integral from 0 to 1 of 18 1/(x^p)

解答

\int_{0}^{1} 18 \frac{1}{x ^{p}} d x

\frac{18}{- p + 1}

求解步骤

\int_{0}^{1} 18 \cdot \frac{1}{x ^{p}} d x

提出常数:

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

= 18 \cdot \int_{0}^{1} \frac{1}{x ^{p}} d x

使用幂法则

= 18 [\frac{x ^{- p + 1}}{- p + 1}]_{0}^{1}

带入上下限计算后得:

\frac{1}{- p + 1}

= 18 \cdot \frac{1}{- p + 1}

化简

= \frac{18}{- p + 1}

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