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

Lösung

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

e - 1 - \frac{e ^{2} - 1}{2 e ^{2}}

Dezimale

1.28594 \dots

Schritte zur Lösung

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= e - 1 - \frac{e ^{2} - 1}{2 e ^{2}}

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

\int_{1}^{4} - x - (2 - x) d x

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

\int_{0}^{π} cos (\frac{1}{2}) x d x

x \to \frac{π}{2} lim (\frac{cos ( x )}{1 - sin ( x )})