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

The integral from 0 to 1 of x(x^2-x) is -1/12

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

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

- \frac{1}{12}

Decimal Notation

- 0.08333 \dots

Solution steps

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

Expand

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

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

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

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

= \frac{1}{4} - \frac{1}{3}

Simplify

\frac{1}{4} - \frac{1}{3} : - \frac{1}{12}

= - \frac{1}{12}

\int \frac{x}{y} d y

n = 0 \sum \infty \frac{- 1 ^{n}}{n !}

y = x^{10} + 1 + 1 - x

\frac{d}{d x} (9 x + 4)

\int (4 sec (x) tan (x) - 2 sec^{2} (x)) d x

What is the integral from 0 to 1 of x(x^2-x) ?
The integral from 0 to 1 of x(x^2-x) is -1/12