integral from 0 to 1 of t*e^{-st}

Lösung

\int_{0}^{1} t \cdot e^{- s t} d t

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

Schritte zur Lösung

\int_{0}^{1} t e^{- s t} d t

Wende U-Substitution an

= \int_{0}^{- s} \frac{e ^{u} u}{s ^{2}} d u

Entferne die Konstante:

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

= \frac{1}{s ^{2}} \cdot \int_{0}^{- s} e^{u} u d u

Wende die partielle Integration an

= \frac{1}{s ^{2}} [e^{u} u - \int e^{u} d u]_{0}^{- s}

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

= \frac{1}{s ^{2}} [e^{u} u - e^{u}]_{0}^{- s}

Vereinfache

\frac{1}{s ^{2}} [e^{u} u - e^{u}]_{0}^{- s} : \frac{[ e ^{u} u - e ^{u} ] _{0}^{- s}}{s ^{2}}

= \frac{[ e ^{u} u - e ^{u} ] _{0}^{- s}}{s ^{2}}

Berechne die Grenzen:

- s e^{- s} - e^{- s} + 1

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

\int_{1.5}^{2} ∣ x - 1 ∣ d x

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

\int_{1}^{2} (4 x^{3} + \frac{6}{x ^{2}}) d x

\int_{- 3}^{3} e^{u + 3} d u

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