integral from 1 to 6 of sqrt(6x+1)

Lösung

\int_{1}^{6} 6 x + 1 d x

\frac{37 37 - 7 7}{9}

Dezimale

22.94910 \dots

Schritte zur Lösung

\int_{1}^{6} 6 x + 1 d x

Wende U-Substitution an

= \int_{7}^{37} \frac{u}{6} d u

Entferne die Konstante:

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

= \frac{1}{6} \cdot \int_{7}^{37} u d u

Wende die Potenzregel an

= \frac{1}{6} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{37}

Berechne die Grenzen:

\frac{74 37 - 14 7}{3}

= \frac{1}{6} \cdot \frac{74 37 - 14 7}{3}

Vereinfache

= \frac{37 37 - 7 7}{9}

\int_{- 1}^{2} x^{2} (x^{3} + 2)^{2} d x

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