integral from 0 to 1 of pi(x-x^3)^2

解答

\int_{0}^{1} π (x - x^{3})^{2} d x

π \frac{8}{105}

十进制

0.23935 \dots

求解步骤

\int_{0}^{1} π (x - x^{3})^{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}^{1} (x - x^{3})^{2} d x

乘开

(x - x^{3})^{2} : x^{2} - 2 x^{4} + x^{6}

= π \cdot \int_{0}^{1} x^{2} - 2 x^{4} + x^{6} 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

= π (\int_{0}^{1} x^{2} d x - \int_{0}^{1} 2 x^{4} d x + \int_{0}^{1} x^{6} d x)

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

\int_{0}^{1} 2 x^{4} d x = \frac{2}{5}

\int_{0}^{1} x^{6} d x = \frac{1}{7}

= π (\frac{1}{3} - \frac{2}{5} + \frac{1}{7})

化简

\frac{1}{3} - \frac{2}{5} + \frac{1}{7} : \frac{8}{105}

= π \frac{8}{105}

t an g e n t 2 x^{3} - 6 x^{2} - 18 - 3

a re a y = - 2 x^{2} + 6, y = - 2

\int \frac{1}{sin ^{2} ( x ) 4 cot ( x )} d x

\int 1 + x^{2} \cdot x^{5} d x

t an g e n t y = - x^{3} + 2 x^{2} - 4, at x = - 1