Hi, there!
When working at a trading company, I often received visitors from abroad.
I would fetch them up at the airport and send them to their hotel, and the
following day, I facilitated a meeting to get a firm order or took them to a
supplier to haggle over purchasing prices from vendors.
One of the most impressive clients was from Colombia, South America.
Its CEO graduated from the Massachusetts Institute of Technology (MIT),
and he was very shrewd and, put it straightforwardly, nimble and agile;
that’s the word.
I took over this client after my seniors became independent and started
their own businesses. I remember, however, it was a very onerous task to
maintain or increase its business volume because he was a tough negoti-
ator, though he was not “lean and agile” as often seen in corporate mission
statements.
Getting the drift above, today, let me give MIT’s math exam a shoot.
Here comes its word problem.
Evaluate the following integral.
\(\int_{0}^{2}\frac{x\,dx}{(1 + x^{2})^{2}}\)
(Source: Massachusetts Institute of Technology Exams)
Even for a novice math learner, it should be an easy try.
Here is my solution to the problem.
\(\int_{0}^{2}\frac{x\,dx}{(1 + x^{2})^{2}}\)
\(u = 1 + x^{2}\)
\(du = 2xdx\)
\(\int_{0}^{2}\frac{x\,dx}{(1 + x^{2})^{2}}\)
\(= \frac{1}{2}\int_{1}^{5}\frac{du}{u^{2}}\)
\(= [-\frac{1}{2u}]_{1}^{5}\)
\(= -\frac{1}{10} – (-\frac{1}{2})\)
\(= \frac{1}{2} – \frac{1}{10}\)
\(= \frac{5 – 1}{10}\)
\(= \frac{4}{10}\)
\(= \frac{2}{5}\)
I’ll be looking forward to solving more challenging word problems from
MIT in due course.
Stay tuned, and expect to see my next post.
Keep well.
Frank Yoshida
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【グローバルビジネスで役立つ数学】でもっと学習する b＾＾)
【参考図書】『もう一度高校数学』（著者：高橋一雄氏）株式会社日本実業出版社
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