integral from 0 to 1 of e^{-x}(4-e^x)

Lösung

\int_{0}^{1} e^{- x} (4 - e^{x}) d x

- \frac{- e + 4}{e} + 2

Dezimale

1.52848 \dots

Schritte zur Lösung

\int_{0}^{1} e^{- x} (4 - e^{x}) d x

Wende die partielle Integration an

= [- e^{- x} (4 - e^{x}) - \int 1 d x]_{0}^{1}

\int 1 d x = x

= [- e^{- x} (4 - e^{x}) - x]_{0}^{1}

Berechne die Grenzen:

- \frac{- e + 4}{e} + 2

= - \frac{- e + 4}{e} + 2

\int_{0}^{3} 0.3 e^{x} d x

\int_{0}^{2} e^{- r^{2}} r d r

\int_{0}^{2} r 4 - r^{2} d r

\int_{- 1}^{1} e^{- 2 x} d x

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