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受欢迎的微积分 >

integral from-1 to 1 of (2-x^2)^{3/2}

解答

\int_{- 1}^{1} (2 - x^{2})^{\frac{3}{2}} d x

解答

\frac{12 π + 32}{16}

+1

十进制

4.35619 \dots

求解步骤

\int_{- 1}^{1} (2 - x^{2})^{\frac{3}{2}} d x

使用换元积分法

= \int_{- 2}^{2} \frac{( 8 - u ^{2} ) ^{\frac{3}{2}}}{16} d u

提出常数:

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

= \frac{1}{16} \cdot \int_{- 2}^{2} (8 - u^{2})^{\frac{3}{2}} d u

使用分布积分法

= \frac{1}{16} [u (8 - u^{2})^{\frac{3}{2}} - \int - 3 u^{2} (8 - u^{2})^{\frac{1}{2}} d u]_{- 2}^{2}

\int - 3 u^{2} (8 - u^{2})^{\frac{1}{2}} d u = - 3 (\frac{1}{4} u^{3} (- u^{2} + 8)^{\frac{1}{2}} + 8 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}}) - 4 sin (2 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}})))

= \frac{1}{16} [u (8 - u^{2})^{\frac{3}{2}} - (- 3 (\frac{1}{4} u^{3} (- u^{2} + 8)^{\frac{1}{2}} + 8 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}}) - 4 sin (2 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}}))))]_{- 2}^{2}

化简

= \frac{1}{16} [u (8 - u^{2})^{\frac{3}{2}} + 3 (\frac{1}{4} u^{3} (- u^{2} + 8)^{\frac{1}{2}} + 8 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}}) - 4 sin (2 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}})))]_{- 2}^{2}

带入上下限计算后得:

12 π + 32

= \frac{1}{16} (12 π + 32)

化简

= \frac{12 π + 32}{16}

作图

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