What is the integral from 0 to 1/2 of 96x^2y^3 ?

The integral from 0 to 1/2 of 96x^2y^3 is 4y^3

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integral from 0 to 1/2 of 96x^2y^3

\int_{0}^{\frac{1}{2}} 96 x^{2} y^{3} d x

4 y^{3}

Solution steps

\int_{0}^{\frac{1}{2}} 96 x^{2} y^{3} d x

Take the constant out:

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

= 96 y^{3} \cdot \int_{0}^{\frac{1}{2}} x^{2} d x

Apply the Power Rule

: [\frac{x ^{3}}{3}]_{0}^{\frac{1}{2}}

= 96 y^{3} [\frac{x ^{3}}{3}]_{0}^{\frac{1}{2}}

Compute the boundaries:

\frac{1}{24}

= 96 y^{3} \frac{1}{24}

Simplify

= 4 y^{3}

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What is the integral from 0 to 1/2 of 96x^2y^3 ?
The integral from 0 to 1/2 of 96x^2y^3 is 4y^3