integral from 0 to 3 of e^{-st}e^{3t}

Lösung

\int_{0}^{3} e^{- s t} e^{3 t} d t

\frac{e ^{- 3 s + 9} - 1}{- s + 3}

Schritte zur Lösung

\int_{0}^{3} e^{- s t} e^{3 t} d t

Vereinfache

e^{- s t} e^{3 t} : e^{- s t + 3 t}

= \int_{0}^{3} e^{- s t + 3 t} d t

Wende U-Substitution an

= \int_{0}^{- 3 s + 9} \frac{e ^{u}}{- s + 3} d u

Entferne die Konstante:

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

= \frac{1}{- s + 3} \cdot \int_{0}^{- 3 s + 9} e^{u} d u

Nutze das gemeinsame Integral :

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

= \frac{1}{- s + 3} [e^{u}]_{0}^{- 3 s + 9}

Vereinfache

\frac{1}{- s + 3} [e^{u}]_{0}^{- 3 s + 9} : \frac{[ e ^{u} ] _{0}^{- 3 s + 9}}{- s + 3}

= \frac{[ e ^{u} ] _{0}^{- 3 s + 9}}{- s + 3}

Berechne die Grenzen:

e^{- 3 s + 9} - 1

= \frac{e ^{- 3 s + 9} - 1}{- s + 3}

\int_{- 4}^{- 1} \frac{4}{x} d x

\int_{0}^{3} 2 x x^{2} + 16 d x

\int_{4}^{5} x (x^{2} - 16)^{3} d x

\int_{0}^{3} 4 x - x^{2} d x

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