Explanation

(c)
The point of intersection of given curves are (0,0)and $(3, - 3)$
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$\therefore Required\ area = area\ of\ curve\ OAB + area\ of\ curve\ OCB$
$= \int_{0}^{2}{\left( 2x - x^{2} \right)dx + \left| \int_{0}^{3}{( - x)dx} \right| - \left| \int_{2}^{3}{\left( 2x - x^{2} \right)dx} \right|}$
$= \left\lbrack x^{2} - \frac{x^{3}}{3} \right\rbrack_{0}^{2} + \left| \left\lbrack - \frac{x^{2}}{2} \right\rbrack_{0}^{3} \right| - \left| \left\lbrack x^{2} - \frac{x^{3}}{3} \right\rbrack_{2}^{3} \right|$
$= \frac{4}{3} + \frac{9}{2} - \frac{4}{3} = \frac{9}{2}\ sq\ units$
Alternate
$Area = \int_{0}^{3}{\left\lbrack \left( 2x - x^{2} \right) - ( - x) \right\rbrack dx}$
$= \int_{0}^{3}{\left( 3x - x^{2} \right)dx = \left\lbrack \frac{{3x}^{2}}{2} - \frac{x^{3}}{3} \right\rbrack_{0}^{3}}$
$= \frac{27}{2} - \frac{27}{3} = \frac{9}{2}\ sq\ units$