What is the integral from 0 to 1 of 3x(x^2-1)^{12} ?

The integral from 0 to 1 of 3x(x^2-1)^{12} is 3/26

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integral from 0 to 1 of 3x(x^2-1)^{12}

\int_{0}^{1} 3 x (x^{2} - 1)^{12} d x

\frac{3}{26}

Decimal

0.11538 \dots

Solution steps

\int_{0}^{1} 3 x (x^{2} - 1)^{12} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{1} x (x^{2} - 1)^{12} d x

Apply u-substitution

: \int_{- 1}^{0} \frac{u ^{12}}{2} d u

= 3 \cdot \int_{- 1}^{0} \frac{u ^{12}}{2} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int_{- 1}^{0} u^{12} d u

Apply the Power Rule

: [\frac{u ^{13}}{13}]_{- 1}^{0}

= 3 \cdot \frac{1}{2} [\frac{u ^{13}}{13}]_{- 1}^{0}

Simplify

3 \cdot \frac{1}{2} [\frac{u ^{13}}{13}]_{- 1}^{0} : \frac{3}{2} [\frac{u ^{13}}{13}]_{- 1}^{0}

= \frac{3}{2} [\frac{u ^{13}}{13}]_{- 1}^{0}

Compute the boundaries:

\frac{1}{13}

= \frac{3}{2} \cdot \frac{1}{13}

Simplify

= \frac{3}{26}

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What is the integral from 0 to 1 of 3x(x^2-1)^{12} ?
The integral from 0 to 1 of 3x(x^2-1)^{12} is 3/26