integral from 0 to 9 of |sqrt(x)-1|

解答

\int_{0}^{9} x - 1 d x

\frac{29}{3}

十进制

9.66666 \dots

求解步骤

\int_{0}^{9} x - 1 d x

拿掉绝对值

= \int_{0}^{1} - x + 1 d x + \int_{1}^{9} x - 1 d x

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

\int_{1}^{9} x - 1 d x = \frac{28}{3}

= \frac{1}{3} + \frac{28}{3}

化简

\frac{1}{3} + \frac{28}{3} : \frac{29}{3}

= \frac{29}{3}

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