integral from 0 to 1 of xsqrt(40x^2+9)

Lösung

\int_{0}^{1} x 40 x^{2} + 9 d x

\frac{79}{30}

Dezimale

2.63333 \dots

Schritte zur Lösung

\int_{0}^{1} x 40 x^{2} + 9 d x

Wende U-Substitution an

= \int_{9}^{49} \frac{u}{80} d u

Entferne die Konstante:

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

= \frac{1}{80} \cdot \int_{9}^{49} u d u

Wende die Potenzregel an

= \frac{1}{80} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{49}

Berechne die Grenzen:

\frac{632}{3}

= \frac{1}{80} \cdot \frac{632}{3}

Vereinfache

= \frac{79}{30}

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