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

The integral from 0 to 1 of pi(x^3-1)^2 is pi 9/14

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

\int_{0}^{1} π (x^{3} - 1)^{2} d x

π \frac{9}{14}

Decimal Notation

2.01959 \dots

Solution steps

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

Expand

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

= π \cdot \int_{0}^{1} x^{6} - 2 x^{3} + 1 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^{6} d x - \int_{0}^{1} 2 x^{3} d x + \int_{0}^{1} 1 d x)

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

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

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

= π (\frac{1}{7} - \frac{1}{2} + 1)

Join

\frac{1}{7} - \frac{1}{2} + 1 : \frac{9}{14}

= π \frac{9}{14}

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What is the integral from 0 to 1 of pi(x^3-1)^2 ?
The integral from 0 to 1 of pi(x^3-1)^2 is pi 9/14