integral from 0 to 1 of x^3(2+x^4)^3

Lösung

\int_{0}^{1} x^{3} (2 + x^{4})^{3} d x

\frac{65}{16}

Dezimale

4.0625

Schritte zur Lösung

\int_{0}^{1} x^{3} (2 + x^{4})^{3} d x

Wende U-Substitution an

= \int_{2}^{3} \frac{u ^{3}}{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_{2}^{3} u^{3} d u

Wende die Potenzregel an

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

Berechne die Grenzen:

\frac{65}{4}

= \frac{1}{4} \cdot \frac{65}{4}

Vereinfache

= \frac{65}{16}

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