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}

\int_{0}^{y} 3 y d y

\int_{2 + c o s (x)}^{x^{2} + 1} \frac{1}{t} d t

\int_{0}^{1} x^{4} (5 + 3 x^{5})^{4} d x

\int_{0}^{1} - 1 - x^{2} d x

\int_{0}^{2} x^{2} - 1 d x

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