What is the integral from 0 to 1 of nx(1-x^2)^n ?

The integral from 0 to 1 of nx(1-x^2)^n is n/(2(n+1))

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integral from 0 to 1 of nx(1-x^2)^n

\int_{0}^{1} n x (1 - x^{2})^{n} d x

\frac{n}{2 ( n + 1 )}

Solution steps

\int_{0}^{1} n x (1 - x^{2})^{n} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= n \cdot \int_{0}^{1} x (1 - x^{2})^{n} d x

Apply u-substitution

: \int_{0}^{1} \frac{( 1 - u ) ^{n}}{2} d u

= n \cdot \int_{0}^{1} \frac{( 1 - u ) ^{n}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= n \frac{1}{2} \cdot \int_{0}^{1} (1 - u)^{n} d u

Apply u-substitution

: \int_{1}^{0} - v^{n} d v

= n \frac{1}{2} \cdot \int_{1}^{0} - v^{n} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= n \frac{1}{2} (- \int_{0}^{1} - v^{n} d v)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= n \frac{1}{2} (- (- \int_{0}^{1} v^{n} d v))

Apply the Power Rule

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

= n \frac{1}{2} (- (- [\frac{v ^{n + 1}}{n + 1}]_{0}^{1}))

Simplify

= \frac{1}{2} n [\frac{v ^{n + 1}}{n + 1}]_{0}^{1}

Compute the boundaries:

\frac{1}{n + 1}

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

Simplify

= \frac{n}{2 ( n + 1 )}

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