Use horizontal strips or a trapezium

Substitute the line into the parabola: $(2x-4)^2=4x$, so $4x^2-20x+16=0$, $x^2-5x+4=0$ and $(x-4)(x-1)=0$. Thus $(x,y)=(4,4)$ or $(1,-2)$.

The line is $x=(y+4)/2$ and the parabola is $x=y^2/4$. For $-2\le y\le4$, the line is to the right. Therefore $A=\int_{-2}^4[(y+4)/2-y^2/4]\,dy=\frac14\int_{-2}^4(2y+8-y^2)\,dy$.

$A=\frac14[y^2+8y-y^3/3]_{-2}^4=\frac14[(16+32-64/3)-(4-16+8/3)]=9$ square units.

The line, y-axis and $y=-2,4$ enclose a trapezium whose parallel horizontal sides have lengths $1,4$ and separation $6$. The source alternative setup is $A=\frac12(4+1)(6)-\frac14\int_{-2}^4y^2\,dy$. Independently evaluating the two terms gives $15-6=9$ square units.