integral from 0 to 5 of (18)/(\sqrt[4]{3x+1)}

Lösung

\int_{0}^{5} \frac{18}{4 3 x + 1} d x

56

Schritte zur Lösung

\int_{0}^{5} \frac{18}{4 3 x + 1} d x

Entferne die Konstante:

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

= 18 \cdot \int_{0}^{5} \frac{1}{4 3 x + 1} d x

Wende U-Substitution an

= 18 \cdot \int_{1}^{16} \frac{1}{3 4 u} d u

Entferne die Konstante:

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

= 18 \cdot \frac{1}{3} \cdot \int_{1}^{16} \frac{1}{4 u} d u

Wende die Potenzregel an

= 18 \cdot \frac{1}{3} [\frac{4}{3} u^{\frac{3}{4}}]_{1}^{16}

Vereinfache

18 \cdot \frac{1}{3} [\frac{4}{3} u^{\frac{3}{4}}]_{1}^{16} : 6 [\frac{4}{3} u^{\frac{3}{4}}]_{1}^{16}

= 6 [\frac{4}{3} u^{\frac{3}{4}}]_{1}^{16}

Berechne die Grenzen:

\frac{28}{3}

= 6 \cdot \frac{28}{3}

Vereinfache

= 56

\int x cos (5 π) x^{2} d x

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