question_answer

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

A)  \[\frac{17}{3}sq\,unit\]       

B)  \[\frac{19}{3}sq\,unit\]

C)  \[9\,sq\,unit\]

D)  \[15\,sq\,unit\]

Correct Answer: C

Solution :

The point of intersection of \[{{y}^{2}}=4x\] and \[y=2x-4\]is                 \[{{(2x-4)}^{2}}=4x\] \[\Rightarrow \]               \[{{x}^{2}}-5x+4=0\] \[\Rightarrow \]               \[(x-1)(x-4)=0\] \[\Rightarrow \]               \[x=1,4\] \[\Rightarrow \]               \[y=-2,4\] \[\therefore \]  Required area                 \[=\int_{-2}^{4}{\left( \frac{y+4}{2} \right)dy-\int_{-2}^{4}{\frac{{{y}^{2}}}{4}}dy}\]                 \[=\frac{1}{2}\left[ \frac{{{y}^{2}}}{2}+4y \right]_{-2}^{4}-\frac{1}{4}\left[ \frac{{{y}^{3}}}{3} \right]_{-2}^{4}\]                 \[=\frac{1}{2}[8+16-(2-8)]-\frac{1}{12}[64+8]\]                 \[=\frac{1}{2}[30]-\frac{1}{12}(72)\]                 \[=15-6=9sq\,unit\]