Explanation

(b)
The point of intersection of the parabola and the line are
$A(2,1)and\ B\left( 1 - \frac{1}{4} \right)$
[image:8_image49.jpeg]
$\therefore The\ requerd\ area = \left\lbrack \int_{- 1}^{2}{y\ dx} \right\rbrack - \left\lbrack \int_{- 1}^{2}{y\ dx} \right\rbrack$
$= \int_{- 1}^{2}{\frac{1}{4}(x + 2)dx - \int_{- 1}^{2}{\frac{1}{4}\ x^{2}\ dx}}$
$= \frac{1}{4}\left\lbrack \frac{x^{2}}{2} + 2x \right\rbrack_{- 1}^{2} - \frac{1}{4}\left\lbrack \frac{x^{3}}{3} \right\rbrack_{- 1}^{2} = \frac{9}{8}\ sq\ units$