# Math Help - Area 3

1. ## Area 3

Calculate the area by integration of the region limited by the curve:

$y=x+2$, $y=3x-3$, $y=1-x$ and $2y-3x+6=0$

And build the graph

$\frac{27}{4}$

2. $y=x+2$ $y=3x-3$
$x+2=3x-3$
$P(\frac{5}{2},\frac{9}{2})$

$x+2=1-x$
$P(-\frac{1}{2},\frac{3}{2})$

$2x+4=3x-6$
$P(10,12)$

$3x-3=1-x$
$P(1,0)$

$6x-6=3x-6$
$P(0,-3)$

$2-2x=3x-6$
$P(\frac{7}{5},-\frac{2}{5}$

For graph. The integral of the -0,5 until 1 the (2x+1) + the integral of the 1 until 2,5 the (-2x+5). So I think the answer $\frac{7}{2}$