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人気のある微分積分 >

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

解

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

解

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

+1

十進法表記

7.53574 \dots

解答ステップ

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

三角関数による置換を適用する

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

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

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

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

簡素化

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

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

グラフ

人気の例

integral from 1 to e of (e^x+2x-1)

\int_{1}^{e} (e^{x} + 2 x - 1) d x

integral from 0 to 5 of (3x+2)

\int_{0}^{5} (3 x + 2) d x

integral from 0 to 3/4 of sqrt(1+4x)

\int_{0}^{\frac{3}{4}} 1 + 4 x d x

integral from-3 to 5 of 1

\int_{- 3}^{5} 1 d x

integral from 0 to 3 of (x+3)

\int_{0}^{3} (x + 3) d x