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

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

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

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

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

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

Take the constant out:

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

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

Expand

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

= 2 \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 (\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 (\frac{1}{3} - \frac{1}{4})

Simplify

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

= \frac{1}{6}

\frac{d}{d x} (2 x (6 x - 18)^{6})

\int \frac{2}{1 - e ^{x}} d x

n = 1 \sum \infty \frac{2 n}{9 n - 1}

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

x \to 99 lim (5)

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