integral from 0 to 2pi of r

Lösung

\int_{0}^{2 π} r d r

2 π^{2}

Dezimale

19.73920 \dots

Schritte zur Lösung

\int_{0}^{2 π} r d r

Wende die Potenzregel an

= [\frac{r ^{2}}{2}]_{0}^{2 π}

Berechne die Grenzen:

2 π^{2}

= 2 π^{2}

\int_{4}^{1} - x d x

\int_{0}^{2} 12 x^{2} d x

\int_{0}^{y^{2}} e^{\frac{x}{y ^{2}}} d x

\int_{2}^{10} (x^{2} + \frac{1}{x}) d x

\int_{0}^{2} (4 x^{2} - x^{4}) d x