The area bounded between the parabola \(y^{2}=4 x\) and the line \(\mathrm{y}=2 \mathrm{x}-4\) is equal to

The point of intersection of \(\mathrm{y}^{2}=4 \mathrm{x}\) and \(y=2 x-4\) is



\(\begin{array}{cc} & (2 x-4)^{2}=4 x \\ \Rightarrow & x^{2}-5 x+4=0 \\ \Rightarrow & \quad(x-1)(x-4)=0 \\ \Rightarrow & x=1,4 \\ \Rightarrow & y=-2,4 \\ \therefore \text { Required area } & \\ = & \int_{-2}^{4}\left(\frac{\mathrm{y}+4}{2}\right) \mathrm{dy}-\int_{-2}^{4} \frac{\mathrm{y}^{2}}{4} \mathrm{dy} \\ = & \frac{1}{2}\left[\frac{\mathrm{y}^{2}}{2}+4 \mathrm{y}\right]_{-2}^{4}-\frac{1}{4}\left[\frac{\mathrm{y}^{3}}{3}\right]_{-2}^{4} \\ = & \frac{1}{2}[8+16-(2-8)]-\frac{1}{12}[64+8] \\ = & \frac{1}{2}[30]-\frac{1}{12}(72) \\ = & 15-6=9 \mathrm{sq} \text { unit }\end{array}\)