integral from 0 to 4 of sqrt(4r^2+1)r

Lösung

\int_{0}^{4} 4 r^{2} + 1 r d r

\frac{65 65 - 1}{12}

Dezimale

43.58722 \dots

Schritte zur Lösung

\int_{0}^{4} 4 r^{2} + 1 r d r

Wende U-Substitution an

= \int_{1}^{65} \frac{u}{8} d u

Entferne die Konstante:

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

= \frac{1}{8} \cdot \int_{1}^{65} u d u

Wende die Potenzregel an

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{65}

Berechne die Grenzen:

\frac{130 65}{3} - \frac{2}{3}

= \frac{1}{8} (\frac{130 65}{3} - \frac{2}{3})

Vereinfache

= \frac{65 65 - 1}{12}

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