Mathematics - Area Under Curves Question with Solution | TestHub

$\begin{array}{rl}& A={\int }_{\frac{1}{2}}^{\frac{1+\sqrt{5}}{2}}\left(1+3x-2x^2-\frac{1}{x}\right)\mathrm{dx}\\ & A={\left[x+\frac{3x^2}{2}-\frac{2x^3}{3}-\ln x\right]}_{\frac{1}{2}}^{1+\sqrt{5}}\\ & A=\frac{1+\sqrt{5}}{2}+\frac{3}{2}{\left(\frac{1+\sqrt{5}}{2}\right)}^2-\frac{2}{3}{\left(\frac{1+\sqrt{5}}{2}\right)}^3-\ln \left(\frac{1+\sqrt{5}}{2}\right)\\ & -\frac{1}{2}-\frac{3}{2}\left(\frac{1}{4}\right)+\frac{2}{3}\left(\frac{1}{8}\right)+\ln \left(\frac{1}{2}\right)\\ & A=\frac{1}{2}+\frac{\sqrt{5}}{2}+\frac{3}{8}+\frac{3}{4}\sqrt{5}+\frac{15}{8}-\frac{4}{3}-\frac{2}{3}\sqrt{5}\\ & -\frac{1}{2}-\frac{3}{8}+\frac{1}{12}-\ln (1+\sqrt{5})\\ & =\sqrt{5}\left(\frac{1}{2}+\frac{3}{4}-\frac{2}{3}\right)+\frac{15}{8}-\frac{4}{3}+\frac{1}{12}-\ln (1+\sqrt{5})\\ & =\frac{14}{24}\sqrt{5}+\frac{15}{24}-\ln (1+\sqrt{5})\end{array}$