Let's integrate: $\underset{0}{\overset{1}{\int }}(x-x^3)dx$

ⅰ $\int (x-x^3)dx$ = $\int xdx-\int x^{3}dx$

$\int xdx-\int x^{3}dx$

ⓘ An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1: $\int xdx$ = $\frac{x^2}{2}$, $-\int x^{3}dx$ = $-\frac{x^4}{4}$

▷ $\frac{x^2}{2}-\frac{x^4}{4}$

Let's evaluate over the interval $[0,1]$: $(\frac{1^2}{2}-\frac{1^4}{4})-(\frac{0^2}{2}-\frac{0^4}{4})$

ⓘ To add fractions, put them over the same denominator, then add the numerators. We choose $4$, the least common multiple of $2$ and $4$. Multiplying the numerator and the denominator by the same number does not change the fraction: $\frac{1}{2}-\frac{1}{4}$ = $\frac{1·2}{2·2}-\frac{1}{4}$ = $\frac{1·2-1}{4}$ = $\frac{2-1}{4}$■A minus sign before a parenthesis changes the sign of each of its terms: $-(\frac{0^2}{2}-\frac{0^4}{4})$ = $-\frac{0^2}{2}+\frac{0^4}{4}$

$\frac{2-1}{4}-\frac{0^2}{2}+\frac{0^4}{4}$

ⓘ The numerator and the denominator are computed before the division: $2-1$ = $1$

$\frac{1}{4}+0+0$

ⓘ Adding 0 changes nothing: $\frac{1}{4}+0+0$ = $\frac{1}{4}$

▶Integral:   $\frac{1}{4}$

The numeric value of: $\frac{1}{4}$ is:

▷ $0.25$