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Question
Find the area of the region included between: y2 = 4x, and y = x
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Solution
The vertex of the parabola y2 = 4x is at the origin O = (0, 0).
Points of intersection of parabola and line are
∴ x2 = 4x
x2 = 4x = 0
∴ x(x - 4) = 0
x = 0 or x = 4
∴ y = x
points are (0, 0) & (4, 4)
Area bounded by parabola and line is = Area (OABCO)
= A (OCBCO) - A (OCBAO)
`int_0^4 2sqrtx*dx - int_0^4x*dx`
= `2 [x^(3/2)/(3/2)]_0^4 - [x^2/2]_0^4`
= `2 xx 2/3 [(4)^(3/2) - (0)^(3/2)] - 1/2 [4^2 - 0^2]`
= `4/3[8 - 0]- 1/2 [16 - 0]`
= `32/3 - 8`
= `(32 - 24)/3`
Area = `8/3` square units.
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