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

The integral from 0 to 1 of (1-x^4)^3 is 128/195

Popular >

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

\int_{0}^{1} (1 - x^{4})^{3} d x

\frac{128}{195}

Decimal

0.65641 \dots

Solution steps

\int_{0}^{1} (1 - x^{4})^{3} d x

Expand

(1 - x^{4})^{3} : 1 - 3 x^{4} + 3 x^{8} - x^{12}

= \int_{0}^{1} 1 - 3 x^{4} + 3 x^{8} - 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} 3 x^{4} d x + \int_{0}^{1} 3 x^{8} d x - \int_{0}^{1} x^{12} d x

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

\int_{0}^{1} 3 x^{4} d x = \frac{3}{5}

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

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

= 1 - \frac{3}{5} + \frac{1}{3} - \frac{1}{13}

Simplify

1 - \frac{3}{5} + \frac{1}{3} - \frac{1}{13} : \frac{128}{195}

= \frac{128}{195}

\int_{2}^{\infty} ln (2 x - 3) d x

\int_{0}^{1} (1 - x^{2})^{7} d x

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

\int_{0}^{1} (1 - x^{4})^{5} d x

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

What is the integral from 0 to 1 of (1-x^4)^3 ?
The integral from 0 to 1 of (1-x^4)^3 is 128/195