integral from 0 to 5 of h(x)

Lösung

\int_{0}^{5} h (x) d x

\frac{25 h}{2}

Schritte zur Lösung

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

Entferne die Konstante:

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

= h \cdot \int_{0}^{5} x d x

Wende die Potenzregel an

= h [\frac{x ^{2}}{2}]_{0}^{5}

Berechne die Grenzen:

\frac{25}{2}

= h \frac{25}{2}

Vereinfache

= \frac{25 h}{2}

\int_{- 1}^{2} (5 x - 5) d x

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

\int_{0}^{\infty} e^{- s} d s

\int_{5}^{8} \frac{1}{x - 2} d x

\int_{1}^{3} t d t