MATH103 April 2009
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Question 05 (b)

A carrot has the shape formed by revolving the region under the graph of
 $f(x)={\sqrt {14x}}$
for $0\leq x\leq 12$ cm about the x axis. The concentration of vitamin A is found to vary in the carrot with x according to
 $c(x)={\frac {1}{12}}e^{x/12}{\frac {\text{mg}}{{\text{cm}}^{3}}}$
What is the total amount of vitamin A in the carrot?

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1

The amount of Vitamin A is
concentration x volume.

Hint 2

You need to multiply the concentration and the diskvolume for every infinitly small disk and then integrate over all disks.

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Solution

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To find the amount A of Vitamin A in the carrot, we need to multiply every infinitly small disk with unit $\ cm^{3}$ with the
passend concentration of Vitamin A and then integrate over all disks:
 ${\begin{aligned}A&=\int _{0}^{12}{\frac {1}{12}}e^{{\frac {x}{12}}}\left({\sqrt {14x}}\right)^{2}\pi dx\\&={\frac {\pi }{12}}\int _{0}^{12}e^{{\frac {x}{12}}}(14x)dx\\&={\frac {\pi }{12}}\left[\int _{0}^{12}14e^{{\frac {x}{12}}}dx\int _{0}^{12}xe^{{\frac {x}{12}}}dx\right]\end{aligned}}$
We let the former integral as is for now and use integration by parts for the latter. Let u = x, du = dx, dv = e^{x/12}dx and v = 12e^{x/12}. Then
${\begin{aligned}A&={\frac {\pi }{12}}\left[\int _{0}^{12}14e^{{\frac {x}{12}}}dx\int _{0}^{12}xe^{{\frac {x}{12}}}dx\right]\\&={\frac {\pi }{12}}\left[\int _{0}^{12}14e^{{\frac {x}{12}}}dx\left(\left.x(12)e^{{\frac {x}{12}}}\right_{0}^{12}\int _{0}^{12}(12)e^{{\frac {x}{12}}}\,dx\right)\right]\\&={\frac {\pi }{12}}\left[14\int _{0}^{12}e^{{\frac {x}{12}}}\,dx12\int _{0}^{12}e^{{\frac {x}{12}}}\,dx+(12)(12)e^{1}0\right]\\&={\frac {\pi }{12}}\left[2\int _{0}^{12}e^{{\frac {x}{12}}}\,dx+(12)(12)e^{1}\right]\\&={\frac {\pi }{12}}\left[\left.2(12)e^{{\frac {x}{12}}}\right_{0}^{12}+(12)(12)e^{1}\right]\\&={\frac {\pi }{12}}\left[2(12)(e^{1}1)+(12)(12)e^{1}\right]\\&=\pi \left[2e^{1}+2+12e^{1}\right]\\&=\pi \left[{\frac {10}{e}}+2\right]\\&={\frac {2\pi (5+e)}{e}}.\end{aligned}}$

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