What is the integral from-1 to 1 of (1-x^2)^3 ?

The integral from-1 to 1 of (1-x^2)^3 is 32/35

Popular >

integral from-1 to 1 of (1-x^2)^3

\int_{- 1}^{1} (1 - x^{2})^{3} d x

\frac{32}{35}

Decimal

0.91428 \dots

Solution steps

\int_{- 1}^{1} (1 - x^{2})^{3} d x

Expand

(1 - x^{2})^{3} : 1 - 3 x^{2} + 3 x^{4} - x^{6}

= \int_{- 1}^{1} 1 - 3 x^{2} + 3 x^{4} - x^{6} 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_{- 1}^{1} 1 d x - \int_{- 1}^{1} 3 x^{2} d x + \int_{- 1}^{1} 3 x^{4} d x - \int_{- 1}^{1} x^{6} d x

\int_{- 1}^{1} 1 d x = 2

\int_{- 1}^{1} 3 x^{2} d x = 2

\int_{- 1}^{1} 3 x^{4} d x = \frac{6}{5}

\int_{- 1}^{1} x^{6} d x = \frac{2}{7}

= 2 - 2 + \frac{6}{5} - \frac{2}{7}

Simplify

2 - 2 + \frac{6}{5} - \frac{2}{7} : \frac{32}{35}

= \frac{32}{35}

\int_{- 8}^{8} ln (x + 9) d x

\int_{0}^{l n (6)} e^{6 x} d x

\int_{- π}^{π} sin^{6} (\frac{x}{2}) d x

\int_{0}^{1} 3 x^{2} + 2 x d x

\int_{0}^{5} (3 x^{2} + 2) d x

What is the integral from-1 to 1 of (1-x^2)^3 ?
The integral from-1 to 1 of (1-x^2)^3 is 32/35