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人気のある微分積分 >

integral from 0 to 2 of sqrt(10+2x-x^2)

解

\int_{0}^{2} 10 + 2 x - x^{2} d x

解

\frac{11 ( 2 arcsin ( \frac{1}{11} ) + sin ( 2 arcsin ( \frac{1}{11} ) ) )}{2}

+1

十進法表記

6.53132 \dots

解答ステップ

\int_{0}^{2} 10 + 2 x - x^{2} d x

平方完成する

10 + 2 x - x^{2} : - (x - 1)^{2} + 11

= \int_{0}^{2} - (x - 1)^{2} + 11 d x

u置換積分法を適用する

= \int_{- 1}^{1} - u^{2} + 11 d u

三角関数による置換を適用する

= \int_{- a r c s i n (\frac{1}{11})}^{a r c s i n (\frac{1}{11})} 11 cos^{2} (v) d v

定数を除く:

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

= 11 \cdot \int_{- a r c s i n (\frac{1}{11})}^{a r c s i n (\frac{1}{11})} cos^{2} (v) d v

三角関数の公式を使用して書き換える

= 11 \cdot \int_{- a r c s i n (\frac{1}{11})}^{a r c s i n (\frac{1}{11})} \frac{1 + cos ( 2 v )}{2} d v

定数を除く:

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

= 11 \cdot \frac{1}{2} \cdot \int_{- a r c s i n (\frac{1}{11})}^{a r c s i n (\frac{1}{11})} 1 + cos (2 v) d v

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

= 11 \cdot \frac{1}{2} (\int_{- a r c s i n (\frac{1}{11})}^{a r c s i n (\frac{1}{11})} 1 d v + \int_{- a r c s i n (\frac{1}{11})}^{a r c s i n (\frac{1}{11})} cos (2 v) d v)

\int_{- a r c s i n (\frac{1}{11})}^{a r c s i n (\frac{1}{11})} 1 d v = 2 arcsin (\frac{1}{11})

\int_{- a r c s i n (\frac{1}{11})}^{a r c s i n (\frac{1}{11})} cos (2 v) d v = 1 \cdot sin (2 arcsin (\frac{1}{11}))

= 11 \cdot \frac{1}{2} (2 arcsin (\frac{1}{11}) + 1 \cdot sin (2 arcsin (\frac{1}{11})))

簡素化

11 \cdot \frac{1}{2} (2 arcsin (\frac{1}{11}) + 1 \cdot sin (2 arcsin (\frac{1}{11}))) : \frac{11 ( 2 arcsin ( \frac{1}{11} ) + sin ( 2 arcsin ( \frac{1}{11} ) ) )}{2}

= \frac{11 ( 2 arcsin ( \frac{1}{11} ) + sin ( 2 arcsin ( \frac{1}{11} ) ) )}{2}

グラフ

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