Question

Find by double integration the area bounded by the parabola 𝒚𝟐=𝟒𝒙 And 𝒚=𝟐𝒙−𝟒 

Answer 1

The parabola 𝑦²=4𝑥 and the line 𝑦=2𝑥−4 intersect where (2𝑥−4)²=4𝑥 

∴ `4x^2-16x+16=4x`          ∴ `4x^2-20x+16=0` 

∴` x^2-5x+4=0`                ∴ `(x-4) (x-1)=0` 

∴` x=1,4`

When x = 1, y = 2 – 4 = -2; and when x = 4, y = 8 - 4 = 4. Thus, the points of intersection are A (1, -2) and B (4, 4).
Now, consider a strip parallel to x-axis. On this strip x varies from x = y²/4 to x = (y+4)/2. The strip then moves parallel to the x-axis from y = -2 to y = 4. 

∴` A= int_-2^4 int_(y^2)"^(y+4)/2 dxdy=int_-2^4 [x]_(y^2/4)^((y+4)/2) dy`    

=`int_-2^4 ((y+4)/2 - y^2/4)dy` 

= `1/4 int_-2^4(2y+8-y^2) dy` 

=`1/4 int_-2^4(2y+8-y^2)dy` 

=` 1/4 [y^2+8y-y^3/3]_-2^4` 

=`1/4[(16+32-64/3)-(4-16+8/3)]` 

=`1/4(60-24)` 

∴` A=9``