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

May 23, 2017

The net area is ${e}^{2} - e - 2 \approx 2.6708$

#### Explanation:

Net area between the graphs is the absolute value of the integral between the limits of the integral:

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

Evaluate the definite
$= | {\left[- {e}^{3 - x} - {x}^{2} + x\right]}_{1}^{2} |$

$= | \left(- {e}^{1} - 4 + 2\right) - \left(- {e}^{2} - 1 + 1\right) |$

$= | - e - 2 + {e}^{2} |$

Because the value inside the absolute value is positive, this can be simplified to:
$= {e}^{2} - e - 2$

$\approx 2.6708$