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

The integral from 0 to 1 of (1-x^2)^4 is 128/315

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integral from 0 to 1 of (1-x^2)^4

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

\frac{128}{315}

Decimal Notation

0.40634 \dots

Solution steps

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

Expand

(1 - x^{2})^{4} : 1 - 4 x^{2} + 6 x^{4} - 4 x^{6} + x^{8}

= \int_{0}^{1} 1 - 4 x^{2} + 6 x^{4} - 4 x^{6} + x^{8} 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} 4 x^{2} d x + \int_{0}^{1} 6 x^{4} d x - \int_{0}^{1} 4 x^{6} d x + \int_{0}^{1} x^{8} d x

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

\int_{0}^{1} 4 x^{2} d x = \frac{4}{3}

\int_{0}^{1} 6 x^{4} d x = \frac{6}{5}

\int_{0}^{1} 4 x^{6} d x = \frac{4}{7}

\int_{0}^{1} x^{8} d x = \frac{1}{9}

= 1 - \frac{4}{3} + \frac{6}{5} - \frac{4}{7} + \frac{1}{9}

Simplify

1 - \frac{4}{3} + \frac{6}{5} - \frac{4}{7} + \frac{1}{9} : \frac{128}{315}

= \frac{128}{315}

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What is the integral from 0 to 1 of (1-x^2)^4 ?
The integral from 0 to 1 of (1-x^2)^4 is 128/315