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

Lösung

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

- e^{- t} + 1

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{0}^{- t} - e^{u} d u

Entferne die Konstante:

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

= - \int_{0}^{- t} e^{u} d u

Nutze das gemeinsame Integral :

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

= - [e^{u}]_{0}^{- t}

Berechne die Grenzen:

e^{- t} - 1

= - (e^{- t} - 1)

Vereinfache

= - e^{- t} + 1

\int (\frac{( x ^{3} + 1 ) ^{2}}{x ^{2}}) d x

\frac{\partial}{\partial x} (x y ln (x^{2} + y^{2}))

\frac{d}{d x} (ln (cosh (3 x)))

x \to \frac{π}{4} lim (\frac{sin ( 2 x ) - 1}{4 x - π})

\frac{d}{d x} (cos (2 x + 3 y))