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

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

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

\int_{0}^{1} π (1 - x^{3})^{2} d x

π \frac{9}{14}

Decimal

2.01959 \dots

Solution steps

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

Expand

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

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

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

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

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

= π (1 - \frac{1}{2} + \frac{1}{7})

Join

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

= π \frac{9}{14}

\frac{\partial}{\partial x} (e^{x} \cdot sin (y))

\int_{0}^{\frac{1}{2}} (4 x + 3) e^{2 x} d x

\frac{\partial}{\partial x} (\frac{5}{x y})

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

x y^{^{'}} + y = x y^{2}

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