integral from-5 to-2 of (7e^x+2/x)

解

\int_{- 5}^{- 2} (7 e^{x} + \frac{2}{x}) d x

7 (\frac{1}{e ^{2}} - \frac{1}{e ^{5}}) + 2 (ln (2) - ln (5))

十進法表記

- 0.93240 \dots

解答ステップ

\int_{- 5}^{- 2} 7 e^{x} + \frac{2}{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_{- 5}^{- 2} 7 e^{x} d x + \int_{- 5}^{- 2} \frac{2}{x} d x

\int_{- 5}^{- 2} 7 e^{x} d x = 7 (\frac{1}{e ^{2}} - \frac{1}{e ^{5}})

\int_{- 5}^{- 2} \frac{2}{x} d x = 2 (ln (2) - ln (5))

= 7 (\frac{1}{e ^{2}} - \frac{1}{e ^{5}}) + 2 (ln (2) - ln (5))