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人気のある微分積分 >

integral from 0 to 2 of sqrt(1+e^{-2x)}

解

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

解

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

+1

十進法表記

2.22141 \dots

解答ステップ

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

u置換積分法を適用する

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{2} \cdot \int_{\frac{1}{e ^{4}}}^{1} \frac{1 + u}{u} d u)

u置換積分法を適用する

= - (- \frac{1}{2} \cdot \int_{1 + \frac{1}{e ^{4}}}^{2} \frac{2 v ^{2}}{v ^{2} - 1} d v)

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{2} \cdot 2 \cdot \int_{1 + \frac{1}{e ^{4}}}^{2} \frac{v ^{2}}{v ^{2} - 1} d v)

因数

v^{2} - 1 : - (- v^{2} + 1)

= - (- \frac{1}{2} \cdot 2 \cdot \int_{1 + \frac{1}{e ^{4}}}^{2} \frac{v ^{2}}{- ( - v ^{2} + 1 )} d v)

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{2} \cdot 2 (- \int_{1 + \frac{1}{e ^{4}}}^{2} \frac{v ^{2}}{- v ^{2} + 1} d v))

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

= - (- \frac{1}{2} \cdot 2 (- \int_{1 + \frac{1}{e ^{4}}}^{2} \frac{1}{- v ^{2} + 1} - 1 d v))

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

= - (- \frac{1}{2} \cdot 2 (- (\int_{1 + \frac{1}{e ^{4}}}^{2} \frac{1}{- v ^{2} + 1} d v - \int_{1 + \frac{1}{e ^{4}}}^{2} 1 d v)))

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

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

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

簡素化

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

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

グラフ

人気の例

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integral from 0 to infinity of 1/(5+x)

\int_{0}^{\infty} \frac{1}{5 + x} d x

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