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人気のある微分積分 >

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

解

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

解

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

+1

十進法表記

2.00349 \dots

解答ステップ

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

u置換積分法を適用する

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

u置換積分法を適用する

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

因数

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

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

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

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

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

簡素化

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

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

グラフ

人気の例

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