 Hi and welcome to the session. I am Purva and I will help you with the following question. Using the method of integration, find the area bounded by the curve mod x plus mod y is equal to 1. Now 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 limit from a to b y dx. So 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 limit from a to b y dx. So this is the key idea which we will use to solve this question. Now we begin with the solution. So the equation of the curve given is mod x plus mod y is equal to 1. Now we know that mod x is equal to x if x is greater than equal to 0 and is equal to minus x if x is less than 0. So mod x is equal to x if x is greater than equal to 0 and mod x is equal to minus x if x is less than 0. Similarly mod y is equal to y if y is greater than equal to 0 and is equal to minus y if y is less than 0. Thus taking all the possibilities mod x plus mod y is equal to 1 is equivalent to the following four equations. First is x plus y is equal to 1, second is x minus y is equal to 1, third is minus x plus y is equal to 1 and fourth is minus x minus y is equal to 1. Thus we have to find the area bounded by these four lines. Now let us first draw the graph and identify the region whose area is to be found out. The line x plus y is equal to 1 passes through 0 comma 1 and 1 comma 0. The line x minus y is equal to 1 passes through 0 comma minus 1 and 1 comma 0. The line minus x plus y is equal to 1 passes through 0 comma 1 and minus 1 comma 0 and the line minus x minus y is equal to 1 passes through 0 comma minus 1 and minus 1 comma 0. So by plotting these points on the graph we get the figure a, b, c, d. Now we have to find the area of this shaded region. So the area of a, b, c, d is equal to area of triangle a, o, b plus area of triangle a, o, d plus area of triangle c, o, d plus area of triangle b, o, c. This is equal to, now by key idea we know that the area of the region bounded by the curve y is equal to x, x axis and the lines x is equal to a and x is equal to b is given by integral limit from a to b y dx. So here area of triangle a, o, b is given by integral limit is from 0 minus 1 to 0 and here y is equal to 1 plus x. So we have 1 plus x dx plus now area of triangle a, o, d. Now area of triangle a, o, d has limits from 0 to 1 and here y is equal to 1 minus x. So we have integral limit is from 0 to 1, 1 minus x dx plus area of triangle c, o, d. Now again we can see that area of triangle c, o, d has limits from 0 to 1 and here y is equal to x minus 1. So we have plus integral limit is from 0 to 1, x minus 1 dx. Now since area can't be negative so we take absolute value plus area of triangle b, o, c. Now here area of triangle b, o, c has limits from minus 1 to 0 and here y is equal to minus x minus 1 that is minus of x plus 1. So we have plus integral limit is from minus 1 to 0 minus x plus 1 dx. Again area can't be negative so we take absolute value and this is equal to now integrating 1 plus x we get x plus x square by 2 and limit is from minus 1 to 0 plus integrating 1 minus x we get x minus x square by 2 and limit is from 0 to 1 plus mod integrating x minus 1 we get x square by 2 minus x and limit is from 0 to 1 plus mod now integrating minus of x plus 1 we get minus x square by 2 plus x and limit is from minus 1 to 0. Now putting the limits we get this is equal to 1 minus 1 upon 2. Now here we have the upper limit as 0 so putting 0 in place of x we get 0 minus putting lower limit as minus 1 in place of x we get minus into minus will become plus so plus 1 minus 1 upon 2 plus again putting the limits we get 1 minus 1 upon 2 because here we have the upper limit as 1 so putting 1 in place of x we get 1 minus 1 upon 2 and lower limit is 0 so we get minus 0 plus again putting the limits we get mod of 1 upon 2 minus 1. Now here again we have the upper limit as 1 so putting 1 in place of x we get 1 upon 2 minus 1 and lower limit is 0 so we get minus 0 plus mod 1 upon 2 minus 1. Now here the upper limit is 0 so putting 0 in place of x we get 0 minus into minus will become plus so putting lower limit minus 1 in place of x here we get 1 upon 2 minus 1. So we get mod of 1 upon 2 minus 1. This is equal to 1 minus 1 upon 2 is 1 upon 2 plus 1 minus 1 upon 2 is 1 upon 2 plus 1 upon 2 minus 1 is minus 1 upon 2 and taking its absolute value we get 1 upon 2 plus 1 upon 2 minus 1 again gives minus 1 upon 2 but taking its absolute value we get plus 1 upon 2 and we get this is equal to 2 so we get area of ABCD is equal to 2. Thus we get our answer as 2. Hope you have understood the solution. Bye and take care.