integral from-2 to 4 of |t^3-9t|

解答

\int_{- 2}^{4} t^{3} - 9 t d t

\frac{93}{2}

十进制

46.5

求解步骤

\int_{- 2}^{4} t^{3} - 9 t d t

拿掉绝对值

= \int_{- 2}^{0} t^{3} - 9 t d t + \int_{0}^{3} - t^{3} + 9 t d t + \int_{3}^{4} t^{3} - 9 t d t

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

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

\int_{3}^{4} t^{3} - 9 t d t = \frac{49}{4}

= 14 + \frac{81}{4} + \frac{49}{4}

化简

14 + \frac{81}{4} + \frac{49}{4} : \frac{93}{2}

= \frac{93}{2}

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