 Hi and welcome to the session. I am Purva and I will help you with the following question. Find the area of the region lined in the first quadrant and bounded by y is equal to 4x square, x is equal to 0, y is equal to 1 and y is equal to 4. Now the area of the region bounded by the curve x is equal to gy, y axis and the lines y is equal to c and y is equal to d is given by integral limit is from c to d, x dy this is equal to integral limit is from c to d gy dy. So the area of the region bounded by the curve x is equal to gy and y axis and the lines y is equal to some say c and y is equal to some d is given by integral limit is from c to d, x dy and this is equal to integral limit is from c to d gy dy. So this is the key idea which we will use to solve this question. Now we start with the solution. Now we have to find the area of the region lined in first quadrant and bounded by y is equal to 4x square, x is equal to 0, y is equal to 1 and y is equal to 4. Now y is equal to 4x square is a parabola whose vertex is 0 0 and is symmetric about y axis. So this is a parabola y is equal to 4x square whose vertex is 0 0 and is symmetric about y axis. And y is equal to 1 and y is equal to 4 are the two lines parallel to x axis. Also this line y is equal to 1 intersects the parabola y is equal to 4x square at the points 1 by 2, 1 and minus 1 by 2, 1 and this line y is equal to 4 intersects the parabola y is equal to 4x square at the points 1, 4 and 1, minus 4. Thus this shaded region lined in the first quadrant and bounded by the parabola y is equal to 4x square, x is equal to 0, y is equal to 1 and y is equal to 4 is the region whose area is to be found out. Therefore required area is equal to integral limit is from 1 to 4 because this shaded region is bounded by the lines y is equal to 1 and y is equal to 4 so we have limit is from 1 to 4. Now equation of parabola is y is equal to 4x square or we can write this as x square is equal to y by 4 or we can write this as x is equal to plus minus under root y by 2. Now we know that the shaded region lies in the first quadrant so in the first quadrant we have only positive sign so neglecting the negative sign we get x is equal to under root y by 2. Thus we get integral limit from 1 to 4 under root y upon 2 dy. Now this is equal to 1 by 2 integrating under root y we get y to the power 1 by 2 plus 1 upon 1 by 2 plus 1 and limit is from 1 to 4 and this is further equal to 1 by 2 into y to the power 3 by 2 upon 3 by 2 and limit is from 1 to 4. This is equal to 1 by 2 into now 1 upon 3 by 2 can be written as 2 by 3 into now putting the limits in place of y we get 4 to the power 3 by 2 minus 1 to the power 3 by 2. Now cancelling out 2 in denominator and numerator we get this is equal to 1 by 3 into now we can write 4 as 2 square so 2 square into 3 by 2. Again here we cancel out 2 in denominator and numerator minus now 1 to the power 3 by 2 is equal to 1 so 1. And this is further equal to 1 by 3 into now 2 to the power 3 is equal to 8 so we have 8 minus 1 and this is equal to 1 by 3 into 8 minus 1 is 7. So we get 7 by 3 as the required area hence the required area is equal to 7 by 3. So this is our required answer hope you have understood the solution take care and God bless you.