# What is the net area between f(x) = 3x^2-4x+2 and the x-axis over x in [1, 3 ]?

##### 1 Answer
Nov 14, 2017

${\int}_{1}^{3} \left(3 {x}^{2} - 4 x + 2\right) \mathrm{dx} = 14$

#### Explanation:

To work this out, we need to evaluate the definite integral between 3 and 1.

Area=${\int}_{1}^{3} \left(3 {x}^{2} - 4 x + 2\right) \mathrm{dx}$
$= {\left[3 \left({x}^{3} / 3\right) - 4 \left({x}^{2} / 2\right) + 2 x\right]}_{1}^{3}$
$= {\left[{x}^{3} - 2 {x}^{2} + 2 x\right]}_{1}^{3}$
$= \left[{\left(3\right)}^{3} - 2 {\left(3\right)}^{2} + 2 \left(3\right)\right] - \left[{\left(1\right)}^{3} - 2 {\left(1\right)}^{2} + 2 \left(1\right)\right]$
$= 15 - 1$
$= 14$