integral from-2 to 5 of |-x^3+9x|

解答

\int_{- 2}^{5} - x^{3} + 9 x d x

\frac{393}{4}

十进制

98.25

求解步骤

\int_{- 2}^{5} - x^{3} + 9 x d x

拿掉绝对值

= \int_{- 2}^{0} x^{3} - 9 x d x + \int_{0}^{3} - x^{3} + 9 x d x + \int_{3}^{5} x^{3} - 9 x d x

\int_{- 2}^{0} x^{3} - 9 x d x = 14

\int_{0}^{3} - x^{3} + 9 x d x = \frac{81}{4}

\int_{3}^{5} x^{3} - 9 x d x = 64

= 14 + \frac{81}{4} + 64

化简

14 + \frac{81}{4} + 64 : \frac{393}{4}

= \frac{393}{4}