Area bounded by y = \(x^2\) - x + 3 and y = 3 The intersection is given as \(x^2\) - x + 3 = 3 \(x^2\) - x = 0 x(x -1) = 0 The graph is sketched between x= 0 and x = 1 \(\int^{1}_{0}\)\(x^2\) - x + 3)dx - \(\int^{1}_{0}\)3dx = \(\int^{1}{0}(x^2\) - x)dx [\(\frac{2x^3}{3} - \frac{x^2}{2}\)] = f(1) - f(0) = [\(\frac{2}{3} - \frac{1}{2}\)] - [0 - 0] = \(\frac{4 - 3}{6}\) = \(\frac{1}{6}\)unit(sq)