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

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

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

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

1

Solution steps

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

Take the constant out:

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

= \frac{3}{2} \cdot \int_{0}^{1} 1 - 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

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

\int_{0}^{1} 1 d y = 1

\int_{0}^{1} y^{2} d y = \frac{1}{3}

= \frac{3}{2} (1 - \frac{1}{3})

Simplify

\frac{3}{2} (1 - \frac{1}{3}) : 1

= 1

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