What is the integral from 0 to 1 of (1-x^2)^5 ?

The integral from 0 to 1 of (1-x^2)^5 is 256/693

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integral from 0 to 1 of (1-x^2)^5

\int_{0}^{1} (1 - x^{2})^{5} d x

\frac{256}{693}

Decimal

0.36940 \dots

Solution steps

\int_{0}^{1} (1 - x^{2})^{5} d x

Expand

(1 - x^{2})^{5} : 1 - 5 x^{2} + 10 x^{4} - 10 x^{6} + 5 x^{8} - x^{10}

= \int_{0}^{1} 1 - 5 x^{2} + 10 x^{4} - 10 x^{6} + 5 x^{8} - x^{10} 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} 5 x^{2} d x + \int_{0}^{1} 10 x^{4} d x - \int_{0}^{1} 10 x^{6} d x + \int_{0}^{1} 5 x^{8} d x - \int_{0}^{1} x^{10} d x

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

\int_{0}^{1} 5 x^{2} d x = \frac{5}{3}

\int_{0}^{1} 10 x^{4} d x = 2

\int_{0}^{1} 10 x^{6} d x = \frac{10}{7}

\int_{0}^{1} 5 x^{8} d x = \frac{5}{9}

\int_{0}^{1} x^{10} d x = \frac{1}{11}

= 1 - \frac{5}{3} + 2 - \frac{10}{7} + \frac{5}{9} - \frac{1}{11}

Simplify

1 - \frac{5}{3} + 2 - \frac{10}{7} + \frac{5}{9} - \frac{1}{11} : \frac{256}{693}

= \frac{256}{693}

What is the integral from 0 to 1 of (1-x^2)^5 ?
The integral from 0 to 1 of (1-x^2)^5 is 256/693