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

Lösung

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

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

Dezimale

1

Schritte zur Lösung

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

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

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

Bestimme unbestimmte (Singularitäts-)Punkte:

x = 1

Wenn

b, a < b < c, f (b) =

unbestimmt

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

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

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

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

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

Vereinfache

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

\int_{0}^{\frac{π}{2}} x cos^{2} (x) d x

\int_{0}^{3} x^{2} - 1 d x

\int_{1}^{7} \frac{1}{x ^{2}} d x

\int_{1}^{4} u d u

\int_{0}^{1} x^{4} (1 + 2 x^{5})^{5} d x