Question

Find the area bounded by the curves $y=x+\sqrt{x^2}$ and xy = 1, The x-axis and the line x=2.

Answer 1

$\begin{array}{c}y=x+\sqrt{x^2},xy=1,x=2\\ Area={\int }_0^{\frac{1}{\sqrt{2}}}\left(x+\sqrt{x^2}\right)dx+{\int }_{\frac{1}{\sqrt{2}}}^2\frac{1}{x}dx\\ ={\left(\frac{x^2}{2}+\frac{x^2}{2}\right)}_0^{\frac{1}{\sqrt{2}}}+{\left(\log \left|x\right|\right)}_{\frac{1}{\sqrt{2}}}^2\\ =\frac{1}{2}+\log 2-\log 1+\log 2^{\frac{1}{2}}\\ =\left(\frac{1}{2}+\frac{3}{2}\log 2\right)\end{array}$