What is the integral from 0 to y of 6(1-y) ?

The integral from 0 to y of 6(1-y) is 6(y-(y^2)/2)

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integral from 0 to y of 6(1-y)

\int_{0}^{y} 6 (1 - y) d y

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

Solution steps

\int_{0}^{y} 6 (1 - y) d y

Take the constant out:

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

= 6 \cdot \int_{0}^{y} 1 - y d y

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} 1 d y - \int_{0}^{y} y d y)

\int_{0}^{y} 1 d y = y

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

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

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What is the integral from 0 to y of 6(1-y) ?
The integral from 0 to y of 6(1-y) is 6(y-(y^2)/2)