Application of Integrals — Class 12 Maths Solution

We have lines

$x + 2y = 2$ …….(i)

$y - x = 1$ …….(ii)

and $2X + y = 7$ …….(iii)

Solving (i) and (ii), we get point of intersection (0,1)

Solving (ii) and (iii), we get point of intersection (2,3)

Solving (i) and (iii), we get point of intersection (4,-1)

These lines are plotted on coordinate plane as shown in the following figure,

$\therefore$ From the figure, area of the shaded region
$A = \int_0^2 {\left( {x + 1 - \frac{{2 - x}}{2}} \right)} dx + \int_2^4 {\left( {7 - 2x - \frac{{2 - x}}{2}} \right)} dx$

$= \int_0^2 {\frac{{3x}}{2}} dx + \int_2^4 {\left( {6 - \frac{3}{2}x} \right)} dx$

$= \left[ {\frac{{3{x^2}}}{4}} \right]_5^2 + \left[ {6x - \frac{{3{x^2}}}{4}} \right]_2^4$

$= 3 + (24 - 12) - (12 - 3) = 6$ sq. units.