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人気のある微分積分 >

integral from 0 to 2 of-sqrt(5-x^2)

解

\int_{0}^{2} - 5 - x^{2} d x

解

- \frac{5 ( 2 arcsin ( \frac{2}{5} ) + sin ( 2 arcsin ( \frac{2}{5} ) ) )}{4}

+1

十進法表記

- 3.76787 \dots

解答ステップ

\int_{0}^{2} - 5 - x^{2} d x

定数を除く:

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

= - \int_{0}^{2} 5 - x^{2} d x

三角関数による置換を適用する

= - \int_{0}^{a r c s i n (\frac{2}{5})} 5 cos^{2} (u) d u

定数を除く:

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

= - 5 \cdot \int_{0}^{a r c s i n (\frac{2}{5})} cos^{2} (u) d u

三角関数の公式を使用して書き換える

= - 5 \cdot \int_{0}^{a r c s i n (\frac{2}{5})} \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

= - 5 \cdot \frac{1}{2} \cdot \int_{0}^{a r c s i n (\frac{2}{5})} 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

= - 5 \cdot \frac{1}{2} (\int_{0}^{a r c s i n (\frac{2}{5})} 1 d u + \int_{0}^{a r c s i n (\frac{2}{5})} cos (2 u) d u)

\int_{0}^{a r c s i n (\frac{2}{5})} 1 d u = arcsin (\frac{2}{5})

\int_{0}^{a r c s i n (\frac{2}{5})} cos (2 u) d u = \frac{sin ( 2 arcsin ( \frac{2}{5} ) )}{2}

= - 5 \cdot \frac{1}{2} arcsin (\frac{2}{5}) + \frac{sin ( 2 arcsin ( \frac{2}{5} ) )}{2}

簡素化

- 5 \cdot \frac{1}{2} arcsin (\frac{2}{5}) + \frac{sin ( 2 arcsin ( \frac{2}{5} ) )}{2} : - \frac{5 ( 2 arcsin ( \frac{2}{5} ) + sin ( 2 arcsin ( \frac{2}{5} ) ) )}{4}

= - \frac{5 ( 2 arcsin ( \frac{2}{5} ) + sin ( 2 arcsin ( \frac{2}{5} ) ) )}{4}

グラフ

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