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

May 2, 2018

The area of f(x) bounded by $x = 0 \mathmr{and} x = 3$ equal

$A = {A}_{1} + {A}_{2} = 13.798 {\left(u n i t e\right)}^{2}$

#### Explanation:

show the steps
The sketch of our function is

To find the point at which the curve intersects the axis of theX-axis

$f \left(x\right) = 0$

${e}^{3 - 2 x} - 2 x + 1 = 0$

$x = 1.279$

$\left(1.279 , 0\right)$

the area equal

$A = {A}_{1} + {A}_{2}$

${A}_{1} = {\int}_{a}^{b} f \left(x\right) - \left(0\right) \cdot \mathrm{dx}$

${A}_{1} = {\int}_{0}^{1.279} {e}^{3 - 2 x} - 2 x + 1 = {\left[- {e}^{3 - 2 \cdot x} / 2 - {x}^{2} + x\right]}_{0}^{1.279} = \frac{9175435 \cdot {e}^{3} - 20823629}{18350870} = 8.908019591376782 {\left(u n i t e\right)}^{2}$

${A}_{2} = {\int}_{b}^{c} \left(0\right) - f \left(x\right)$

${A}_{2} = {\int}_{1.279}^{3} \left(0\right) - \left({e}^{3 - 2 x} - 2 x + 1\right) = {\left[{e}^{3 - 2 \cdot x} / 2 + {x}^{2} - x\right]}_{1.279}^{3} = \frac{{e}^{- 3} \cdot \left(89281591 \cdot {e}^{3} + 9175435\right)}{18350870} = 4.89014466396688 {\left(u n i t e\right)}^{2}$

The area of f(x) bounded by $x = 0 \mathmr{and} x = 3$ equal

$A = {A}_{1} + {A}_{2} = 13.798 {\left(u n i t e\right)}^{2}$