What is the integral from 0 to 4 of (x^2)/(16) ?

The integral from 0 to 4 of (x^2)/(16) is 4/3

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integral from 0 to 4 of (x^2)/(16)

\int_{0}^{4} \frac{x ^{2}}{16} d x

\frac{4}{3}

Solution steps

\int_{0}^{4} \frac{x ^{2}}{16} d x

Take the constant out:

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

= \frac{1}{16} \cdot \int_{0}^{4} x^{2} d x

Apply the Power Rule

: [\frac{x ^{3}}{3}]_{0}^{4}

= \frac{1}{16} [\frac{x ^{3}}{3}]_{0}^{4}

Compute the boundaries:

\frac{64}{3}

= \frac{1}{16} \cdot \frac{64}{3}

Simplify

= \frac{4}{3}

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What is the integral from 0 to 4 of (x^2)/(16) ?
The integral from 0 to 4 of (x^2)/(16) is 4/3