integral from 1 to 2 of (4/x-e^{-x})

解

\int_{1}^{2} (\frac{4}{x} - e^{- x}) d x

4 ln (2) - \frac{1}{e} + \frac{1}{e ^{2}}

十進法表記

2.54004 \dots

解答ステップ

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

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} \frac{4}{x} d x - \int_{1}^{2} e^{- x} d x

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

\int_{1}^{2} e^{- x} d x = \frac{1}{e} - \frac{1}{e ^{2}}

= 4 ln (2) - (\frac{1}{e} - \frac{1}{e ^{2}})

簡素化

= 4 ln (2) - \frac{1}{e} + \frac{1}{e ^{2}}