integral from 0.75 to 1 of 630x^4(1-x)^4

解答

\int_{0.75}^{1} 630 x^{4} (1 - x)^{4} d x

0.04892 \dots

求解步骤

\int_{0.75}^{1} 630 x^{4} (1 - x)^{4} d x

提出常数:

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

= 630 \cdot \int_{0.75}^{1} x^{4} (1 - x)^{4} d x

乘开

x^{4} (1 - x)^{4} : x^{4} - 4 x^{5} + 6 x^{6} - 4 x^{7} + x^{8}

= 630 \cdot \int_{0.75}^{1} x^{4} - 4 x^{5} + 6 x^{6} - 4 x^{7} + x^{8} d x

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

= 630 (\int_{0.75}^{1} x^{4} d x - \int_{0.75}^{1} 4 x^{5} d x + \int_{0.75}^{1} 6 x^{6} d x - \int_{0.75}^{1} 4 x^{7} d x + \int_{0.75}^{1} x^{8} d x)

\int_{0.75}^{1} x^{4} d x = 0.1525390625

\int_{0.75}^{1} 4 x^{5} d x = 0.54801 \dots

\int_{0.75}^{1} 6 x^{6} d x = 0.74272 \dots

\int_{0.75}^{1} 4 x^{7} d x = 0.44994 \dots

\int_{0.75}^{1} x^{8} d x = 0.10276 \dots

= 630 (0.1525390625 - 0.54801 \dots + 0.74272 \dots - 0.44994 \dots + 0.10276 \dots)

化简

= 0.04892 \dots

\int_{0}^{a} 1 - x^{2} d x

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

\int_{0}^{y} x e^{y^{2}} d x

\int_{- 1}^{1} 2 - x^{2} - x^{2} d x

\int_{0}^{4} ∣ 1 - 2 x ∣ d x