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

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

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

\int_{0}^{1} 2 π x (x - x^{2}) d x

\frac{π}{6}

Decimal Notation

0.52359 \dots

Solution steps

\int_{0}^{1} 2 π x (x - x^{2}) d x

Take the constant out:

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

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

Expand

x (x - x^{2}) : x^{2} - x^{3}

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

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

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

\int_{0}^{1} x^{3} d x = \frac{1}{4}

= 2 π (\frac{1}{3} - \frac{1}{4})

Simplify

2 π (\frac{1}{3} - \frac{1}{4}) : \frac{π}{6}

= \frac{π}{6}

y^{^{''}} + 10 y = 0, y (0) = 2, y^{^{'}} (0) = 10

\int tan^{7} (x) sec^{17} (x) d x

s (t) = \frac{1}{t ^{2} + 6 t - 7}

\int \frac{ln ( 4 x )}{x} d x

\frac{\partial}{\partial x} (- \frac{2 y}{x ^{2} + y ^{2} + 1})

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