integral from 0 to 6 of sqrt(4-2x)

Lösung

\int_{0}^{6} 4 - 2 x d x

\frac{8}{3} + i \frac{16 2}{3}

Schritte zur Lösung

\int_{0}^{6} 4 - 2 x d x

Wende U-Substitution an

= \int_{4}^{- 8} - \frac{u}{2} d u

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

= - \int_{- 8}^{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_{- 8}^{4} u d u)

Wende die Potenzregel an

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{- 8}^{4})

Vereinfache

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{- 8}^{4}

Berechne die Grenzen:

\frac{16}{3} + i \frac{32 2}{3}

= \frac{1}{2} (\frac{16}{3} + i \frac{32 2}{3})

Vereinfache

= \frac{8}{3} + i \frac{16 2}{3}

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