 Hello and welcome to the session I am Deepika here. Let's discuss the question which says using the method of integration find the area of the region bounded by lines 2x plus 5 is equal to 4, 3x minus 2y is equal to 6 and x minus 3y plus 5 is equal to 0. Now we know that the area of the region bounded by the curve y is equal to fx, x axis and the lines x is equal to a and x is equal to b where b is greater than a is given by integral from a to b y dx. So using this key idea let's start the solution. In this question we have to find the area of the region bounded by the lines 2x plus y is equal to 4. Let us give this equation as number 1, 3x minus 2y is equal to 6. Let us give this as number 2 and 3y plus 5 is equal to 0. Let us give this as number 3. First we will find the points of intersection of these lines. For this we have to solve these equations simultaneously by taking them in pairs. Now from equation 1 and 2 we have an equation 1 by 2 we have 4x plus 2y is equal to 8 and equation 2 is 3x minus 2y is equal to 6. On adding we have 7x is equal to 14 or x is equal to 2. Now from equation 1 is equal to 2 implies y is equal to 0. Let a is the point of intersection of lines 1 and 2. So the point of intersection whose coordinates are 2, 0. Let c is the point of intersection of lines 2 and 3. So we will get the coordinates of c by solving equation 2 and 3. Now from equation 2 we have 3x minus 2y is equal to 6, 35 equation 3 by 3 we get 3x minus 9y is equal to minus 15. So on subtraction we have y is equal to 21 or y is equal to y is equal to 3 implies x is equal to 4. So the point of intersection whose coordinates are 4, 3. Let b is the point of intersection of lines 1 and 3. So from equation 1 and 3 we have from 1 we have 2x plus y is equal to 4 and from 3 we have x minus 3y is equal to minus 5, multiply equation 3 by 2 we have 2x minus 6y is equal to minus 10. So on subtraction we have 7y is equal to 14 or y is equal to 2. Now from equation 1 y is equal to 2 implies x is equal to 1 and it is the point of intersection of lines 1 and 3 is b whose coordinates are 1 and 2. These points to identify the region whose area we have to determine. So we get a triangle abc. So we have to find the area of this shaded region. Therefore required area is trapezium bn, bn area of triangle acm plus area of this is equal to. Now area of trapezium bnmc is integral from 1 to 4 y of line bc that is it is equal to integral from 1 to 4 y of line bc dx which is equal to integral from 1 to 2 y of line integral from 1 to 2 y which is integral from 2 to 4 y of line is equal to integral from 1 to 4 y of line bc. Now y of line bc is x plus 5 over 3 dx minus integral from 1 to 2 y of line ab Now line ab is 2x plus y is equal to 4 so y is equal to 4 minus 2x integral from 2 to 4 y of line ac now line ac is 3x minus 2 y is equal to 6 so y is equal to 3x minus 6 upon 2. Now this is equal to 1 over 3 into x square upon 2 plus 5x and the limits are from 1 to 4 minus 4x minus 2x square upon 2 and the limits are from 1 to 2 minus 1 by 2 into 3x square upon 2 minus 6x and the limits are from 2 to 4 and this is equal to 1 over 3 into 4 square upon 2 that is 16 upon 2 plus 5 into 4 20 minus 1 over 2 plus 5 minus 4 into 2 8 minus 2 into 2 square 8 over 2 minus 4 into 1 4 minus 1 square upon 2 which is 2 over 2 1 over 2 into 3 into 4 square which is 3 into 16 that is 48 upon 2 minus 6 into 4 24 minus 3 into 2 square that is 12 over 2 minus 6 into 2 12 and this is equal to 1 over 3 into 8 plus 20 that is 28 minus 11 over 2 minus 4 which is 4 1 by 2 into 24 minus 24 which is 0 minus 6 minus 12 which is minus 6 and so this is equal to minus 1 by 2 into 6 and this is equal to 1 over 3 into 45 over 2 minus 1 minus 3 and this is equal to 15 over 2 minus 4 and this is again equal to 7 over 2 hence the required area is 7 over 2 so the answer for the above question is 7 over 2 I hope the solution is clear to you bye and check you