What is the integral from 0 to 1 of [x^{1/n}-x] ?

The integral from 0 to 1 of [x^{1/n}-x] is n/(n+1)-1/2

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integral from 0 to 1 of [x^{1/n}-x]

\int_{0}^{1} [x^{\frac{1}{n}} - x] d x

\frac{n}{n + 1} - \frac{1}{2}

Solution steps

\int_{0}^{1} x^{\frac{1}{n}} - x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int_{0}^{1} x d x = \frac{1}{2}

= \frac{n}{n + 1} - \frac{1}{2}

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What is the integral from 0 to 1 of [x^{1/n}-x] ?
The integral from 0 to 1 of [x^{1/n}-x] is n/(n+1)-1/2