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

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

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

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

1

Solution steps

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

Take the constant out:

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

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

Expand

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

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

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 x - \int_{0}^{1} 3 x d x + \int_{0}^{1} 3 x^{2} d x - \int_{0}^{1} x^{3} d x)

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

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

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

\int_{0}^{1} x^{3} d x = \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-x)^3 ?
The integral from 0 to 1 of 4(1-x)^3 is 1