Application of Integrals — Class 12 Maths Solution

$y = 4x + 5$ …..(i)
$y = 5 - x$ …...(ii)
$4y = x + 5$ ….(iii)

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Solving (i) and (ii), we get point of intersection (0,5)

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

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

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

From the figure, area of the shaded region
$A = \int_{ - 1}^0 {(4x + 5)} dx + \int_0^3 {\left[ {(5 - x)dx = \int_{ - 1}^3 {\frac{{x + 5}}{4}} dx} \right.}$

$= \left[ {\frac{{4{x^2}}}{2} + 5x} \right]_{ - 1}^0 + \left[ {5x - \frac{{{x^2}}}{2}} \right]_0^3 - \frac{1}{4}\left[ {\frac{{{x^2}}}{2} + 5x} \right]_{ - 1}^3$

$= [0 - 2 + 5] + \left[ {15 - \frac{9}{2} - 0} \right] - \frac{1}{4}\left[ {\frac{9}{2} + 15 - \frac{1}{2} + 5} \right]$

$= 3 + \frac{{21}}{2} + \frac{1}{4} \cdot 24 = \frac{{15}}{2}$ sq.units