integral from 0 to 1 of sqrt(4+x)

Lösung

\int_{0}^{1} 4 + x d x

\frac{10 5 - 16}{3}

Dezimale

2.12022 \dots

Schritte zur Lösung

\int_{0}^{1} 4 + x d x

Wende U-Substitution an

= \int_{4}^{5} u d u

Wende die Potenzregel an

= [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{5}

Berechne die Grenzen:

\frac{10 5}{3} - \frac{16}{3}

= \frac{10 5}{3} - \frac{16}{3}

Vereinfache

= \frac{10 5 - 16}{3}

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