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

解

\int_{0}^{r} (r^{2} - y^{2}) d y

\frac{π}{4} r^{2}

解答ステップ

\int_{0}^{r} r^{2} - y^{2} d y

三角関数による置換を適用する

= \int_{0}^{\frac{π}{2}} r^{2} cos^{2} (u) d u

定数を除く:

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

= r^{2} \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

三角関数の公式を使用して書き換える

= r^{2} \cdot \int_{0}^{\frac{π}{2}} \frac{1 + cos ( 2 u )}{2} d u

定数を除く:

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

= r^{2} \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 + cos (2 u) d u

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

= r^{2} \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u + \int_{0}^{\frac{π}{2}} cos (2 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 u) d u = 0

= r^{2} \frac{1}{2} (\frac{π}{2} + 0)

簡素化

r^{2} \frac{1}{2} (\frac{π}{2} + 0) : \frac{π}{4} r^{2}

= \frac{π}{4} r^{2}