integral from 0 to 4 of pi((12-3x)/4)^2

解答

\int_{0}^{4} π (\frac{12 - 3 x}{4})^{2} d x

π 12

十进制

37.69911 \dots

求解步骤

\int_{0}^{4} π (\frac{12 - 3 x}{4})^{2} d x

提出常数:

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

= π \cdot \int_{0}^{4} (\frac{12 - 3 x}{4})^{2} d x

使用分布积分法

= π [\frac{1}{16} x (12 - 3 x)^{2} - \int - \frac{3}{8} x (- 3 x + 12) d x]_{0}^{4}

\int - \frac{3}{8} x (- 3 x + 12) d x = - \frac{3}{8} (- x^{3} + 6 x^{2})

= π [\frac{1}{16} x (12 - 3 x)^{2} - (- \frac{3}{8} (- x^{3} + 6 x^{2}))]_{0}^{4}

化简

π [\frac{1}{16} x (12 - 3 x)^{2} - (- \frac{3}{8} (- x^{3} + 6 x^{2}))]_{0}^{4} : π [\frac{1}{16} (3 x^{3} - 36 x^{2} + 144 x)]_{0}^{4}

= π [\frac{1}{16} (3 x^{3} - 36 x^{2} + 144 x)]_{0}^{4}

带入上下限计算后得:

12

= π 12

\int_{0}^{1} (3 x + 2)^{4} d x

\int_{0}^{1} 3 x e^{- x^{2}} d x

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

\int_{- \frac{π}{2}}^{\frac{π}{2}} 9 cos (x) d x

\int_{1}^{3} \frac{z ^{2} + 2}{z} d z