Science:Math Exam Resources/Courses/MATH101/April 2010/Question 01 (e)

MATH101 April 2010

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Question 01 (e)
Find the $x$-coordinate of the centroid of the region below $y=1/x^{2}$ and above the $x$-axis, from $x=1$ to $x=2.$ Simplify your answer as much as possible.

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
Recall the centroid ${\bar {x}}$ for a shape with area $A$ is given by $\displaystyle {\bar {x}}={\frac {1}{A}}\int _{a}^{b}xf(x)\,dx$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First, we compute the area of the curve in the region specified. $A=\int _{1}^{2}{\frac {1}{x^{2}}}\,dx=-x^{-1}\mid _{1}^{2}=-{\frac {1}{2}}+{\frac {1}{1}}={\frac {1}{2}}$ Next, using the formula for the x-coordinate of the centroid, we have ${\begin{aligned}{\bar {x}}&={\frac {1}{\tfrac {1}{2}}}\int _{1}^{2}x\cdot {\frac {1}{x^{2}}}\,dx\\&=2\int _{1}^{2}{\frac {1}{x^{1}}}\,dx\\&=2\ln(x)\mid _{1}^{2}\\&=2(\ln(2)-\ln(1))\\&=2\ln(2)\\&=\ln(4)\end{aligned}}$ completing the problem.

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