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

The integral from 0 to 1 of (x^2-x)^2 is 1/30

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

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

\frac{1}{30}

Decimal Notation

0.03333 \dots

Solution steps

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

Expand

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

= \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

= \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}

= \frac{1}{5} - \frac{1}{2} + \frac{1}{3}

Simplify

\frac{1}{5} - \frac{1}{2} + \frac{1}{3} : \frac{1}{30}

= \frac{1}{30}

\frac{d}{d x} (\frac{tan ( x )}{x})

n = 0 \sum \infty \frac{n}{3 ^{n}}

f (x) = arccos (2 x)

x \to 0 + lim (\frac{e ^{\frac{1}{x}}}{x})

(3 x + 4)^{2}

What is the integral from 0 to 1 of (x^2-x)^2 ?
The integral from 0 to 1 of (x^2-x)^2 is 1/30