integral from 3 to 6 of 1/(sqrt(8x+1))

Lösung

\int_{3}^{6} \frac{1}{8 x + 1} d x

\frac{1}{2}

Dezimale

0.5

Schritte zur Lösung

\int_{3}^{6} \frac{1}{8 x + 1} d x

Wende U-Substitution an

= \int_{25}^{49} \frac{1}{8 u} d u

Entferne die Konstante:

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

= \frac{1}{8} \cdot \int_{25}^{49} \frac{1}{u} d u

Wende die Potenzregel an

= \frac{1}{8} [2 u^{\frac{1}{2}}]_{25}^{49}

Wende Radikal Regel an:

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

= \frac{1}{8} [2 u]_{25}^{49}

Berechne die Grenzen:

4

= \frac{1}{8} \cdot 4

Vereinfache

= \frac{1}{2}

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