# What is the net area between f(x) = e^(3-x)-2x+1 and the x-axis over x in [0, 3 ]?

Jul 24, 2017

The area is $= 13.09 {u}^{2}$

#### Explanation:

The area is

$A = {\int}_{0}^{3} \left({e}^{3 - x} - 2 x + 1\right) \mathrm{dx}$

$= {\left[- {e}^{3 - x} - {x}^{2} + x\right]}_{0}^{3}$

$= \left(- 1 - 9 + 3\right) - \left(- {e}^{3} - 0 + 0\right)$

$= {e}^{3} - 7$

$= 13.09 {u}^{2}$

graph{e^(3-x)-2x-1 [-28.86, 28.9, -14.44, 14.41]}

Jul 24, 2017

${\int}_{0}^{3} \left({e}^{3 - x} - 2 x + 1\right) \mathrm{dx} = {e}^{3} - 7$

#### Explanation:

Let us write $u = 3 - x$ hence $\mathrm{du} = - \mathrm{dx}$,

hence $\int {e}^{3 - x} \mathrm{dx} = - \int {e}^{u} \mathrm{du} = - {e}^{u} = - {e}^{3 - x}$

The net area between $f \left(x\right) = {e}^{3 - x} - 2 x + 1$ and the x-axis over $x \in \left[0 , 3\right]$ is derived by

${\int}_{0}^{3} \left({e}^{3 - x} - 2 x + 1\right) \mathrm{dx}$

= ${\int}_{0}^{3} {e}^{3 - x} \mathrm{dx} - {\int}_{0}^{3} 2 x \mathrm{dx} + {\int}_{0}^{3} \mathrm{dx}$

= ${\left[- {e}^{3 - x} - {x}^{2} + x\right]}_{0}^{3}$

= $\left(- {e}^{0} - {3}^{2} + 3\right) - \left(- {e}^{3} - {0}^{2} + 0\right)$

= $- 1 - 9 + 3 + {e}^{3}$

= ${e}^{3} - 7$