total area of shape \( = 2 + 2\int_0^1 {\left( {1 - {x^2}} \right)dx = 2 + 2\left[ {x - \frac{1}{3}{x^3}} \right]} _0^1 = 2 + 2\left[ {\left( {1 - \frac{1}{3}} \right) - 0} \right] = 2 + \frac{4}{3} = \frac{{10}}{3}\)thus, the area of the shaded region in the figure below needs to be \(\frac{5}{3}\)find equation of the line passing through the points A and C:gradient of AC \( = \frac{{\left( {1 - {c^2}} \right) - \left( { - 1} \right)}}{{c - \left( { - 1} \right)}} = \frac{{2 - {c^2}}}{{c + 1}}\)gradient and point \(\left( { - 1,\; - 1} \right)\) in gradient-slope form of line:\(y + 1 = \left( {\frac{{2 - {c^2}}}{{c + 1}}} \right)\left( {x + 1} \right)\)\(y = \left( {\frac{{2 - {c^2}}}{{c + 1}}} \right)x + \frac{{2 - {c^2}}}{{c + 1}} - \frac{{c + 1}}{{c + 1}}\)equation of line AC is \(y = \left( {\frac{{2 - {c^2}}}{{c + 1}}} \right)x + \frac{{1 - c - {c^2}}}{{c + 1}}\)area of shaded region \( = \int_{ - 1}^c {\left[ {\left( {1 - {x^2}} \right) - \left( {\frac{{2 - {c^2}}}{{c + 1}}x + \frac{{1 - c - {c^2}}}{{c + 1}}} \right)} \right]} \,dx = \frac{5}{3}\)solve for c:\(\int_{ - 1}^c {\left[ { - {x^2} + \frac{{{c^2} - 2}}{{c + 1}}x + \frac{{{c^2} + c - 1}}{{c + 1}} + \frac{{c + 1}}{{c + 1}}} \right]} \,dx = \frac{5}{3}\)