integral from 0 to 3 of sqrt(9-r^2)r

Lösung

\int_{0}^{3} 9 - r^{2} r d r

9

Schritte zur Lösung

\int_{0}^{3} 9 - r^{2} r d r

Wende U-Substitution an

= \int_{9}^{0} - \frac{u}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{9} - \frac{u}{2} d u

Entferne die Konstante:

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

= - (- \frac{1}{2} \cdot \int_{0}^{9} u d u)

Wende die Potenzregel an

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9})

Vereinfache

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

Berechne die Grenzen:

18

= \frac{1}{2} \cdot 18

Vereinfache

= 9

\int_{1}^{e^{x}} 7 ln (x) d x

\int_{3}^{4} e^{x} d x

\int_{1}^{3} 4^{3 x} d x

\int_{3}^{5} sin (I n x) d x

\int_{0}^{4} x + 1 d x