What is the integral from 0 to 1 of x^n ?

The integral from 0 to 1 of x^n is 1/(n+1)

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integral from 0 to 1 of x^n

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

\frac{1}{n + 1}

Solution steps

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

Apply the Power Rule

: [\frac{x ^{n + 1}}{n + 1}]_{0}^{1}

= [\frac{x ^{n + 1}}{n + 1}]_{0}^{1}

Compute the boundaries:

\frac{1}{n + 1}

= \frac{1}{n + 1}

\frac{4}{e ^{x} + e ^{- x}}

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0

\frac{x}{( x ^{2} + 1 ) ( x ^{2} + 9 )}

\int t (t^{2} + 1)^{8} d t

What is the integral from 0 to 1 of x^n ?
The integral from 0 to 1 of x^n is 1/(n+1)