 Hello students let's work out the following problem it says find the area of the region bounded by the parabola y is equal to x square and the lines y is equal to mod x. Let us now move on to the solution. The given parabola is y is equal to x square and the lines are y is equal to mod x. Now we know that the mod function is written as mod x is written as x if x is greater than 0 and it is minus x if x is less than 0. So we have to find the area of the region bounded by the parabola y is equal to x square the line y is equal to x and the line y is equal to minus x. So this is the line y is equal to x this is the line y is equal to minus x and we have to find the area bounded by the parabola and these two lines. So we have to find this area. Now we will first find area between the parabola and the line. So y is equal to x is equal to x square so this implies y is equal to x minus x square. Now to find the area of this region we need to find the pointer of intersection of this line and the parabola. So here y is equal to x so we have x square is equal to x from this. So this implies x square minus x is equal to 0 and this implies x into x minus 1 is equal to 0 and this again implies x is equal to 0 and 1. So in this region x takes value from 0 to 1. Now the area let us call this as a1 it is integral 0 to 1 the function is x minus x square dx let us now integrate this function integral of x with respect to x is x square by 2 minus integral of x square is x cube by 3 and here the lower limit of the integral is 0 and the upper limit is 1. Now we will apply the fundamental theorem. So we will put x is equal to 1 first minus now we put x is equal to 0 so this is 0 by 2 minus 0 by 3 so this is equal to 1 by 6 1 by 2 minus 1 by 3 is 1 by 6. Now we find the area of the region bounded by the parabola y is equal to x square and the line y is equal to minus x so here y is equal to minus x is equal to x square. So this implies y is equal to minus x minus x square. Now we have to find the point of intersection of the line y is equal to minus x and the parabola so here y is equal to minus x so this implies minus x is equal to x square and this implies minus x minus x square is equal to 0 and this implies minus x into x plus 1 is equal to 0 and this implies x is equal to 0 and x is equal to minus 1. So now the area let us call this as A2 is integral minus 1 to 0 minus x minus x square dx. Now we integrate this integral of minus x is minus x square by 2 and the integral of minus x square is minus x cube by 3 and here the lower limit is minus 1 and the upper limit is 0. Again we will apply the fundamental theorem so we will put x is equal to 0 first so it becomes 0 by 2 minus 0 by 3 minus. Now we will quote x is equal to minus 1 so it becomes minus 1 by 2 minus minus 1 cube by 3 so this is equal to minus of minus 1 by 2 plus 1 by 3 and this is equal to minus of minus 1 by 6 which is equal to 1 by 6 so the required area is equal to A1 plus A2 now A1 is 1 by 6 A2 is also 1 by 6 so the area is 2 by 6 which is equal to 1 by 3 hence the required area is 1 by 3. So this completes the question and the session life for now take care have a good day.