integral from 1/2 to 5 of sqrt(2x-1)

Lösung

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

9

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{0}^{9} \frac{u}{2} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{0}^{9} u d u

Wende die Potenzregel an

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

Berechne die Grenzen:

18

= \frac{1}{2} \cdot 18

Vereinfache

= 9

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