# How do you evaluate the definite integral int (x^3-x^2)dx from [1,3]?

Feb 8, 2017

${\int}_{1}^{3} \left({x}^{3} - {x}^{2}\right) \mathrm{dx} = 11. \overline{3}$

#### Explanation:

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = F \left(b\right) - F \left(a\right)$, where $F ' \left(x\right) = f \left(x\right)$

In this case $a = 1$ and $b = 3$

and $f \left(x\right) = {x}^{3} - {x}^{2}$

Also, since $\int f \left(x\right) + g \left(x\right) \mathrm{dx} = \int f \left(x\right) \mathrm{dx} + \int g \left(x\right) \mathrm{dx}$

${\int}_{1}^{3} \left({x}^{3} - {x}^{2}\right) \mathrm{dx} = {\int}_{1}^{3} {x}^{3} \mathrm{dx} - {\int}_{1}^{3} {x}^{2} \mathrm{dx}$

and since $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right)$ we get

int_1^3(x^3-x^2)dx=x^4/4]_1^3-x^3/3]_1^3=(81/4-1/4)-(27/3-1/3)

$= \frac{80}{4} - \frac{26}{3} = 20 - 8. \overline{6} = 11. \overline{3}$