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

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

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

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

1

Solution steps

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

Take the constant out:

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

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

Expand

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

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

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

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

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

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

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

Simplify

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

= 1

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