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Question
Draw a rough sketch of the curve y2 = 4x and find the area of region enclosed by the curve and the line y = x.
Solution
y2 = 4x is a right-handed parabola with vertex at O(0,0) and axis of parabola is x- axis.
y = x is a line passing through origin O(0,0)
Now, Finding their intersection
y2 = 4x
⇒ x2 = 4x
⇒ x2 - 4x = 0
⇒ x(x - 4) = 0
⇒ x = 0 and x = 4
Also y = x ⇒ y = 0 and y = 4
∴ Points of intersections are (0,0) and (4,4)
Required area = `2 int_0^4 sqrt"x" "dx" - int_0^4 "x" "dx"`
`= 2["x"^(3/2)/(3/2)]_0^4 - ["x"^2/2]_0^4`
`= 2xx2/3 |(4)^(3/2) - 0| - |4^2/2 -0 |`
`= 4/3 xx 8 - 8`
`= (32 - 24)/3 = 8/3`
`= 2 8/3 "sq.units"`
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