Explanation

(c)
$\therefore$ Required area
$= \int_{0}^{1}{\left\lbrack \left( 2x - 2x^{2} \right) - \left( x\log x \right) \right\rbrack dx}$
[image:8_image19.jpeg]
$= \left\lbrack x^{2} - \frac{{2x}^{3}}{3} - \left( \frac{x^{2}}{2}\log{x - \frac{x^{2}}{4}} \right) \right\rbrack_{0}^{1}$
$= \left\lbrack 1 - \frac{2}{3} - \left( 0 - \frac{1}{4} \right) \right\rbrack = \frac{7}{12}\ sq\ unit$