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受欢迎的微积分 >

integral from-2 to 2 of sqrt(8-x^2)

解答

\int_{- 2}^{2} 8 - x^{2} d x

解答

4 (\frac{π}{2} + 1)

+1

十进制

10.28318 \dots

求解步骤

\int_{- 2}^{2} 8 - x^{2} d x

使用三角换元法

= \int_{- \frac{π}{4}}^{\frac{π}{4}} 8 cos^{2} (u) d u

提出常数:

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

= 8 \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} cos^{2} (u) d u

使用三角恒等式改写

= 8 \cdot \int_{- \frac{π}{4}}^{\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

= 8 \cdot \frac{1}{2} \cdot \int_{- \frac{π}{4}}^{\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

= 8 \cdot \frac{1}{2} (\int_{- \frac{π}{4}}^{\frac{π}{4}} 1 d u + \int_{- \frac{π}{4}}^{\frac{π}{4}} cos (2 u) d u)

\int_{- \frac{π}{4}}^{\frac{π}{4}} 1 d u = \frac{π}{2}

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

= 8 \cdot \frac{1}{2} (\frac{π}{2} + 1)

化简

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

= 4 (\frac{π}{2} + 1)

作图

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