integral from 4 to 5 of e^{4-x}

Lösung

\int_{4}^{5} e^{4 - x} d x

1 - \frac{1}{e}

Dezimale

0.63212 \dots

Schritte zur Lösung

\int_{4}^{5} e^{4 - x} d x

Wende U-Substitution an

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Entferne die Konstante:

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

= - (- \int_{- 1}^{0} e^{u} d u)

Nutze das gemeinsame Integral :

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

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

Vereinfache

= [e^{u}]_{- 1}^{0}

Berechne die Grenzen:

1 - \frac{1}{e}

= 1 - \frac{1}{e}

\int_{1}^{e^{5}} \frac{6 ln ( x )}{x} d x

\int_{x^{3}}^{x^{4}} (2 t - 6)^{6} d t

\int_{- 5}^{0} x d x

\int_{0}^{1} 3 t e^{t^{2} - 1} d t

\int_{0}^{2} \frac{1}{2 x + 5} d x