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

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

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integral from 0 to 2 of 4^{-θ}

\int_{0}^{2} 4^{- θ} d θ

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

Decimal Notation

0.67626 \dots

Solution steps

\int_{0}^{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_{0}^{2} 4^{- θ} d θ = - \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 )}

\int (3 x^{2} - 1) d x

f (x) = 2 x^{3} + 4 x^{2} + 6 x

x \to 0 + lim (sin (\frac{5 π}{6} e^{x}))

\int (x + 1) [ln (x + 1) + c] d x

y^{2} y^{^{'}} = y^{2}

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