Let $y^2 = 4x ---- (1)$ Parabola

And $y = 2x-4 ---- (2)$ Line

Substituting eq (2) in (1)

We get

$(2x - 4)^2 = 4x\\ 4x^2 – 16x + 16 = 4x \\ 4x^2 – 20x + 16 = 0 \\ X = 1 \space or\space x = 4 \\ When \space x = 1 ,y = 2 – 4 = -2\\ When \space x = 4 ,y = 8 - 4 = 4\\ $

Parabola and line intersect at $(1, -2)$ and $(4, 4)$

The area bounded by parabola and line = $\int\limits_{y_1}^{y_2}\int\limits_{x_1}^{x_2}dxdy$

Now consider horizontal strip as the limits will remain constant.

1) Outer limit y:- Horizontal strip will slide from $y = -2$ to $y = 4$

So $y = -2$ to $y = 4$

2) Inner limit x:-

(a)Upper limit is equation of line

$Y= 2x – 4\\ \therefore x=\dfrac {y+4}2$

(b) Lower limit x is equation of parabola i.e $y^2 = 4x \\ \therefore x=\dfrac {y^2}4\\ Area= \int\limits_{y=-2}^{y=4}\int\limits_{x=\frac {y^2}4}^{x=\frac {y+4}2}dxdy \\ =\int\limits_{y=-2}^{y=4}[x] ^{\frac {y+4}2}_{\frac {y^2}4}dy\\ =\int\limits_{-2}^4 \Bigg[\dfrac {y+4}2 -\dfrac {y^2}4\Bigg]dy \\ = \int\limits^4_{-2}\Bigg(\dfrac y2+2-\dfrac {y^2}4\Bigg)dy\\ =\Bigg[\dfrac {y^2}{2\times 2}+2y-\dfrac {y^3}{4\times 3}\Bigg]^4_{-2}\\ =\Bigg[\dfrac {4^2}4+2\times 4-\dfrac {4^3}{4\times 3}\Bigg]-\Bigg[\dfrac 44-4+\dfrac {2^3}{4\times 3}\Bigg]\\ Area =\Bigg[4+8-\dfrac{16}3\Bigg] -\Bigg[1-4+\dfrac 23\Bigg] \\ =\dfrac {20}3-\Bigg[\dfrac {-7}3\Bigg] \\ =\dfrac {27}3 \\ =9$