integral from 0 to 4 of e^{-st}t

Lösung

\int_{0}^{4} e^{- s t} t d t

\frac{- 4 s e ^{- 4 s} - e ^{- 4 s} + 1}{s ^{2}}

Schritte zur Lösung

\int_{0}^{4} e^{- s t} t d t

Wende U-Substitution an

= \int_{0}^{- s \cdot 4} \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 \cdot 4} e^{u} u d u

Wende die partielle Integration an

= \frac{1}{s ^{2}} [e^{u} u - \int e^{u} d u]_{0}^{- 4 s}

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

= \frac{1}{s ^{2}} [e^{u} u - e^{u}]_{0}^{- 4 s}

Vereinfache

\frac{1}{s ^{2}} [e^{u} u - e^{u}]_{0}^{- 4 s} : \frac{[ e ^{u} u - e ^{u} ] _{0}^{- 4 s}}{s ^{2}}

= \frac{[ e ^{u} u - e ^{u} ] _{0}^{- 4 s}}{s ^{2}}

Berechne die Grenzen:

- 4 s e^{- 4 s} - e^{- 4 s} + 1

= \frac{- 4 s e ^{- 4 s} - e ^{- 4 s} + 1}{s ^{2}}

\int_{1}^{2} \frac{3 x ^{3} + 2}{x ^{3}} d x

\int_{- 3}^{3} 3 - x^{2} d x

\int_{- 1}^{1} x^{2} - \frac{1}{3} d x

\int_{10}^{\infty} \frac{1}{x ^{3}} d x

\int_{- 2}^{- 1} \frac{1}{x ^{3} + x} d x