MATH105 April 2011
• Q1 (a) • Q1 (b) • Q2 • Q3 • Q4 • Q5 (a) • Q5 (b) • Q5 (c) • Q6 • Q7 • Q8 (a) • Q8 (b) • Q8 (c) • Q9 (a) • Q9 (b) • Q9 (c) • Q9 (d) • Q9 (e) • Q9 (f) • Q9 (g) •
Question 02

Sketch the region in the first quadrant bounded by the curves
$y=x,\qquad y=4x,\qquad {\text{and}}\qquad xy=1$
and find the area of this region.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

The area between the curves f(x) and g(x) is given by
 $\int _{a}^{b}f(x)g(x)\,dx.$

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Solution

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

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