integral from 0 to 1 of x^2sqrt(4+x^3)

Lösung

\int_{0}^{1} x^{2} 4 + x^{3} d x

\frac{10 5 - 16}{9}

Dezimale

0.70674 \dots

Schritte zur Lösung

\int_{0}^{1} x^{2} 4 + x^{3} d x

Wende U-Substitution an

= \int_{4}^{5} \frac{u}{3} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int_{4}^{5} u d u

Wende die Potenzregel an

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{5}

Berechne die Grenzen:

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

= \frac{1}{3} (\frac{10 5}{3} - \frac{16}{3})

Vereinfache

= \frac{10 5 - 16}{9}

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