Question

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.

Answer 1

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 1^st 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 1^st 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,

x = \(\frac{1}{3}\) \(\sqrt{y}\)

Area of the required region = Area of ABCD.

The Area of the required region  = \(\frac{14}{9}\)sq.units