What is the integral from 0 to 1 of pi(x-x^3)^2 ?

The integral from 0 to 1 of pi(x-x^3)^2 is pi 8/105

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integral from 0 to 1 of pi(x-x^3)^2

\int_{0}^{1} π (x - x^{3})^{2} d x

π \frac{8}{105}

Decimal Notation

0.23935 \dots

Solution steps

\int_{0}^{1} π (x - x^{3})^{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} (x - x^{3})^{2} d x

Expand

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

= π \cdot \int_{0}^{1} x^{2} - 2 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

= π (\int_{0}^{1} x^{2} d x - \int_{0}^{1} 2 x^{4} d x + \int_{0}^{1} x^{6} d x)

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

\int_{0}^{1} 2 x^{4} d x = \frac{2}{5}

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

= π (\frac{1}{3} - \frac{2}{5} + \frac{1}{7})

Join

\frac{1}{3} - \frac{2}{5} + \frac{1}{7} : \frac{8}{105}

= π \frac{8}{105}

2 x^{3} - 6 x^{2} - 18 - 3

y = - 2 x^{2} + 6, y = - 2

\int 1 + x^{2} \cdot x^{5} d x

y = - x^{3} + 2 x^{2} - 4, at x = - 1

What is the integral from 0 to 1 of pi(x-x^3)^2 ?
The integral from 0 to 1 of pi(x-x^3)^2 is pi 8/105