Question

Using integration, find the area bounded by the lines $x+2y=2,y-x=1$ and $2x+y=7$

Answer 1

Given lines are $x+2y=2$ ...(1)

$y+2x=7$ ...(2)

$y-x=1$ ...(3)

I) $x+2y=2$

$x$	$0$	$2$
$y$	$1$	$0$

$(0,1);(2,0)$

II) $2x+y=7$

$x$	$0$	$\frac{7}{2}$
$y$	$7$	$0$

$(0,7);(\frac{7}{2},0)$

III) $y-x=1$

$x$	$0$	$-1$
$y$	$1$	$0$

$(0,1);(-1,0)$

Now, we find points of intersection

1) (I) and (II)

$x+2y=2,y+2x=7$

$y=7-2x$

$\Rightarrow x+2[7-2x]=2$

$x+14-4x=2$

$-3x=2-14=-12$

$x=\frac{12}{3}=4$

$y=7-2\left(4\right)=7-8=-1$

$\therefore$ point $A(4,-1)$

2) (II) and (III)

$y+2x=7$, $y-x=1$

$y=x+1$

$\Rightarrow x+1+2x=7$

$3x=6$

$x=2$

$\Rightarrow y=2+1=3$

Point $B(2,3)$

3) (I) and (III)

$x+2y=2$, $y-x=1$

$y=x+1$

$\Rightarrow x+2[x+1]=2$

$x+2x+2=2$

$3x=0$

$x=0$

$\Rightarrow y=0+1=1$

Point $C(0,1)$

We draw graph [Ref. image]

To find shaded tregion

$ar\left(△ABC\right)$

$={\int }_0^2{y}_{CB}dx+{\int }_4^2{y}_{BA}.dx-{\int }_4^0{y}_{AC}.dx$

$={\int }_0^2(x+1)dx+{\int }_0^2(7-2x)dx-{\int }_4^0\left(\frac{2-x}{2}\right)dx$

$={[\frac{x^2}{2}+x]}_0^2+{[7x-\frac{2x^2}{2}]}_4^2-\frac{1}{2}{[2x-\frac{x^2}{2}]}_4^0$

$=[\frac{2^2}{2}+2]-[\frac{o^2}{2}+0]+\left[7\right(2)-\frac{2(2{)}^2}{2}]-\left[7\right(4)-\frac{2(4{)}^2}{2}]-\frac{1}{2}\left[2\right(0)-\frac{0^2}{2}]+\frac{1}{2}\left[2\right(4)-\frac{4^2}{2}]$

$=[2+2]-\left[0\right]+[14-4]-[28-16]-\frac{1}{2}\left[0\right]+\frac{1}{2}[8-8]$

$=4+10-12+0$

$=14-12$

$=2sq.units$