integral from 1 to 16 of 1/(2sqrt(v))

Lösung

\int_{1}^{16} \frac{1}{2 v} d v

3

Schritte zur Lösung

\int_{1}^{16} \frac{1}{2 v} d v

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{1}^{16} \frac{1}{v} d v

Wende die Potenzregel an

= \frac{1}{2} [2 v^{\frac{1}{2}}]_{1}^{16}

Wende Radikal Regel an:

a^{\frac{1}{n}} = n a

= \frac{1}{2} [2 v]_{1}^{16}

Berechne die Grenzen:

6

= \frac{1}{2} \cdot 6

Vereinfache

= 3

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