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Beliebt Rechnen >

integral from 0 to 4 of sqrt(1+e^{-2y)}

Lösung

\int_{0}^{4} 1 + e^{- 2 y} d y

Lösung

- \frac{ln ( 2 + 1 ) - ln ( ( 2 - 1 ) ( 1 + \frac{1}{e ^{8}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{8}} - 1 )}{2} + 2 - 1 + \frac{1}{e ^{8}}

+1

Dezimale

4.22590 \dots

Schritte zur Lösung

\int_{0}^{4} 1 + e^{- 2 y} d y

Wende U-Substitution an

= \int_{1}^{\frac{1}{e ^{8}}} - \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 ^{8}}}^{1} - \frac{1 + u}{2 u} d u

Entferne die Konstante:

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

= - (- \frac{1}{2} \cdot \int_{\frac{1}{e ^{8}}}^{1} \frac{1 + u}{u} d u)

Wende U-Substitution an

= - (- \frac{1}{2} \cdot \int_{1 + \frac{1}{e ^{8}}}^{2} \frac{2 v ^{2}}{v ^{2} - 1} d v)

Entferne die Konstante:

\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 ^{8}}}^{2} \frac{v ^{2}}{v ^{2} - 1} d v)

Faktorisiere

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

= - (- \frac{1}{2} \cdot 2 \cdot \int_{1 + \frac{1}{e ^{8}}}^{2} \frac{v ^{2}}{- ( - v ^{2} + 1 )} d v)

Entferne die Konstante:

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

= - (- \frac{1}{2} \cdot 2 (- \int_{1 + \frac{1}{e ^{8}}}^{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 ^{8}}}^{2} \frac{1}{- v ^{2} + 1} - 1 d v))

Wende die Summenregel an:

\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 ^{8}}}^{2} \frac{1}{- v ^{2} + 1} d v - \int_{1 + \frac{1}{e ^{8}}}^{2} 1 d v)))

\int_{1 + \frac{1}{e ^{8}}}^{2} \frac{1}{- v ^{2} + 1} d v = \frac{ln ( 2 + 1 ) - ln ( ( 2 - 1 ) ( 1 + \frac{1}{e ^{8}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{8}} - 1 )}{2}

\int_{1 + \frac{1}{e ^{8}}}^{2} 1 d v = 2 - 1 + \frac{1}{e ^{8}}

= - - \frac{1}{2} \cdot 2 - \frac{ln ( 2 + 1 ) - ln ( ( 2 - 1 ) ( 1 + \frac{1}{e ^{8}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{8}} - 1 )}{2} - (2 - 1 + \frac{1}{e ^{8}})

Vereinfache

- - \frac{1}{2} \cdot 2 - \frac{ln ( 2 + 1 ) - ln ( ( 2 - 1 ) ( 1 + \frac{1}{e ^{8}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{8}} - 1 )}{2} - (2 - 1 + \frac{1}{e ^{8}}) : - \frac{ln ( 2 + 1 ) - ln ( ( 2 - 1 ) ( 1 + \frac{1}{e ^{8}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{8}} - 1 )}{2} + 2 - 1 + \frac{1}{e ^{8}}

= - \frac{ln ( 2 + 1 ) - ln ( ( 2 - 1 ) ( 1 + \frac{1}{e ^{8}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{8}} - 1 )}{2} + 2 - 1 + \frac{1}{e ^{8}}

Graph

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