Find the area bounded by the parabola y2 = 4x and the line y = 2x – 4.

(i) By using horizontal strips


(ii) By using vertical strips.

Given: - Two curves are y2 = 4x and y = 2x – 4



Now to find the area between these two curves, we have to find common area i.e. Shaded portion


Intersection of parabola y2 = 4x with line y = 2x – 4


Putting the value of y from the equation of a line in parabola equation, we get,


y2 = 4x


(2x – 4)2 = 4x


4x2 – 16x + 16 = 4x


4x2 – 20x + 16 = 0


4x2 – 16x – 4x + 16 = 0


4x(x – 4) – 4(x – 4) = 0


4(x – 1)(x – 4) = 0


(x – 1)(x – 4) = 0


x = 1,4


When x = 1, y = √4x


y = + 2, – 2; we take – 2 as the intersection is in the 4th quadrant and when x = 4, y = √4x


y = + 4, – 4; we take + 4 as the intersection is in 1st quadrant


Therefore intersection points are B(4,4) and C(1, – 2)


Area of the bounded region, taking strips


i) By using horizontal strips


Therefore, limits are for y and integrating with respect to y


Area bounded by region = {Area under line from – 2 to 4} –{Area under parabola from – 2 to 4}






Putting limits, we get







ii) By using vertical strips.


Therefore, limits are for x, and integrating with respect to x


Area bounded by region = {2(Area under parabola from 0 to 1) + (Area under parabola from 1 to 4)} – {Area under line from 1 to 4}


Tip: - Parabola is symmetrical about x - axis therefore its area is twice the area above x - axis. So, till its latus rectum i.e here a = 1, area is twice the area above x - axis.






Putting limits, we get,









Hence from both methods we get same answer


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