 Hello students, let's solve the following question. It says, find the area bounded by the curve x square is equal to 4y and the straight line x is equal to 4y minus 2. Let us now move on to the solution. Here we have to find the area enclosed by the parabola x square is equal to 4y and the line x is equal to 4y minus 2. So this is the line x is equal to 4y minus 2 and this is the parabola x square is equal to 4y and we have to find the area enclosed by the parabola and the line. So we have to find this area, this is the common area. Now to find this area we first need to find these two point of intersection, this and this. So let's first find out the point of intersection. Now from here we see that x is equal to 4y minus 2. So we will put x is equal to 4y minus 2 in this. So we have 4y minus 2 whole square is equal to 4y. Now apply the formula of a minus b whole square. So this becomes 16y square plus 4 minus 16y is equal to 4y and this implies 16y square minus 20y plus 4 is equal to 0. Let's now factorize, factorizing we get the factors as 4y minus 1 into y minus 1 is equal to 0. So we have y is equal to 1y 4 and y is equal to 1. Now if y is equal to 1y 4, x is equal to 4 into 1y 4 minus 2 which is equal to 1 minus 2 which is minus 1 and if x is 1, y is 1 then x is 4 into 1 minus 2 which is equal to 2. So the points of intersection parabola and the line x is equal to 4y minus 2 are minus 1, 1 by 4 that is when x is minus 1, y is 1 by 4 and 2, 1 that is when x is 2, y is 1. Now we have x square is equal to 4y and x is equal to 4y minus 2. So from this we have y is equal to x square by 4 and from this we have y is equal to x plus 2 by 4. So we have y is equal to x plus 2 by 4 is equal to x square by 4. So this implies y is equal to x plus 2 by 4 minus x square by 4. So this is the function which we have to integrate from minus 1 to 2 because this gives us the area of the shaded region in which x is going from minus 1 to 2 and we know that the area of the region bounded by the curve y is equal to fx and the ordinate x is equal to ax is equal to b is given by the integral a to b fx dx and integrating this function fx it gives us another function of x and here the local limit is a and the upper limit is b which is equal to fb minus fa. So now the required area is equal to the integral minus 1 by minus 1 to 2 because x is going from minus 1 to 2 and here fx is this that is 1 by 4 into x plus 2 minus x square dx that is now integrate this function. The integral of x with respect to x is x square by 2 plus the integral of 2 with respect to x is 2x minus the integral of x square is x cube by 3 and here the lower limit is minus 1 and the upper limit is 2 and here we have used the formula for the integral of x to the power n dx which is given by x to the power n plus 1 upon n plus 1. Now we will apply the second fundamental theorem so we will put x is equal to 2 first so it becomes 2 square by 2 plus 2 into 2 minus 2 cube by 3 minus now we will put x is equal to minus 1. So this becomes 1 by 2 plus 2 into minus 1 minus minus 1 cube upon 3 now again this is equal to 1 by 4 into 2 plus 4 minus 8 by 3 minus 1 by 2 minus 2. It becomes plus 1 by 3 which is 1 by 4 6 minus 8 by 3 now we open the bracket so it becomes minus 1 by 2 plus 2 minus 1 by 3 this is again equal to 1 by 4 into 6 plus 2 is 8 minus 9 by 3 plus minus 1 by 2 which is again equal to 1 by 4 into 8 minus 3 minus 1 by 2 and simply find this we get the area as 9 by 8 hence the required area is 9 by 8. So this completes the question and the session by for now take care have a good day.