Using integration, find the area of region bounded by the line y-1=x, the x-axis, and the ordinates x=-2 and x=3.
Given the boundaries of the area to be found are,
• The line equation is y = x +1
• The y= 0, x-axis
• x = -2 (a line parallel toy-axis)
• x = 3 (a line parallel toy-axis)
Thus the given boundaries are,
• The line y = x+1.
• x=-2 is parallel toy-axis at 2 units away from the y-axis.
• x=3 is parallel toy-axis at 3 units away from the y-axis.
• y = 0, the x-axis.
The four vertices of the region are,
•Point A, where the line y = x+ and x=3 meet i.e. A(3,4).
•Point B, where the line y = x +1 and x=-1 meet i.e.
•Point C, where the x-axis and x=-2 meet i.e. C (-2,0).
•Point D, where the x-axis and x=3 meet i.e. D(3,0).
Area of the required region = Area of ABCD.
From (1) we can clearly say that, the area of ABCD has to be divided into twopieces i.e. area under CBE and ADE as the line equations changes the sign of x.
So the equation AB becomes negative between after it crosses the point E.
[Using the formula and ]
The Area of the required region = 8.5 sq. units.
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