Explanation

(c)
The point of instruction of $y^{2} = 4x\ and\ y = 2x - 4\ is$
$(2x - {4)}^{2} = 4x$
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$\Longrightarrow \ x^{2} - 5x + 4 = 0$
$\Longrightarrow (x - 1)(x - 4) = 0$
$\Longrightarrow x = 1,4$
$\Longrightarrow y = - 2,4$
$\therefore$ Required area
$\int_{- 2}^{4}{\left( \frac{y + 4}{2} \right)dy - \int_{- 2}^{4}{\frac{y^{2}}{4}dy}}$
$= \frac{1}{2}\left\lbrack \frac{y^{2}}{2} + 4y \right\rbrack_{- 2}^{4} - \frac{1}{4}\left\lbrack \frac{y^{3}}{3} \right\rbrack_{- 2}^{4}$
$= \frac{1}{2}\left\lbrack 8 + 16 - (2 - 8) \right\rbrack - \frac{1}{12}\lbrack 64 + 8\rbrack$
$= 15 - 6 = 9\ sq\ units$