What is the integral from 0 to (1-u)/4 of 2(1-y) ?

The integral from 0 to (1-u)/4 of 2(1-y) is 2((-u+1)/4-((-u+1)^2)/(32))

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integral from 0 to (1-u)/4 of 2(1-y)

\int_{0}^{\frac{1 - u}{4}} 2 (1 - y) d y

2 (\frac{- u + 1}{4} - \frac{( - u + 1 ) ^{2}}{32})

Solution steps

\int_{0}^{\frac{1 - u}{4}} 2 (1 - y) d y

Take the constant out:

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

= 2 \cdot \int_{0}^{\frac{1 - u}{4}} 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

= 2 (\int_{0}^{\frac{1 - u}{4}} 1 d y - \int_{0}^{\frac{1 - u}{4}} y d y)

\int_{0}^{\frac{1 - u}{4}} 1 d y = \frac{- u + 1}{4}

\int_{0}^{\frac{1 - u}{4}} y d y = \frac{( - u + 1 ) ^{2}}{32}

= 2 (\frac{- u + 1}{4} - \frac{( - u + 1 ) ^{2}}{32})

What is the integral from 0 to (1-u)/4 of 2(1-y) ?
The integral from 0 to (1-u)/4 of 2(1-y) is 2((-u+1)/4-((-u+1)^2)/(32))