integral from-2 to 2 of sqrt(4x+8)

Lösung

\int_{- 2}^{2} 4 x + 8 d x

\frac{32}{3}

Dezimale

10.66666 \dots

Schritte zur Lösung

\int_{- 2}^{2} 4 x + 8 d x

Wende U-Substitution an

= \int_{0}^{16} \frac{u}{4} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int_{0}^{16} u d u

Wende die Potenzregel an

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

Berechne die Grenzen:

\frac{128}{3}

= \frac{1}{4} \cdot \frac{128}{3}

Vereinfache

= \frac{32}{3}

\int_{0}^{2} \frac{3}{2} (1 - x)^{2} d x

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