The area enclosed between the parabola $y^2=4 x$ and the line $y=2 x-4$ is

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Aicharya · Accepted Answer

Putting $x=\frac{y^2}{4}$ in $y=2 x-4$, we get $\begin{aligned} & y=2\left(\frac{y^2}{4}\right)-4 \\ & \Rightarrow y^2-2 y-8=0 \\ & \Rightarrow(y-4)(y+2)=0 \\ & \Rightarrow y=4,-2 \end{aligned}$ $\begin{aligned} \therefore \quad & \text { Required area }=\int_{-2}^4\left(\frac{y+4}{2}-\frac{y^2}{4}\right) \mathrm{d} y \\ & =\frac{1}{2}\left[\frac{y^2}{2}+4 y\right]_{-2}^4-\frac{1}{4}\left[\frac{y^3}{3}\right]_{-2}^4 \\ & =\frac{1}{2}[8+16-(2-8)]-\frac{1}{12}[64-(-8)] \\ & =15-6 \\ & =9 \text { sq. units } \end{aligned}$

The area enclosed between the parabola $y^2=4 x$ and the line $y=2 x-4$ is

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