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

The integral from 0 to 1 of 2pix(5x(x-1)^2) is pi/3

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

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

\frac{π}{3}

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 π 5 \cdot \int_{0}^{1} xx (x - 1)^{2} d x

Simplify

= 10 π \cdot \int_{0}^{1} xx (x - 1)^{2} d x

Expand

xx (x - 1)^{2} : x^{4} - 2 x^{3} + x^{2}

= 10 π \cdot \int_{0}^{1} x^{4} - 2 x^{3} + 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

= 10 π (\int_{0}^{1} x^{4} d x - \int_{0}^{1} 2 x^{3} d x + \int_{0}^{1} x^{2} d x)

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

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

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

= 10 π (\frac{1}{5} - \frac{1}{2} + \frac{1}{3})

Simplify

10 π (\frac{1}{5} - \frac{1}{2} + \frac{1}{3}) : \frac{π}{3}

= \frac{π}{3}

\frac{5 - sec ( x )}{tan ( x )}

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

x \to 1 + lim (3)

\frac{\partial}{\partial x} (\frac{- x - 3}{y ^{2}})

(2 x + 1)^{^{'}}

What is the integral from 0 to 1 of 2pix(5x(x-1)^2) ?
The integral from 0 to 1 of 2pix(5x(x-1)^2) is pi/3