integral from 0 to 1 of 6x^2sqrt(1-x^3)

Lösung

\int_{0}^{1} 6 x^{2} 1 - x^{3} d x

\frac{4}{3}

Dezimale

1.33333 \dots

Schritte zur Lösung

\int_{0}^{1} 6 x^{2} 1 - x^{3} d x

Entferne die Konstante:

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

= 6 \cdot \int_{0}^{1} x^{2} 1 - x^{3} d x

Wende U-Substitution an

= 6 \cdot \int_{1}^{0} - \frac{u}{3} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 6 (- \int_{0}^{1} - \frac{u}{3} d u)

Entferne die Konstante:

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

= 6 (- (- \frac{1}{3} \cdot \int_{0}^{1} u d u))

Wende die Potenzregel an

= 6 (- (- \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}))

Vereinfache

6 (- (- \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1})) : 2 [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}

= 2 [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}

Berechne die Grenzen:

\frac{2}{3}

= 2 \cdot \frac{2}{3}

Vereinfache

= \frac{4}{3}

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