 Hello and welcome to the session. Let us discuss the following question. It says, find the area of the region given by those x and y for which y square is less than equal to 4x and 4x square plus 4y square is less than equal to 9. Let us now move on to the solution. Here we have to find the area enclosed by the region y square less than equal to 4x and 4x square plus 4y square less than equal to 9. Now y square is equal to 4x is the parabola and y 4x square plus 4y square is equal to 9 is a circle and we have to find area enclosed by this parabola and this circle that is in the interior of these two regions. So this is the region for which we have to find the area and this is the circle which centre 0 0 and radius 3 by 2 and from here we can say that y is equal to 2 root x and from here we say that y is equal to under the root 3 by 2 square minus x square and we are taking just the positive square root. Now to find this area we will divide the whole region into two different regions and here we see that one of the region is covered by the parabola this and the line which passes through the point of intersection of the parabola and the circle and the other one is covered by this line and the circle. So we first need to find the point of intersection of the parabola and the circle because this line passes through the point of intersection of the circle and the parabola. Now from here we know that y square is equal to 4x so we will substitute y square is equal to 4x in this. So we have 4x square plus 4 into 4x minus 9 is equal to 0 so again equal to 4x square plus 16x minus 9 is equal to 0. Now solving this quadratic equation reprising 18x minus 2x minus 9 is equal to 0 and the factors are 2x minus 1 into 2x plus 9 is equal to 0 and from this we can say that x is equal to 1 by 2 or x is equal to minus 9 by 2 but we know that the area which we need to find lies in the first and the fourth quadrant where x is positive. So x is equal to minus 9 by 2 is rejected and we have x is equal to 1 by 2. So this is the line x is equal to 1 by 2 which is parallel to the y axis. Now the required area is given by the integral 0 to 1 by 2 to root x dx. This is an area covered by the parabola where x goes from 0 to 1 by 2 plus the integral 1 by 2 to 3 by 2 under the root 3 by 2 square minus x square dx and this is the area covered by the circle where x goes from 1 by 2 to 3 by 2 because the radius of the circle is 3 by 2 so this whole length is 3 by 2 this point is 3 by 2 0 but since we have taken just the positive square root and the region is symmetric about the x axis this is the area given in the first quadrant only. So to find the complete area this complete area we need to multiply this area by 2 because this is the area in the first quadrant only and the region is symmetric about x axis so we will multiply this with 2 to get the complete area. Now this is again equal to 2 into 2 is 4 integral 0 to 1 by 2 root x dx plus 2 into integral 1 by 2 to 3 by 2 under the root 3 by 2 square minus x square dx. Now we will integrate this the integral of under the root x that is x to the power 1 by 2 is x to the power 1 by 2 plus 1 that is 3 by 2 upon 3 by 2 and here the lower limit is 0 and the upper limit is 1 by 2 plus 2 into now here we will apply the formula for the integral of under the root a square minus x square which is given by x by 2 into under the root a square minus x square plus a square by 2 here a is 3 by 2 so it is 3 by 2 square by 2 sin inverse x upon a that is x upon 3 by 2 and here the lower limit is 1 by 2 and the upper limit is 3 by 2 now this is again equal to 4 by 3 x to the power 3 by 2 plus multiplying 2 with this expression we have x into under the root 3 by 2 square minus x square plus 9 by 4 sin inverse x upon 3 by 2 now we will apply the second fundamental theorem so we will put x is equal to 1 by 2 in this so it becomes 1 by 2 to the power 3 by 2 minus 0 to the power 3 by 2 plus now we will put x is equal to 3 by 2 first so it is 3 by 2 under the root 3 by 2 square minus 3 by 2 square plus 9 by 4 sin inverse 3 by 2 upon 3 by 2 minus now we will put x is equal to 1 by 2 so this becomes 1 by 2 under the root 3 by 2 square minus 1 by 2 square plus 9 by 4 into sin inverse 1 by 2 upon 3 by 2 here we see that when we multiply 2 with 4 it gives us 8 so here and here we have 8 by 3 now we simplify this expression this 8 by 3 into 1 by 2 to the power 3 by 2 it can be written as 1 by 2 to the power 1 into 1 by 2 to the power 1 by 2 plus 3 by 2 into 0 3 by 2 square minus 3 by 2 square is 0 plus 9 by 4 sin inverse 1 is 5 by 2 so it is 9 by 4 into i by 2 minus 1 by 2 into under the root 3 by 2 square minus 1 by 2 square is root 2 plus 9 by 4 sin inverse 1 by 2 upon 3 by 2 is 1 by 3 now again this is equal to 4 by 3 into 1 by root 2 plus 9 by by 8 minus root 2 by 2 minus 9 by 4 sin inverse 1 by 3 now again 4 can be written as 2 into 2 upon 3 into root 2 minus root 2 by 2 plus 9 by by 8 minus 9 by 4 sin inverse 1 by 3 now this can be written as 2 root 2 by 3 minus root 2 by 2 plus 9 by 9 by by 8 minus 9 by 4 sin inverse 1 by 3 acting root 2 by 2 from 2 root 2 by 3 we have root 2 by 6 plus 9 by by 8 minus 9 by 4 sin inverse 1 by 3 so this is the required area root 2 by 6 plus 9 by by 8 minus 9 by 4 sin inverse 1 by 3 so this completes the question and the session bye for now take care have a good day