Explanation

(d)
Given curves are
$y^{2} = 2x + 1$
$and\ x - y = 1$
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Points of intersection are $A(0, - 1)and\ \ B(4,3)$
$Area = \int_{- 1}^{3}{(1 + y)dy - \int_{- 1}^{3}{\left( \frac{y^{2} - 1}{2} \right)dy}}$
$= \left\lbrack y + \frac{y^{2}}{2} \right\rbrack_{1}^{3} - \left\lbrack \frac{1}{2}\left( \frac{y^{3}}{3} - y \right) \right\rbrack_{- 1}^{3}$
$= \left\lbrack 3 + \frac{9}{2} - \left( - 1 + \frac{1}{2} \right) \right\rbrack - \frac{1}{2}\left\lbrack 9 - 3 - \left( - \frac{1}{3} + 1 \right) \right\rbrack$
$= 8 - \frac{8}{3} = \frac{16}{3}$