MATH105 April 2010
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Question 03

Find the area of the region bounded by
 $y={\frac {2}{x}},\quad y=4{\sqrt {x}}\quad {\text{and}}\quad y=x.$

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

What does this region look like? How can you find the area of such a region?

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Solution

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First, we sketch each of the three curves to see what is the region that we are talking about.
We will need to find the coordinates of points A and B which are the intersection of curves. For A we have
 ${\begin{aligned}x&={\frac {2}{x}}\\x^{2}&=2\\x&=\pm {\sqrt {2}}\end{aligned}}$
We can see from the picture that we are interested in the point with a positive x coordinate, so
 $\displaystyle A=({\sqrt {2}},{\sqrt {2}})$
For B we have
 ${\begin{aligned}4{\sqrt {x}}&={\frac {2}{x}}\\x^{3/2}&={\frac {1}{2}}\\x&=\left({\frac {1}{2}}\right)^{2/3}={\frac {1}{\sqrt[{3}]{4}}}\end{aligned}}$
And so since y = 2/x we have
 $\displaystyle B=\left({\frac {1}{\sqrt[{3}]{4}}},2{\sqrt[{3}]{4}}\right)$
This allows us to write the desired area as the sum of 2 integrals
 ${\begin{aligned}{\text{area}}&=\int _{0}^{\frac {1}{\sqrt[{3}]{4}}}\left((x)(4{\sqrt {x}})\right)\,dx+\int _{\frac {1}{\sqrt[{3}]{4}}}^{\sqrt {2}}\left((x)({\frac {2}{x}})\right)\,dx\\&=\int _{0}^{\frac {1}{\sqrt[{3}]{4}}}\left(x+4{\sqrt {x}}\right)\,dx+\int _{\frac {1}{\sqrt[{3}]{4}}}^{\sqrt {2}}\left(x+{\frac {2}{x}}\right)\,dx\\&=\left[{\frac {x^{2}}{2}}+{\frac {4\cdot 2}{3}}x^{3/2}\right]_{0}^{\frac {1}{\sqrt[{3}]{4}}}+\left[{\frac {x^{2}}{2}}+2\ln(x)\right]_{\frac {1}{\sqrt[{3}]{4}}}^{\sqrt {2}}\\&={\frac {(2^{2/3})^{2}}{2}}+{\frac {8}{3}}(2^{2/3})^{3/2}+00{\frac {({\sqrt {2}})^{2}}{2}}+2\ln({\sqrt {2}})+{\frac {(2^{2/3})^{2}}{2}}2\ln(2^{2/3})\\&={\frac {4}{3}}1+\ln(2)+{\frac {4}{3}}\ln(2)\\&={\frac {1}{3}}+{\frac {7}{3}}\ln(2)\end{aligned}}$

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