integral from 1 to 2 of 16\sqrt[3]{2x-3}

Lösung

\int_{1}^{2} 16 3 2 x - 3 d x

0

Schritte zur Lösung

\int_{1}^{2} 16 3 2 x - 3 d x

Entferne die Konstante:

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

= 16 \cdot \int_{1}^{2} 3 2 x - 3 d x

Wende U-Substitution an

= 16 \cdot \int_{- 1}^{1} \frac{3 u}{2} d u

Entferne die Konstante:

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

= 16 \cdot \frac{1}{2} \cdot \int_{- 1}^{1} 3 u d u

Wende die Potenzregel an

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

Vereinfache

16 \cdot \frac{1}{2} [\frac{3}{4} u^{\frac{4}{3}}]_{- 1}^{1} : 8 [\frac{3}{4} u^{\frac{4}{3}}]_{- 1}^{1}

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

Berechne die Grenzen:

0

= 8 \cdot 0

Vereinfache

= 0

\frac{\partial}{\partial x} (x sin (v))

\frac{d}{d x} (cosh^{2} (x) - sinh^{2} (x))

d er i v a t i v e f (x) = x^{5} - x^{3} + 3

\int_{- 27}^{1} \frac{1}{3 x} d x

d er i v a t i v e (2 x + 1) ln (x)