What is the integral from 0 to 1 of (x^2-x^3) ?

The integral from 0 to 1 of (x^2-x^3) is 1/12

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integral from 0 to 1 of (x^2-x^3)

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

\frac{1}{12}

Decimal Notation

0.08333 \dots

Solution steps

\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

= \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}

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

Simplify

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

= \frac{1}{12}

x \to 0 lim (\frac{1 - cos ( 7 x )}{4 x ^{2}})

\frac{d}{d x} (A (2 x e^{x} + e^{x} x^{2}))

x \to 2 lim (\frac{1}{2})

\frac{d}{d x} (- x^{\frac{2}{3}} (x - 4))

\int \frac{3}{2} x d x

What is the integral from 0 to 1 of (x^2-x^3) ?
The integral from 0 to 1 of (x^2-x^3) is 1/12