What is the integral from 0 to (3q)/2 of 6y(1-y) ?

The integral from 0 to (3q)/2 of 6y(1-y) is 6((9q^2)/8-(9q^3)/8)

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integral from 0 to (3q)/2 of 6y(1-y)

\int_{0}^{\frac{3 q}{2}} 6 y (1 - y) d y

6 (\frac{9 q ^{2}}{8} - \frac{9 q ^{3}}{8})

Solution steps

\int_{0}^{\frac{3 q}{2}} 6 y (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}^{\frac{3 q}{2}} y (1 - y) d y

Expand

y (1 - y) : y - y^{2}

= 6 \cdot \int_{0}^{\frac{3 q}{2}} y - y^{2} 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}^{\frac{3 q}{2}} y d y - \int_{0}^{\frac{3 q}{2}} y^{2} d y)

\int_{0}^{\frac{3 q}{2}} y d y = \frac{9 q ^{2}}{8}

\int_{0}^{\frac{3 q}{2}} y^{2} d y = \frac{9 q ^{3}}{8}

= 6 (\frac{9 q ^{2}}{8} - \frac{9 q ^{3}}{8})

What is the integral from 0 to (3q)/2 of 6y(1-y) ?
The integral from 0 to (3q)/2 of 6y(1-y) is 6((9q^2)/8-(9q^3)/8)