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受欢迎的微积分 >

integral from 0 to 1 of sqrt(2-x^2)

解答

\int_{0}^{1} 2 - x^{2} d x

解答

\frac{π}{4} + \frac{1}{2}

+1

十进制

1.28539 \dots

求解步骤

\int_{0}^{1} 2 - x^{2} d x

使用三角换元法

= \int_{0}^{\frac{π}{4}} 2 cos^{2} (u) d u

提出常数:

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

= 2 \cdot \int_{0}^{\frac{π}{4}} cos^{2} (u) d u

使用三角恒等式改写

= 2 \cdot \int_{0}^{\frac{π}{4}} \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

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} 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

= 2 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{4}} 1 d u + \int_{0}^{\frac{π}{4}} cos (2 u) d u)

\int_{0}^{\frac{π}{4}} 1 d u = \frac{π}{4}

\int_{0}^{\frac{π}{4}} cos (2 u) d u = \frac{1}{2}

= 2 \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2})

化简

2 \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2}) : \frac{π}{4} + \frac{1}{2}

= \frac{π}{4} + \frac{1}{2}

作图

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