integral from 0 to 4.4 of \sqrt[3]{3x+4}

Lösung

\int_{0}^{4.4} 3 3 x + 4 d x

9.51229 \dots

Schritte zur Lösung

\int_{0}^{4.4} 3 3 x + 4 d x

Wende U-Substitution an

= \int_{4}^{17.2} \frac{3 u}{3} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int_{4}^{17.2} 3 u d u

Wende die Potenzregel an

= \frac{1}{3} [\frac{3}{4} u^{\frac{4}{3}}]_{4}^{17.2}

Vereinfache

= 0.33333 \dots [0.75 u^{\frac{4}{3}}]_{4}^{17.2}

Berechne die Grenzen:

28.53689 \dots

= 0.33333 \dots \cdot 28.53689 \dots

Vereinfache

= 9.51229 \dots

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

\int_{0}^{\frac{π}{4}} 6 x^{2} + tan (x) d x

\int_{- 1}^{1} x^{3} e^{x} d x

\int_{0}^{2} (x^{3} + x) d x

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