integral from 0 to 1 of e^{-(x+y)}

Lösung

\int_{0}^{1} e^{- (x + y)} d x

- e^{- y - 1} + e^{- y}

Schritte zur Lösung

\int_{0}^{1} e^{- (x + y)} d x

Wende U-Substitution an

= \int_{- y}^{- y - 1} - e^{u} d u

Entferne die Konstante:

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

= - \int_{- y}^{- y - 1} e^{u} d u

Nutze das gemeinsame Integral :

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

= - [e^{u}]_{- y}^{- y - 1}

Berechne die Grenzen:

e^{- y - 1} - e^{- y}

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

Vereinfache

= - e^{- y - 1} + e^{- y}

\int_{0}^{4} \frac{x}{16 + 5 x} d x

\int_{- 4}^{4} e^{x} d x

\int_{0}^{1} (5 x^{e} + 3 e^{x}) d x

\int_{0}^{1} sin (π) x cos (π) x d x

\int_{t - 2}^{0} (x + 2) d x