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

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

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

\int_{0}^{1} π (1 - x^{2})^{2} d x

π \frac{8}{15}

Decimal Notation

1.67551 \dots

Solution steps

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

Expand

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

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

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

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

\int_{0}^{1} 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}

(3 + 4 x) + (5 y - 1) \frac{d y}{d x} = 0

f (x) = x^{2} + 5 x - 2, at x = 4

x \to 0 lim (2 x^{2} + 3 x)

\int \frac{1 + 4 x}{1 + x + 2 x ^{2}} d x

\frac{d}{d x} (\frac{2 x ^{2} - 4 x + 3}{2 - 3 x})

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