What is the integral from 0 to y/3 of (y^2-9x^2)/2 ?

The integral from 0 to y/3 of (y^2-9x^2)/2 is (y^3)/9

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integral from 0 to y/3 of (y^2-9x^2)/2

\int_{0}^{\frac{y}{3}} \frac{y ^{2} - 9 x ^{2}}{2} d x

\frac{y ^{3}}{9}

Solution steps

\int_{0}^{\frac{y}{3}} \frac{y ^{2} - 9 x ^{2}}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{y}{3}} y^{2} - 9 x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{\frac{y}{3}} y^{2} d x - \int_{0}^{\frac{y}{3}} 9 x^{2} d x)

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

\int_{0}^{\frac{y}{3}} 9 x^{2} d x = \frac{y ^{3}}{9}

= \frac{1}{2} (\frac{y ^{3}}{3} - \frac{y ^{3}}{9})

Simplify

\frac{1}{2} (\frac{y ^{3}}{3} - \frac{y ^{3}}{9}) : \frac{y ^{3}}{9}

= \frac{y ^{3}}{9}

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What is the integral from 0 to y/3 of (y^2-9x^2)/2 ?
The integral from 0 to y/3 of (y^2-9x^2)/2 is (y^3)/9