What is the integral from 0 to x of 8/81 xy ?

The integral from 0 to x of 8/81 xy is (4x^3)/(81)

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integral from 0 to x of 8/81 xy

\int_{0}^{x} \frac{8}{81} x y d y

\frac{4 x ^{3}}{81}

Solution steps

\int_{0}^{x} \frac{8}{81} x y d y

Take the constant out:

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

= \frac{8}{81} x \cdot \int_{0}^{x} y d y

Multiply fractions:

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

= \frac{8 x}{81} \cdot \int_{0}^{x} y d y

Apply the Power Rule

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

= \frac{8 x}{81} [\frac{y ^{2}}{2}]_{0}^{x}

Compute the boundaries:

\frac{x ^{2}}{2}

= \frac{8 x}{81} \cdot \frac{x ^{2}}{2}

Simplify

= \frac{4 x ^{3}}{81}

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What is the integral from 0 to x of 8/81 xy ?
The integral from 0 to x of 8/81 xy is (4x^3)/(81)