - (- \frac{1}{8 ln ( 2 )})

Simplify

= \frac{3}{32 ln ( 2 )}

y^{^{'}} + 2 y = 2 t

\int 2 x + 7 e^{x} d x

x \to 2 lim (x)

\frac{d}{d x} ((10 x - 2 x^{2}) (16 - 2 x))

y^{^{'}} + 2 x y^{2} = 0

What is the integral from 1 to 2 of 4^{-x} ?
The integral from 1 to 2 of 4^{-x} is 3/(32ln(2))