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

Lösung

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

\frac{2 + e ^{3} - e}{2 e} - 1

Dezimale

2.56240 \dots

Schritte zur Lösung

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

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

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

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

Vereinfache

\frac{e ^{2} - 1}{2} - (1 - \frac{1}{e}) : \frac{2 + e ^{3} - e}{2 e} - 1

= \frac{2 + e ^{3} - e}{2 e} - 1

\frac{\partial}{\partial x} (x^{4} - y^{3})

\frac{d ^{2}}{d x ^{2}} (x^{2} + 12)

a re a x^{3} + x^{2}, 2 x^{2} + 2 x

in v erse l a pl a ce cos (w)

\frac{d}{d x} (- \frac{4 x}{( x - 2 ) ^{3}})