integral from 0 to 2 of x\sqrt[3]{4+x^2}

Lösung

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

\frac{12 - 3 3 4}{2}

Dezimale

3.61889 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{4}^{8} \frac{3 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_{4}^{8} 3 u d u

Wende die Potenzregel an

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

Berechne die Grenzen:

12 - 334

= \frac{1}{2} (12 - 334)

Vereinfache

= \frac{12 - 3 3 4}{2}

\frac{\partial}{\partial x} (\frac{3}{3 x + 5 y})

a re a 6 x^{2} ln (x), 24 ln (x)

\frac{d y}{d x} = \frac{1}{2} x + 2

x^{2} y^{^{''}} + 3 x y^{^{'}} + 2 y = 0, y (1) = 0.2, y^{^{'}} (1) = 0.3

\frac{d}{d x} (cos (sin (tan (π) x)))