What is the integral from 0 to 1 of 4(1-y)^3 ?

The integral from 0 to 1 of 4(1-y)^3 is 1

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integral from 0 to 1 of 4(1-y)^3

\int_{0}^{1} 4 (1 - y)^{3} d y

1

Solution steps

\int_{0}^{1} 4 (1 - y)^{3} d y

Take the constant out:

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

= 4 \cdot \int_{0}^{1} (1 - y)^{3} d y

Expand

(1 - y)^{3} : 1 - 3 y + 3 y^{2} - y^{3}

= 4 \cdot \int_{0}^{1} 1 - 3 y + 3 y^{2} - y^{3} 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

= 4 (\int_{0}^{1} 1 d y - \int_{0}^{1} 3 y d y + \int_{0}^{1} 3 y^{2} d y - \int_{0}^{1} y^{3} d y)

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

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

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

\int_{0}^{1} y^{3} d y = \frac{1}{4}

= 4 (1 - \frac{3}{2} + 1 - \frac{1}{4})

Simplify

4 (1 - \frac{3}{2} + 1 - \frac{1}{4}) : 1

= 1

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What is the integral from 0 to 1 of 4(1-y)^3 ?
The integral from 0 to 1 of 4(1-y)^3 is 1