integral from 0 to 1 of e^{-kx}

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{0}^{- k} - \frac{e ^{u}}{k} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{k} \cdot \int_{0}^{- k} e^{u} d u

Nutze das gemeinsame Integral :

\int e^{u} d u = e^{u}

= - \frac{1}{k} [e^{u}]_{0}^{- k}

Vereinfache

- \frac{1}{k} [e^{u}]_{0}^{- k} : - \frac{[ e ^{u} ] _{0}^{- k}}{k}

= - \frac{[ e ^{u} ] _{0}^{- k}}{k}

Berechne die Grenzen:

e^{- k} - 1

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

\int_{0}^{2} (2 x^{2} - 3 x) d x

\int_{x}^{0} e^{- 8 t} d t

\int_{1}^{\infty} \frac{1}{2 ^{n}}

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

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