 Hello and welcome to the session. In this session we discussed the following question which says, using the method of integration, find the area of the region bounded by the lines 2x plus y equal to 4, 3x minus 2y equal to 6, and x minus 3y plus 5 equal to 0. Let's move on to the solution now. We are given the lines 2x plus y equal to 4, let this be equation 1, 3x minus 2y equal to 6, let this be equation 2, x minus 3y plus 5 equal to 0, let this be equation 3. We have to find the area of the region bounded by these three lines. First let's draw the line 2x plus y equal to 4. For x equal to 0 in this equation we have y is equal to 4, then for x equal to 1 in this equation we get y is equal to 2, for x equal to 2 in this equation we get y equal to 0. So let's plot these three points on the graph so as to get the equation 2x plus y equal to 4 on the graph. This is the line 2x plus y equal to 4. Now consider the second line 3x minus 2y equal to 6. Now we will draw this line. For x equal to 2 in this equation we get y equal to 0, for x equal to 4 in this equation we get y as 3, for x equal to 6 we get y equal to 6. Now we plot these points on the graph so as to get the line 3x minus 2y equal to 6 on the graph. So this line is the line 3x minus 2y equal to 6. Consider the next line x minus 3y plus 5 equal to 0. For x equal to 1 in this equation we get y equal to 2 and for x equal to 4 we get y equal to 3. So now we plot these two points on the graph so as to get the line x minus 3y plus 5 equal to 0. So this is the line x minus 3y plus 5 equal to 0. Now the line 2x plus y equal to 4 and the line 3x minus 2y equal to 6 meet at the point c which coordinates 2,0. So we have point of intersection of lines 1 and 2 is the point 2,0 that is the point c which coordinates 2,0. Now as you can see from the graph the point of intersection of the line x minus 3y plus 5 equal to 0 and 3x minus 2y equal to 6 is the point e which coordinates 4,3. Thus now we have point of intersection of lines 2 and 3 is the point e which coordinates 4,3. You can see that the point of intersection of the lines 2x plus y equal to 4 and x minus 3y plus 5 equal to 0 is the point b which coordinates 1,2. So point of intersection of lines 1 and 3 is the point b which coordinates 1,2. We have to find the area of this shaded region that is the region bounded by the three given lines. So we have to find the area of the triangle bce. Thus area of the triangle bce is equal to area under the segment be. So this is equal to area under segment be minus the area under segment bc minus the area under segment ce. Now area under the segment be would be given by integral. Limit goes from the x coordinate of the point b that is 1 to the x coordinate of the point e that is 4. So integral 1 to 4 the value of y from this equation which would be y is equal to x plus 5 upon 3. So this is integral 1 to 4 x plus 5 upon 3 dx minus area under the segment bc which would be integral. Limit goes from x coordinate of point b that is 1 to the x coordinate of the point c that is 2. So integral 1 to 2 and the value of y from this equation which would be y is equal to 4 minus 2x. So here we have 4 minus 2x dx minus area under segment ce which would be integral. Limit goes from x coordinate of the point c that is 2 to the x coordinate of the point e that is 4. So integral 2 to 4 and value of y from this equation which would be y is equal to 3x minus 6 upon 2. So 3x minus 6 upon 2 dx. So solving this further we get this is equal to 1 upon 3 into x square upon 2 plus 5x limit goes from 1 to 4 minus 4x minus 2x square upon 2. The limit goes from 1 to 2 minus 3x square upon 4 minus 6 upon 2x and the limit goes from 2 to 4. So now putting the limits that is upper limits and the lower limits we get this is equal to 1 upon 3 into 16 upon 2 plus 20 minus 1 upon 2 minus 5. 4 into 2 that is 8 minus 2 into 4 upon 2 minus 4 plus 2 upon 2 minus 3 into 16 upon 4 minus 6 upon 2 into 4 minus 3 upon 4 into 4 plus. 6 upon 2 into 2. So further 2 8 times is 16 then 2 cancels with 2 2 cancels with 2 4 4 times is 16 2 2 times is 4 4 cancels with 4 and 2 cancels with 2. So this is equal to 1 upon 3 into 8 plus 20 28 minus 1 upon 2 minus 5 minus 8 minus 4 minus 4 plus 1 minus 12 minus 12 minus 3 plus 6. So further we get this is equal to 1 upon 3 into 23 minus 1 upon 2 minus now 8 minus 8 is 0 and here we have just 1 so it is minus 1 minus 12 minus 12 is 0 and 6 minus 3 is 3 so this is minus 3. So further we get this is equal to 1 upon 3 into 46 minus 1 that is 45 upon 2 minus 4 now 3 15 times is 45 so here we have 15 upon 2 minus 4. So further this is equal to 15 minus 8 upon 2 thus we get this is equal to 7 upon 2 thus we get area of triangle BCE is equal to 7 upon 2 square units. So this is the required area that is the area bounded by the 3 given lines is 7 upon 2 square units. So this is our final answer this completes the session hope you have understood the solution of this question.