Science:Math Exam Resources/Courses/MATH105/April 2011/Question 02/Solution 1

To compute the area, we need to know the points of intersection of the curves. For y=4x and xy=1, i.e. point P, this is when x(4x) = 1, i.e. $x^{2}=1/4$. Equivalently x=1/2, and y=2.

For y=x and xy=1, i.e. point Q, this is (x,y) = (1,1).

Thus, the total area is

${\begin{aligned}A&=\int _{0}^{1/2}(4x-x)\,dx+\int _{1/2}^{1}\left({\frac {1}{x}}-x\right)\,dx\\&=3\int _{0}^{1/2}x\,dx+\int _{1/2}^{1}{\frac {1}{x}}\,dx-\int _{1/2}^{1}x\,dx\\&=\left.{\frac {3}{2}}x^{2}\right|_{0}^{1/2}+\left.\ln x\right|_{1/2}^{1}-\left.{\frac {x^{2}}{2}}\right|_{1/2}^{1}\\&={\frac {3}{8}}+\ln 1-\ln(1/2)-\left({\frac {1}{2}}-{\frac {1}{8}}\right)\\&={\frac {3}{8}}+\ln 2-{\frac {1}{2}}+{\frac {1}{8}}=\color {blue}{\ln 2}\end{aligned}}$