https://gyazo.com/92d834eb3c6b3c500bda5cf7e2d68fc2
I just used the definition.
$$\bigg|\lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*)\Delta x\bigg| \leq \lim_{n\to\infty} \sum_{i=1}^{n} |f(x_i^*)|\Delta x$$ by def
idk what to do. Hints?
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Basically, you are done. – Jacky Chong Feb 14 '17 at 04:32
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This question is taken from a past midterm question which is out of like 70. I'm having a hard time believing that this answer would give me 10% of that exam. :/. Btw do you know how to paste images like without having to click on it and it just shows. – user349557 Feb 14 '17 at 04:33
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Your answer is not rigorous, but heuristically correct. – Jacky Chong Feb 14 '17 at 04:34
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I just applied the definition :O . Is there nothing more I should do? An explanation maybe – user349557 Feb 14 '17 at 04:34
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I don't think the above would make the cut in a slightly more strict school. Perhaps it was expected something like: using the triangle inequality, we know that $$\left|\sum_{i=1}^n f(x_i)\Delta_{x_i}\right|\le\sum_{i=1}^n |f(x_i)|\Delta_{x_i}$$ so when taking the limit when $;n\to\infty;$ while $;\sup\left|\Delta_x\right|\to 0;$, the inequality still is true as we know from sequences and their limits. Q.E.D. – DonAntonio Feb 14 '17 at 09:42
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$− |f(x)| ≤ f(x) ≤ |f(x)|$ holds for all x ∈ dom(f). I was thinking showing that – user349557 Feb 14 '17 at 16:47