Sketch the region lying in the first quadrant and bounded by y=9x2, x=0, y=1 and y=4. Find the area of the region, using integration.

Given the boundaries of the area to be found are,


• The curve y = 9x2


• x = 0, (y-axis)


• y = 1 (a line parallel to x-axis)


• y = 4 (a line parallel to x-axis)


• The area which is occurring in the 1st quadrant is required.



As per the given boundaries,


• The curve y = 9x2, has only the positive numbers as x has even power, so it is about the y-axis equally distributed on both sides.


• y= 1 and y=4 are parallel to x-axis at of 1 and 4 units respectively from the x-axis.


As the area should be in the 1st quadrant, the four boundaries of the region to be found are,


•Point A, where the curve y = 9x2 and y-axis meet i.e. A(0,4)


•Point B, where the curve y = 9x2 and y=4 meet


•Point C, where the curve y = 9x2 and y=1 meet


•Point D, where the y-axis and y=1 meet i.e. D(0,1).


Consider the curve, y = 9x2


Now,



Area of the required region = Area of ABCD.




[Using the formula]



The Area of the required region


1