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人気のある微分積分 >

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

解

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

解

- ln (1 + e + 1) + ln ((1 + e - 1) (2 + 1)) - ln (2 - 1) + 2 1 + e - 22

+1

十進法表記

1.64205 \dots

解答ステップ

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

u置換積分法を適用する

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

u置換積分法を適用する

= \int_{2}^{1 + e} \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

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

因数

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

= 2 \cdot \int_{2}^{1 + e} \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

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

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

= 2 (- \int_{2}^{1 + e} \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

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

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

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

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

簡素化

2 (- (\frac{ln ( 1 + e + 1 ) - ln ( ( 1 + e - 1 ) ( 2 + 1 ) ) + ln ( 2 - 1 )}{2} - (1 + e - 2))) : - ln (1 + e + 1) + ln ((1 + e - 1) (2 + 1)) - ln (2 - 1) + 2 1 + e - 22

= - ln (1 + e + 1) + ln ((1 + e - 1) (2 + 1)) - ln (2 - 1) + 2 1 + e - 22

グラフ

人気の例

integral from-2 to 2 of 5x^4

\int_{- 2}^{2} 5 x^{4} d x

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

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

integral from 0 to 8 of 8+\sqrt[3]{x}

\int_{0}^{8} 8 + 3 x d x

integral from-1 to 3 of 1/(x^4)

\int_{- 1}^{3} \frac{1}{x ^{4}} d x

integral from 0 to pi/2 of 1+cos(2y)

\int_{0}^{\frac{π}{2}} 1 + cos (2 y) d y