What is the integral from 0 to y of 6(y-x) ?

The integral from 0 to y of 6(y-x) is 3y^2

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integral from 0 to y of 6(y-x)

\int_{0}^{y} 6 (y - x) d x

3 y^{2}

Solution steps

\int_{0}^{y} 6 (y - x) d x

Take the constant out:

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

= 6 \cdot \int_{0}^{y} y - x 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

= 6 (\int_{0}^{y} y d x - \int_{0}^{y} x d x)

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

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

= 6 (y^{2} - \frac{y ^{2}}{2})

Simplify

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

= 3 y^{2}

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What is the integral from 0 to y of 6(y-x) ?
The integral from 0 to y of 6(y-x) is 3y^2