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

Lösung

\int_{0}^{1} 4 - x^{2} x d x

\frac{8 - 3 3}{3}

Dezimale

0.93461 \dots

Schritte zur Lösung

\int_{0}^{1} 4 - x^{2} x d x

Wende U-Substitution an

= \int_{4}^{3} - \frac{u}{2} d u

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

= - \int_{3}^{4} - \frac{u}{2} d u

Entferne die Konstante:

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

= - (- \frac{1}{2} \cdot \int_{3}^{4} u d u)

Wende die Potenzregel an

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{3}^{4})

Vereinfache

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

Berechne die Grenzen:

\frac{16}{3} - 23

= \frac{1}{2} (\frac{16}{3} - 23)

Vereinfache

= \frac{8 - 3 3}{3}

\int_{0}^{3} 3 1 + 36 x^{2} d x

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