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Beliebt Rechnen >

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

Lösung

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

Lösung

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

+1

Dezimale

0.57995 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

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

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

Entferne die Konstante:

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

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

Wende U-Substitution an

= - (- \int_{1 - \frac{1}{e}}^{0} - \frac{2 v ^{2}}{- v ^{2} + 1} d v)

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

= - (- (- \int_{0}^{1 - \frac{1}{e}} - \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

= - (- (- (- 2 \cdot \int_{0}^{1 - \frac{1}{e}} \frac{v ^{2}}{- v ^{2} + 1} d v)))

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

= - (- (- (- 2 \cdot \int_{0}^{1 - \frac{1}{e}} \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

= - (- (- (- 2 (\int_{0}^{1 - \frac{1}{e}} \frac{1}{- v ^{2} + 1} d v - \int_{0}^{1 - \frac{1}{e}} 1 d v))))

\int_{0}^{1 - \frac{1}{e}} \frac{1}{- v ^{2} + 1} d v = \frac{1}{2} (ln (- \frac{1}{e} + 1 + 1) - ln (- - \frac{1}{e} + 1 + 1))

\int_{0}^{1 - \frac{1}{e}} 1 d v = - \frac{1}{e} + 1

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

Vereinfache

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

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

Graph

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