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

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

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integral from 1 to 2 of 4^{-θ}

\int_{1}^{2} 4^{- θ} d θ

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

Decimal Notation

0.13525 \dots

Solution steps

\int_{1}^{2} 4^{- θ} d θ

Compute the indefinite integral:

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

Compute the boundaries:

\int_{1}^{2} 4^{- θ} d θ = - \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 )}

f (x) = 2 (2)^{4} - 16 (2)^{3} + 8

y^{^{'}} = - y + sin (x) + cos (x)

f (x) = 3 x + 2 z + 4 x z

\int_{1}^{7} \frac{ln ( x )}{x} d x

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

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