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受欢迎的微积分 >

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

解答

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

解答

\frac{π}{2}

+1

十进制

1.57079 \dots

求解步骤

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

对

2 x - x^{2}

配方:

- (x - 1)^{2} + 1

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

使用换元积分法

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

使用三角换元法

= \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (v) d v

使用三角恒等式改写

= \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 + cos ( 2 v )}{2} d v

提出常数:

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

= \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} 1 + cos (2 v) d v

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

= \frac{1}{2} (\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d v + \int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 v) d v)

\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d v = π

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 v) d v = 0

= \frac{1}{2} (π + 0)

化简

\frac{1}{2} (π + 0) : \frac{π}{2}

= \frac{π}{2}

作图

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