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

The integral from 0 to 1 of 1/2 (1-x^2)^2 is 4/15

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

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

\frac{4}{15}

Decimal Notation

0.26666 \dots

Solution steps

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

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} (1 - x^{2})^{2} d x

Expand

(1 - x^{2})^{2} : 1 - 2 x^{2} + x^{4}

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

= \frac{1}{2} (\int_{0}^{1} 1 d x - \int_{0}^{1} 2 x^{2} d x + \int_{0}^{1} x^{4} d x)

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

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

\int_{0}^{1} x^{4} d x = \frac{1}{5}

= \frac{1}{2} (1 - \frac{2}{3} + \frac{1}{5})

Simplify

\frac{1}{2} (1 - \frac{2}{3} + \frac{1}{5}) : \frac{4}{15}

= \frac{4}{15}

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What is the integral from 0 to 1 of 1/2 (1-x^2)^2 ?
The integral from 0 to 1 of 1/2 (1-x^2)^2 is 4/15