What is the integral from 0 to 1 of 8/3 x^3y ?

The integral from 0 to 1 of 8/3 x^3y is (2y)/3

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integral from 0 to 1 of 8/3 x^3y

\int_{0}^{1} \frac{8}{3} x^{3} y d x

\frac{2 y}{3}

Solution steps

\int_{0}^{1} \frac{8}{3} x^{3} y d x

Take the constant out:

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

= \frac{8}{3} y \cdot \int_{0}^{1} x^{3} d x

Multiply fractions:

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

= \frac{8 y}{3} \cdot \int_{0}^{1} x^{3} d x

Apply the Power Rule

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

= \frac{8 y}{3} [\frac{x ^{4}}{4}]_{0}^{1}

Compute the boundaries:

\frac{1}{4}

= \frac{8 y}{3} \cdot \frac{1}{4}

Simplify

= \frac{2 y}{3}

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What is the integral from 0 to 1 of 8/3 x^3y ?
The integral from 0 to 1 of 8/3 x^3y is (2y)/3