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受欢迎的微积分 >

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

解答

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

解答

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

+1

十进制

7.53574 \dots

求解步骤

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

使用三角换元法

= \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

提出常数:

\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

使用三角恒等式改写

= 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

提出常数:

\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

使用积分加法定则:

\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})))

化简

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}

作图

流行的例子

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