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Question
Find the area of the region bound by the curves y = 6x – x2 and y = x2 – 2x
Solution
y = 6x - x2, y = x2 - 2x
x2 - 6x = - y
⇒ x2 - 6x + 9 = - y + 9
⇒ ( x - 3 )2 = - ( y - 9 )
This represents a downward parabola with vertex (3, 9) ...(i)
y = x2 - 2x
x2 - 2x + 1 = y + 1
⇒ ( x - 1 )2 = y + 1
⇒ ( x -1 )2 = y - ( -1 ) ...(ii)
This represents an upward parabola with vertex (1, -1)
Point of intersection is given by:
6x - x2 = x2 - 2x
⇒ 2x2 = 8x
⇒ x2 - 4x = 0
⇒ x (x - 4) = 0
x = 0, x = 4
y = 6 x 4 - 42
= 24 - 16 = 8
Point of intersection is (0, 0) ; (4, 8)
Reqd. area = ` int_0^4 (6x - x^2) - (x^2 - 2x) dx`
= `int_0^4 8x - 2x^2 dx = [ (8x^2)/(2) - (2)/(3) x^3]_0^4`
= 4 x 42 - `(2)/(3) (4)^3 = (64)/(3)` sq. units.
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