What is the integral from 0 to 3 of 4/81 xy ?

The integral from 0 to 3 of 4/81 xy is (2x)/9

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integral from 0 to 3 of 4/81 xy

\int_{0}^{3} \frac{4}{81} x y d y

\frac{2 x}{9}

Solution steps

\int_{0}^{3} \frac{4}{81} x y d y

Take the constant out:

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

= \frac{4}{81} x \cdot \int_{0}^{3} y d y

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{4 x}{81} \cdot \int_{0}^{3} y d y

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{0}^{3}

= \frac{4 x}{81} [\frac{y ^{2}}{2}]_{0}^{3}

Compute the boundaries:

\frac{9}{2}

= \frac{4 x}{81} \cdot \frac{9}{2}

Simplify

= \frac{2 x}{9}

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What is the integral from 0 to 3 of 4/81 xy ?
The integral from 0 to 3 of 4/81 xy is (2x)/9