What is the integral from 0 to 1 of [x-x^k] ?

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

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integral from 0 to 1 of [x-x^k]

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

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

Solution steps

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

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

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

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

\frac{d y}{d x} - \frac{y}{2} = \frac{5}{2}

\frac{x}{1 - x}, 0

h \to 0 lim (h + 2)

\int e^{- \frac{x}{2}} cos (3 x) d x

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