Science:Math Exam Resources/Courses/MATH101/April 2007/Question 02 (d)

MATH101 April 2007

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Question 02 (d)
Let R be the region under the curve $y=e^{-x}$ and above the x-axis, for $0\leq x\leq 1$. Find the x-coordinate of the centroid (centre of mass) of R.

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 x coordinate of the centroid is computed using the formulas $\displaystyle {{\bar {x}}={\frac {1}{A}}\int _{a}^{b}xf(x)dx}$ where $A=\int _{a}^{b}f(x)\,dx$

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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. $A=\int _{0}^{1}e^{-x}\,dx=\left.-e^{-x}\right|_{0}^{1}=-e^{-1}+e^{0}=-{\frac {1}{e}}+1={\frac {e-1}{e}}$ Next, to compute the x-coordinate, we want to compute ${\bar {x}}={\frac {1}{A}}\int _{a}^{b}xf(x)\,dx={\frac {e}{e-1}}\int _{0}^{1}xe^{-x}\,dx$ To do this, we use integration by parts. Let ${\begin{aligned}u=x\quad &\quad v=-e^{-x}\\du=dx\quad &\quad dv=e^{-x}\end{aligned}}$ So that ${\begin{aligned}{\bar {x}}&={\frac {e}{e-1}}\int _{0}^{1}xe^{-x}\,dx\\&={\frac {e}{e-1}}\left(\left.-xe^{-x}\right|_{0}^{1}+\int _{0}^{1}e^{-x}\,dx\right)\\&={\frac {e}{e-1}}\left(-e^{-1}-\left.e^{-x}\right|_{0}^{1}\right)\\&={\frac {e}{e-1}}\left(-e^{-1}-e^{-1}+e^{0}\right)\\&={\frac {e}{e-1}}\left(1-{\frac {2}{e}}\right)\\&={\frac {e}{e-1}}\left({\frac {e-2}{e}}\right)\\&={\frac {e-2}{e-1}}\end{aligned}}$ completing the question.

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