What is the integral from y to 1 of 6y ?

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

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integral from y to 1 of 6y

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

3 (- y^{2} + 1)

Solution steps

\int_{y}^{1} 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_{y}^{1} y d y

Apply the Power Rule

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

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

Compute the boundaries:

\frac{1}{2} - \frac{y ^{2}}{2}

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

Simplify

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

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