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人気のある微分積分 >

integral from 0 to 1 of ((e^x+2))/(e^x)

解

\int_{0}^{1} \frac{( e ^{x} + 2 )}{e ^{x}} d x

解

- \frac{2}{e} + 3

+1

十進法表記

2.26424 \dots

解答ステップ

\int_{0}^{1} \frac{e ^{x} + 2}{e ^{x}} d x

u置換積分法を適用する

= \int_{1}^{e} \frac{u + 2}{u ^{2}} d u

拡張

\frac{u + 2}{u ^{2}} : \frac{1}{u} + \frac{2}{u ^{2}}

= \int_{1}^{e} \frac{1}{u} + \frac{2}{u ^{2}} d u

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}^{e} \frac{1}{u} d u + \int_{1}^{e} \frac{2}{u ^{2}} d u

\int_{1}^{e} \frac{1}{u} d u = 1

\int_{1}^{e} \frac{2}{u ^{2}} d u = 2 (- \frac{1}{e} + 1)

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

簡素化

1 + 2 (- \frac{1}{e} + 1) : - \frac{2}{e} + 3

= - \frac{2}{e} + 3

グラフ

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