integral from 0 to 1 of 2ksqrt(x)

Lösung

\int_{0}^{1} 2 k x d x

\frac{4 k}{3}

Schritte zur Lösung

\int_{0}^{1} 2 k x d x

Entferne die Konstante:

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

= 2 k \cdot \int_{0}^{1} x d x

Wende die Potenzregel an

= 2 k [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{1}

Berechne die Grenzen:

\frac{2}{3}

= 2 k \frac{2}{3}

Vereinfache

= \frac{4 k}{3}

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