Q93 / 180 · KCET 2009 · Mathematics
The area bounded between the parabola $y^{2}=4 x$ and the line $\mathrm{y}=2 \mathrm{x}-4$ is equal to
- A$\frac{17}{3}$ sq unit
- B$\frac{19}{3}$ sq unit
- Correct answerC9 sq unit
- D15 sq unit
Solution
The point of intersection of $\mathrm{y}^{2}=4 \mathrm{x}$ and $y=2 x-4$ is
$\begin{array}{cc} & (2 x-4)^{2}=4 x \\ \Rightarrow & x^{2}-5 x+4=0 \\ \Rightarrow & \quad(x-1)(x-4)=0 \\ \Rightarrow & x=1,4 \\ \Rightarrow & y=-2,4 \\ \therefore \text { Required area } & \\ = & \int_{-2}^{4}\left(\frac{\mathrm{y}+4}{2}\right) \mathrm{dy}-\int_{-2}^{4} \frac{\mathrm{y}^{2}}{4} \mathrm{dy} \\ = & \frac{1}{2}\left[\frac{\mathrm{y}^{2}}{2}+4 \mathrm{y}\right]_{-2}^{4}-\frac{1}{4}\left[\frac{\mathrm{y}^{3}}{3}\right]_{-2}^{4} \\ = & \frac{1}{2}[8+16-(2-8)]-\frac{1}{12}[64+8] \\ = & \frac{1}{2}[30]-\frac{1}{12}(72) \\ = & 15-6=9 \mathrm{sq} \text { unit }\end{array}$
$\begin{array}{cc} & (2 x-4)^{2}=4 x \\ \Rightarrow & x^{2}-5 x+4=0 \\ \Rightarrow & \quad(x-1)(x-4)=0 \\ \Rightarrow & x=1,4 \\ \Rightarrow & y=-2,4 \\ \therefore \text { Required area } & \\ = & \int_{-2}^{4}\left(\frac{\mathrm{y}+4}{2}\right) \mathrm{dy}-\int_{-2}^{4} \frac{\mathrm{y}^{2}}{4} \mathrm{dy} \\ = & \frac{1}{2}\left[\frac{\mathrm{y}^{2}}{2}+4 \mathrm{y}\right]_{-2}^{4}-\frac{1}{4}\left[\frac{\mathrm{y}^{3}}{3}\right]_{-2}^{4} \\ = & \frac{1}{2}[8+16-(2-8)]-\frac{1}{12}[64+8] \\ = & \frac{1}{2}[30]-\frac{1}{12}(72) \\ = & 15-6=9 \mathrm{sq} \text { unit }\end{array}$
Hence, the correct answer is option: 'C'.