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

Lösung

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

\frac{17 17 - 1}{12}

Dezimale

5.75773 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{17} \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}^{17} u d u

Wende die Potenzregel an

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

Berechne die Grenzen:

\frac{34 17}{3} - \frac{2}{3}

= \frac{1}{8} (\frac{34 17}{3} - \frac{2}{3})

Vereinfache

= \frac{17 17 - 1}{12}

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