integral of rsqrt(4-r^2)-r^2

解

\int r 4 - r^{2} - r^{2} d r

- \frac{1}{3} (4 - r^{2})^{\frac{3}{2}} - \frac{r ^{3}}{3} + C

解答ステップ

\int r 4 - r^{2} - r^{2} d r

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int r 4 - r^{2} d r - \int r^{2} d r

\int r 4 - r^{2} d r = - \frac{1}{3} (4 - r^{2})^{\frac{3}{2}}

\int r^{2} d r = \frac{r ^{3}}{3}

= - \frac{1}{3} (4 - r^{2})^{\frac{3}{2}} - \frac{r ^{3}}{3}

定数を解答に追加する

= - \frac{1}{3} (4 - r^{2})^{\frac{3}{2}} - \frac{r ^{3}}{3} + C