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_{- 2}^{2} 4 + x^{2} d x

\int_{0}^{\infty} 2 e^{- 5 x} d x

\int_{0}^{5} x e^{2 x} d x

\int_{1}^{2} (\frac{ln ( x )}{x ^{2}}) d x

\int_{- x - 2}^{x} x^{2} d x