What is the integral from 0 to 2 of 4^{-x} ?

The integral from 0 to 2 of 4^{-x} is (15)/(32ln(2))

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integral from 0 to 2 of 4^{-x}

\int_{0}^{2} 4^{- x} d x

\frac{15}{32 ln ( 2 )}

Decimal Notation

0.67626 \dots

Solution steps

\int_{0}^{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_{0}^{2} 4^{- x} d x = - \frac{1}{32 ln ( 2 )} - (- \frac{1}{2 ln ( 2 )})

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

Simplify

= \frac{15}{32 ln ( 2 )}

y^{^{'}} = \frac{1}{x + y}

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

x^{2} + 3 x + 1

f (t) = 7 t + 3

What is the integral from 0 to 2 of 4^{-x} ?
The integral from 0 to 2 of 4^{-x} is (15)/(32ln(2))