integral from 0 to 1 of sqrt(1x+6)

Lösung

\int_{0}^{1} 1 x + 6 d x

\frac{14 7}{3} - 46

Dezimale

2.54888 \dots

Schritte zur Lösung

\int_{0}^{1} 1 \cdot x + 6 d x

Vereinfache

= \int_{0}^{1} x + 6 d x

Wende U-Substitution an

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

Wende die Potenzregel an

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

Berechne die Grenzen:

\frac{14 7}{3} - 46

= \frac{14 7}{3} - 46

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