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

Lösung

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

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

Dezimale

6.78865 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Wende U-Substitution an

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

Faktorisiere

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

= \int_{2}^{1 + e^{4}} \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

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

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

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

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

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

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

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

Vereinfache

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