integral of sqrt(16-r^2)r

Lösung

\int 16 - r^{2} r d r

- \frac{1}{3} (16 - r^{2})^{\frac{3}{2}} + C

Schritte zur Lösung

\int 16 - r^{2} r d r

Wende U-Substitution an

= \int - \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 u d u

Wende die Potenzregel an

= - \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Setze in

u = 16 - r^{2}

ein

= - \frac{1}{2} \cdot \frac{2}{3} (16 - r^{2})^{\frac{3}{2}}

Vereinfache

- \frac{1}{2} \cdot \frac{2}{3} (16 - r^{2})^{\frac{3}{2}} : - \frac{1}{3} (16 - r^{2})^{\frac{3}{2}}

= - \frac{1}{3} (16 - r^{2})^{\frac{3}{2}}

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{3} (16 - r^{2})^{\frac{3}{2}} + C

d er i v a t i v e \frac{7}{4 x}

y^{^{''}} + 7 y^{^{'}} + 12 = 0

x \to 2 + lim (\frac{1}{x})

\int 6 sin^{5} (x) d x

(csc (x))^{^{'}}