Question

Sketch the graphs of the curves y² = x and y² = 4 – 3x and find the area enclosed between them. 

Answer 1

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