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

Lösung

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

\frac{5}{8}

Dezimale

0.625

Schritte zur Lösung

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

Wende die partielle Integration an

= [\frac{1}{4} x^{4} (2 + x^{4}) - \int x^{7} d x]_{0}^{1}

\int x^{7} d x = \frac{x ^{8}}{8}

= [\frac{1}{4} x^{4} (2 + x^{4}) - \frac{x ^{8}}{8}]_{0}^{1}

Vereinfache

[\frac{1}{4} x^{4} (2 + x^{4}) - \frac{x ^{8}}{8}]_{0}^{1} : [\frac{1}{8} (4 x^{4} + x^{8})]_{0}^{1}

= [\frac{1}{8} (4 x^{4} + x^{8})]_{0}^{1}

Berechne die Grenzen:

\frac{5}{8}

= \frac{5}{8}

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