integral from 1 to 5 of (6n^{-2}-n^{-3})

解答

\int_{1}^{5} (6 n^{- 2} - n^{- 3}) d n

\frac{108}{25}

十进制

4.32

求解步骤

\int_{1}^{5} 6 n^{- 2} - n^{- 3} d n

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

= \int_{1}^{5} 6 n^{- 2} d n - \int_{1}^{5} n^{- 3} d n

\int_{1}^{5} 6 n^{- 2} d n = \frac{24}{5}

\int_{1}^{5} n^{- 3} d n = \frac{12}{25}

= \frac{24}{5} - \frac{12}{25}

化简

\frac{24}{5} - \frac{12}{25} : \frac{108}{25}

= \frac{108}{25}

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