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受欢迎的微积分 >

integral from 0 to 2 of sqrt(9-x^2)

解答

\int_{0}^{2} 9 - x^{2} d x

解答

\frac{9}{2} arcsin (\frac{2}{3}) + 5

+1

十进制

5.51984 \dots

求解步骤

\int_{0}^{2} 9 - x^{2} d x

使用三角换元法

= \int_{0}^{a r c s i n (\frac{2}{3})} 9 cos^{2} (u) d u

提出常数:

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

= 9 \cdot \int_{0}^{a r c s i n (\frac{2}{3})} cos^{2} (u) d u

使用三角恒等式改写

= 9 \cdot \int_{0}^{a r c s i n (\frac{2}{3})} \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

= 9 \cdot \frac{1}{2} \cdot \int_{0}^{a r c s i n (\frac{2}{3})} 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

= 9 \cdot \frac{1}{2} (\int_{0}^{a r c s i n (\frac{2}{3})} 1 d u + \int_{0}^{a r c s i n (\frac{2}{3})} cos (2 u) d u)

\int_{0}^{a r c s i n (\frac{2}{3})} 1 d u = arcsin (\frac{2}{3})

\int_{0}^{a r c s i n (\frac{2}{3})} cos (2 u) d u = \frac{2 5}{9}

= 9 \cdot \frac{1}{2} (arcsin (\frac{2}{3}) + \frac{2 5}{9})

化简

9 \cdot \frac{1}{2} (arcsin (\frac{2}{3}) + \frac{2 5}{9}) : \frac{9}{2} arcsin (\frac{2}{3}) + 5

= \frac{9}{2} arcsin (\frac{2}{3}) + 5

作图

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