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Question
Sketch the graphs of the curves y2 = x and y2 = 4 – 3x and find the area enclosed between them.
Solution
y2 = -3x + 4
= - 3 `( x - (4)/(3))`
The vertex L of this parabola is `((4)/(3),0)`.
It cuts the y-axis at A (0, 2) and B (0, - 2).
The points of intersection of these two parabolas are given by the equation
y2 = x and y2 = 4 - 3x as x = - 3x + 4 ⇒ x = 1
Then y2 = 1 ⇒ y = ± 1
Thus, the points of intersection are P (1, 1) and Q (1, - 1). Let PQ cut the x-axis at R.
∴ Total area of POQLP = 2 area of OPRQO
= 2 `[ int_0^1sqrtx dx + int_1^(4/3) sqrt(4 - 3x) dx]`
= 2`[ (( x^(3/2)) /(3/2))_0^1 + ((2(4 - 3x))/((-3) xx 3))_1^(4/3)]`
= 2`[ ((2)/(3) - 0) - (2)/(9) (0 - 1)]`
= 2`[(2)/(3) + (2)/(9)] = 2 [ (6 + 2)/(9)]`
= `(16)/(9)` sq. units
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