integral from 0 to 1 of ((-x+1)^3)/6

解答

\int_{0}^{1} \frac{( - x + 1 ) ^{3}}{6} d x

\frac{1}{24}

十进制

0.04166 \dots

求解步骤

\int_{0}^{1} \frac{( - x + 1 ) ^{3}}{6} d x

提出常数:

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

= \frac{1}{6} \cdot \int_{0}^{1} (- x + 1)^{3} d x

乘开

(- x + 1)^{3} : - x^{3} + 3 x^{2} - 3 x + 1

= \frac{1}{6} \cdot \int_{0}^{1} - x^{3} + 3 x^{2} - 3 x + 1 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

= \frac{1}{6} (- \int_{0}^{1} x^{3} d x + \int_{0}^{1} 3 x^{2} d x - \int_{0}^{1} 3 x d x + \int_{0}^{1} 1 d x)

\int_{0}^{1} x^{3} d x = \frac{1}{4}

\int_{0}^{1} 3 x^{2} d x = 1

\int_{0}^{1} 3 x d x = \frac{3}{2}

\int_{0}^{1} 1 d x = 1

= \frac{1}{6} (- \frac{1}{4} + 1 - \frac{3}{2} + 1)

化简

\frac{1}{6} (- \frac{1}{4} + 1 - \frac{3}{2} + 1) : \frac{1}{24}

= \frac{1}{24}

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