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

The integral from 1 to 5 of (6n^{-2}-n^{-3}) is 108/25

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

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

\frac{108}{25}

Decimal Notation

4.32

Solution steps

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

Apply the Sum Rule:

\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}

Simplify

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

= \frac{108}{25}

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What is the integral from 1 to 5 of (6n^{-2}-n^{-3}) ?
The integral from 1 to 5 of (6n^{-2}-n^{-3}) is 108/25