Hi, there!
In my EIKEN® Grade 1 class, I am determined to ask several questions to all
of my students in English and let them evaluate their performance and wheth-
er the talks are organized.
――Are there any areas for improvement in her talk?
Interestingly, in most if not all cases, they tend to hesitate to feedback mutu-
ally because they do not want to be offensive to others by giving a critique.
Japanese people are often considered “inscrutable” partly because they are
unwilling to be assertive.
Conversely, deep down inside, they are not happy to receive any comments
on their performance even though they ostensibly say they appreciate any
feedback. How unmanageable they are! Honestly, I feel more comfortable
talking with non-Japanese people than the Japanese.
I might have polished myself with ages; however, I still have a lot of areas
for improvement, so let me scrutinize today an “area” in a math word prob-
lem.
Determine the area of the region bounded by \(y = 3x^{2} + 4\) and \(y = -3x + 10\).
Here is my solution to the problem.
In this case I can get the intersection points by setting the two equations
equal.
\(3x^{2} + 4 = -3x + 10\)
\(3^{2} + 3x – 6 = 0\)
\(3(x + 2)(x – 1) = 0\)
They will intersect at \(x = －2\) and \(x = 1\).
With the graph I can now identify the upper and lower function and so I can
now find the enclosed area.
\(A = \int_{a}^{b}(_{function}^{upper}) – (_{function}^{lower})dx\)
\(= \int_{-2}^{1}-3x + 10 – (3x^{2} + 4)dx\)
\(= \int_{-2}^{1}-3x^{2} – 3x + 6\,dx\)
\(= [-x^{3} – \frac{3}{2}x^{2} + 6x]_{-2}^{1}\)
\(= \frac{27}{2}\)
If I miss a figure, I will have an area for improvement. If so, there will be a
long way to reach my being impeccable.
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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