Determine the area enclosed by y =x and
y =x^2
y=x....(1)y=x.... (1)y=x....(1)
y=x2.....(2)y=x^2.....(2)y=x2.....(2)
From equation (1) and (2)
x=x2x=x^2x=x2
or x−x2=0x-x^2=0x−x2=0
x=0,1x=0,1x=0,1
But y=xy=xy=x
y=0,1y=0,1y=0,1
(0,0)(0,0)(0,0) and (1,1)(1, 1)(1,1) are points of intersection
A=∫01(x−x2)dxA=\int_0^1(x-x^2) dxA=∫01(x−x2)dx
=[x2/2−x3/3]01=[{x^2/2} - {x^3/3}]_0^1=[x2/2−x3/3]01
=[1/2−1/3]−[0−0]=[1/2-1/3]-[0-0]=[1/2−1/3]−[0−0]
=1/6=1/6=1/6 square units
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