integral from 0 to x of e^{-2t}

Lösung

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

- \frac{1}{2} (e^{- 2 x} - 1)

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

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

Berechne die Grenzen:

e^{- 2 x} - 1

= - \frac{1}{2} (e^{- 2 x} - 1)

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

\int_{5}^{2} 1

\int_{0}^{\frac{9}{8}} \frac{1}{64 x ^{2} + 81} d x

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

\int_{x}^{y} 1 d x