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

The integral from-1 to 0 of pi(1-x^2)^2 is pi 8/15

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

\int_{- 1}^{0} π (1 - x^{2})^{2} d x

π \frac{8}{15}

Decimal Notation

1.67551 \dots

Solution steps

\int_{- 1}^{0} π (1 - x^{2})^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \cdot \int_{- 1}^{0} (1 - x^{2})^{2} d x

Expand

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

= π \cdot \int_{- 1}^{0} 1 - 2 x^{2} + x^{4} 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

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

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

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

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

= π (1 - \frac{2}{3} + \frac{1}{5})

Join

1 - \frac{2}{3} + \frac{1}{5} : \frac{8}{15}

= π \frac{8}{15}

f (x) = 2 x^{3} - x^{2} + 3, (2, 15)

n = 1 \sum \infty \frac{x ^{n}}{n !}

\int \frac{30}{1 - cos ( 5 x )} d x

\int \frac{x ^{6}}{x ^{14} - 1} d x

(15 y^{2} - x y) d x + x^{2} d y = 0

What is the integral from-1 to 0 of pi(1-x^2)^2 ?
The integral from-1 to 0 of pi(1-x^2)^2 is pi 8/15