Sketch the region lying in the first quadrant and bounded by y=4x2, x=0,y=2 and y=4. Find the area of the region using integration.
Given the boundaries of the area to be found are,
• The curve y = 4x2
• y = 0, (x-axis)
• y = 2 (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 = 4x2, has only the positive numbers as x has even power, so it is about the y-axis equally distributed on both sides.
• y= 2 and y=4 are parallel to x-axis at of 2 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 = 4x2 and y-axis meet i.e. A(0,4)
•Point B, where the curve y = 4x2 and y=4 meet i.e. B(1,4)
•Point C, where the curve y = 4x2 and y=2 meet
•Point D, where the y-axis and y=2 meet i.e. D(0,2).
Consider the curve, y = 4x2
Area of the required region = Area of ABCD.
[Using the formula]
The Area of the required region
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