What is the integral from 0 to 1 of (x^2)/3 ?

The integral from 0 to 1 of (x^2)/3 is 1/9

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integral from 0 to 1 of (x^2)/3

\int_{0}^{1} \frac{x ^{2}}{3} d x

\frac{1}{9}

Solution steps

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

Apply the Power Rule

: [\frac{x ^{3}}{3}]_{0}^{1}

= \frac{1}{3} [\frac{x ^{3}}{3}]_{0}^{1}

Compute the boundaries:

\frac{1}{3}

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

Simplify

= \frac{1}{9}

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What is the integral from 0 to 1 of (x^2)/3 ?
The integral from 0 to 1 of (x^2)/3 is 1/9