What is the integral from 0 to 1 of-((x-1)^3)/3 ?

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

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integral from 0 to 1 of-((x-1)^3)/3

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

\frac{1}{12}

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{- 1}^{0} u^{3} d u

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

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{- 1}^{0}

= - \frac{1}{3} [\frac{u ^{4}}{4}]_{- 1}^{0}

Compute the boundaries:

- \frac{1}{4}

= - \frac{1}{3} (- \frac{1}{4})

Simplify

= \frac{1}{12}

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