 Good morning friends, I am Poojwa and today I will help you with the following question. Find the area bounded by curves x comma y such that y is greater than equal to x square and y is equal to mod x. Let us now begin with the solution. Now here we are required to find the area bounded by curves x comma y such that y is greater than equal to x square and y is equal to mod x which means we have to find the area of the region formed by the set of points x comma y satisfying y greater than equal to x square and y is equal to mod x. So to identify the region let us draw the graph of y is equal to x square y is equal to x if x is greater than equal to 0 and y is equal to minus x if x is less than 0. Now y is equal to x square is a parabola with vertex at 0 0 and symmetric about y axis. So this is a parabola y is equal to x square with vertex at 0 0 and it is symmetric about y axis. Also y is equal to x is a line for x greater than equal to 0 passing through 0 comma 0 and 1 comma 1 and y is equal to minus x is a line passing through 0 comma 0 and minus 1 comma 1 and this shaded region is the region whose area is to be found out. Therefore we have required area is equal to now we will get the area of this shaded region by subtracting the equation of parabola from the equation of line OA and here limit is from 0 to 1 so we get therefore required area is equal to integral limit is from 0 to 1 now the equation of line here is y is equal to x so we get x dx minus integral limit is from 0 to 1 now equation of parabola is y is equal to x square so we get x square dx and here we get the area of this shaded region by subtracting equation of parabola OQB from the equation of line OB and here limit is from minus 1 to 0 so we get plus integral limit is from minus 1 to 0 now here equation of line is y is equal to minus x so we get minus x dx minus integral limit is from minus 1 to 0 and equation of parabola is y is equal to x square so we get x square dx and this is equal to now integrating x we get x square by 2 and here limit is from 0 to 1 minus integrating x square we get x cube by 3 and here limit is from 0 to 1 plus integrating minus x we get minus x square by 2 and here limit is from minus 1 to 0 minus integrating x square we get x cube by 3 and here again limit is from minus 1 to 0 and this is equal to now putting the limits we get here upper limit is 1 so putting 1 in place of x we get 1 upon 2 minus here lower limit is 0 so putting 0 in place of x we get 0 minus putting limits here we get upper limit is 1 so putting 1 in place of x we get 1 upon 3 minus lower limit here is 0 so putting 0 we get 0 plus here we have upper limit is 0 so putting 0 in place of x we get minus 0 minus here lower limit is minus 1 so putting minus 1 in place of x we get minus 1 upon 2 minus here upper limit is 0 so putting 0 we get 0 minus lower limit is minus 1 so putting minus 1 in place of x we get minus 1 upon 3 and this is equal to 1 by 2 minus 0 is 1 by 2 minus 1 by 3 minus 0 is 1 by 3 plus here minus into minus will become plus so plus 1 by 2 minus 0 gives plus 1 by 2 and here minus into minus will become plus and plus into minus will become minus so minus 1 by 3 plus 0 gives minus 1 by 3 and this is equal to 1 by 2 plus 1 by 2 is equal to 1 and minus 1 by 3 and minus 1 by 3 is equal to minus 2 by 3 and this is equal to 1 upon 3 therefore we get the required area is 1 upon 3 thus we write our answer as 1 upon 3 hope you have understood the solution by and take care