What is the integral from 0 to 1 of (1-x^2)^6 ?

The integral from 0 to 1 of (1-x^2)^6 is 1024/3003

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integral from 0 to 1 of (1-x^2)^6

\int_{0}^{1} (1 - x^{2})^{6} d x

\frac{1024}{3003}

Decimal

0.34099 \dots

Solution steps

\int_{0}^{1} (1 - x^{2})^{6} d x

Expand

(1 - x^{2})^{6} : 1 - 6 x^{2} + 15 x^{4} - 20 x^{6} + 15 x^{8} - 6 x^{10} + x^{12}

= \int_{0}^{1} 1 - 6 x^{2} + 15 x^{4} - 20 x^{6} + 15 x^{8} - 6 x^{10} + x^{12} 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} 6 x^{2} d x + \int_{0}^{1} 15 x^{4} d x - \int_{0}^{1} 20 x^{6} d x + \int_{0}^{1} 15 x^{8} d x - \int_{0}^{1} 6 x^{10} d x + \int_{0}^{1} x^{12} d x

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

\int_{0}^{1} 6 x^{2} d x = 2

\int_{0}^{1} 15 x^{4} d x = 3

\int_{0}^{1} 20 x^{6} d x = \frac{20}{7}

\int_{0}^{1} 15 x^{8} d x = \frac{5}{3}

\int_{0}^{1} 6 x^{10} d x = \frac{6}{11}

\int_{0}^{1} x^{12} d x = \frac{1}{13}

= 1 - 2 + 3 - \frac{20}{7} + \frac{5}{3} - \frac{6}{11} + \frac{1}{13}

Simplify

1 - 2 + 3 - \frac{20}{7} + \frac{5}{3} - \frac{6}{11} + \frac{1}{13} : \frac{1024}{3003}

= \frac{1024}{3003}

What is the integral from 0 to 1 of (1-x^2)^6 ?
The integral from 0 to 1 of (1-x^2)^6 is 1024/3003