What is the integral from 0 to y of (6(y-x))/8 ?

The integral from 0 to y of (6(y-x))/8 is (3y^2)/8

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integral from 0 to y of (6(y-x))/8

\int_{0}^{y} \frac{6 ( y - x )}{8} d x

\frac{3 y ^{2}}{8}

Solution steps

\int_{0}^{y} \frac{6 ( y - x )}{8} d x

Take the constant out:

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

= \frac{6}{8} \cdot \int_{0}^{y} y - x d x

Cancel the common factor:

2

= \frac{3}{4} \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

= \frac{3}{4} (\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}

= \frac{3}{4} (y^{2} - \frac{y ^{2}}{2})

Simplify

\frac{3}{4} (y^{2} - \frac{y ^{2}}{2}) : \frac{3 y ^{2}}{8}

= \frac{3 y ^{2}}{8}

What is the integral from 0 to y of (6(y-x))/8 ?
The integral from 0 to y of (6(y-x))/8 is (3y^2)/8