integral from 1 to 2 of 4^{-θ}

解答

\int_{1}^{2} 4^{- θ} d θ

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

十进制

0.13525 \dots

求解步骤

\int_{1}^{2} 4^{- θ} d θ

计算不定积分:

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

带入上下限计算后得:

\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 )})

化简

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

d er i v a t i v e f (x) = 2 (2)^{4} - 16 (2)^{3} + 8

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

e x t re m e 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})