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

The integral from 0 to 1-y of (6y) is 3(-y+1)^2

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

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

3 (- y + 1)^{2}

Solution steps

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

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{0}^{1 - y}

= 6 [\frac{y ^{2}}{2}]_{0}^{1 - y}

Compute the boundaries:

\frac{1}{2} (- y + 1)^{2}

= 6 \cdot \frac{1}{2} (- y + 1)^{2}

Simplify

= 3 (- y + 1)^{2}

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What is the integral from 0 to 1-y of (6y) ?
The integral from 0 to 1-y of (6y) is 3(-y+1)^2