integral from 0 to 2 of x^3sqrt(9x^4+1)

Lösung

\int_{0}^{2} x^{3} 9 x^{4} + 1 d x

\frac{145 145 - 1}{54}

Dezimale

32.31539 \dots

Schritte zur Lösung

\int_{0}^{2} x^{3} 9 x^{4} + 1 d x

Wende U-Substitution an

= \int_{1}^{145} \frac{u}{36} d u

Entferne die Konstante:

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

= \frac{1}{36} \cdot \int_{1}^{145} u d u

Wende die Potenzregel an

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{145}

Berechne die Grenzen:

\frac{290 145}{3} - \frac{2}{3}

= \frac{1}{36} (\frac{290 145}{3} - \frac{2}{3})

Vereinfache

= \frac{145 145 - 1}{54}

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