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

解答

\int_{- B}^{B} (- B^{2} - x^{2}) d x

- \frac{π}{2} B^{2}

求解步骤

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

提出常数:

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

= - \int_{- B}^{B} B^{2} - x^{2} d x

使用三角换元法

: \int_{- \frac{π}{2}}^{\frac{π}{2}} B^{2} cos^{2} (u) d u

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

提出常数:

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

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

使用三角恒等式改写

= - B^{2} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} \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

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

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

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

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

= - B^{2} \frac{1}{2} (π + 0)

化简

- B^{2} \frac{1}{2} (π + 0) : - \frac{π}{2} B^{2}

= - \frac{π}{2} B^{2}

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