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

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

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

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

\frac{1}{4}

Decimal

0.25

Solution steps

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

Expand

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

= \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

= \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}

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

Simplify

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

= \frac{1}{4}

\int_{0}^{2} 2 x + 1 d x

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

\int_{- 1}^{1} y d y

\int_{0}^{2} x^{2} + 1 d x

\int_{1}^{2} \frac{1 + x}{x ^{3}} d x

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