integral from 0 to b of te^{-st}

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Wende die partielle Integration an

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

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

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

Vereinfache

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

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

Berechne die Grenzen:

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

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

\int_{0}^{\frac{π}{2}} e^{2 x} sin (3 x) d x

\int_{2}^{5} x + 5 d x

\int_{0}^{π} 3 x sin (x^{2}) d x

\int_{- 1}^{1} \frac{7}{1 + 49 x ^{2}} d x

\int_{0}^{4} 4 - y d y