What is the integral from 0 to 1 of pi(5x^6-5x)^2 ?

The integral from 0 to 1 of pi(5x^6-5x)^2 is pi 625/156

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integral from 0 to 1 of pi(5x^6-5x)^2

\int_{0}^{1} π (5 x^{6} - 5 x)^{2} d x

π \frac{625}{156}

Decimal Notation

12.58650 \dots

Solution steps

\int_{0}^{1} π (5 x^{6} - 5 x)^{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} (5 x^{6} - 5 x)^{2} d x

Expand

(5 x^{6} - 5 x)^{2} : 25 x^{12} - 50 x^{7} + 25 x^{2}

= π \cdot \int_{0}^{1} 25 x^{12} - 50 x^{7} + 25 x^{2} 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} 25 x^{12} d x - \int_{0}^{1} 50 x^{7} d x + \int_{0}^{1} 25 x^{2} d x)

\int_{0}^{1} 25 x^{12} d x = \frac{25}{13}

\int_{0}^{1} 50 x^{7} d x = \frac{25}{4}

\int_{0}^{1} 25 x^{2} d x = \frac{25}{3}

= π (\frac{25}{13} - \frac{25}{4} + \frac{25}{3})

Join

\frac{25}{13} - \frac{25}{4} + \frac{25}{3} : \frac{625}{156}

= π \frac{625}{156}

\int_{0}^{10} x e^{- x} d x

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

\int_{0}^{b} e^{ux} a (b - x) d x

\int \frac{1}{x e ^{l n (x)}} d x

x \to - 1 lim (1 x^{2})

What is the integral from 0 to 1 of pi(5x^6-5x)^2 ?
The integral from 0 to 1 of pi(5x^6-5x)^2 is pi 625/156