# Calculate the area bounded by the curves  y=x^2-4x+5 and y=-2x+8 ?

Jan 2, 2018

$\frac{32}{3}$

#### Explanation:

graph{(y-(x^2-4x+5))(y-(-2x+8))=0 [-5, 5, -5, 12]}

First we solve the simultaneous equations

$\left\{\begin{matrix}y = {x}^{2} - 4 x + 5 \\ y = - 2 x + 8\end{matrix}\right.$

Which requires that:

${x}^{2} - 4 x + 5 = - 2 x + 8$
$\therefore {x}^{2} - 2 x - 3 = 0$
$\therefore \left(x + 1\right) \left(x - 3\right) = 0$
$\therefore x = - 1 , = 3$

Hence, the area sought is given by:

$A = {\int}_{- 1}^{3} \setminus \left(- 2 x + 8\right) - \left({x}^{2} - 4 x + 5\right) \setminus \mathrm{dx}$
$\setminus \setminus = {\int}_{- 1}^{3} \setminus - 2 x + 8 - {x}^{2} + 4 x - 5 \setminus \mathrm{dx}$
$\setminus \setminus = {\int}_{- 1}^{3} \setminus - {x}^{2} + 2 x + 3 \setminus \mathrm{dx}$
$\setminus \setminus = {\left[- {x}^{3} / 3 + {x}^{2} + 3 x\right]}_{- 1}^{3}$
$\setminus \setminus = \left(- 9 + 9 + 9\right) - \left(\frac{1}{3} + 1 - 3\right)$
$\setminus \setminus = 9 - \left(- \frac{5}{3}\right)$
$\setminus \setminus = \frac{32}{3}$