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The weird thing is, the last sentence says, for the case where $x_1$, "since $\Delta x_1$ approaches zero as $\max \Delta x_k \rightarrow 0...$". But given that $\Delta x_1$ is formed between $x_0$ and $x_1$, and if $x_1$ is equal to $0$ and the question says $[0,1]$, wouldn't $\Delta x_1 = 0$ at all times?

Michael Hardy
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most venerable sir
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No one ever hypothesized that $x_1=0$. The only thing the author is worried about is if $x_1^*=0$ - that is, is the representative of the interval $[x_0,x_1]$ equal to $0$? This can certainly happen without $x_1$ or $\Delta x_1$ being zero.

Milo Brandt
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  • Yeah, I didn't see the star in the superscript. But what is the difference between $x_1^*$ and just $x_1$? – most venerable sir Jan 10 '16 at 13:47
  • From what I read, $x_1^*$ is just a chosen x value on the first interval to determine the height of the rectangle. And $x_1$ is the actually end point of the first interval. – most venerable sir Jan 10 '16 at 13:57
  • But still, how can the exact area under graph be 1? f(0)=0 – most venerable sir Jan 10 '16 at 14:04
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    @Doeser Right, $x_1^$ is the point used to determine the height of the first interval. The proof is saying that, if we take a Riemann sum using $x_1^$ as the representative of the first interval, then the Riemann sum is less than $1$. However, as every other interval is okay, we find the exact sum is $1-\Delta x_1$. The $-\Delta x_1$ term that having $f(0)=0$ introduces goes away in the limit, since the integral is defined as the limit of finer sums (i.e. sending $\Delta x_n\rightarrow 0$) - so the area of the sums might never reach $1$, but it approaches it from below. – Milo Brandt Jan 10 '16 at 17:19
  • Ok my concept of calculus is a bit messed up here. The integral is the EXACT AREA under the curve, even though the Riemann sum may never be 1. – most venerable sir Jan 12 '16 at 02:20
  • Ok, I underarms off how the rectangular method may never get the exact area under the curve, no matter how fine it is. Because the curve is curvy and no amount of rectangles and completely fill out the space. But we are taking around an area under the curve that's shaped EXACTLY like a rectangle!!!. – most venerable sir Jan 12 '16 at 02:22
  • So it's possible that the approximating rectangles have a total area that is EQUAL to the exact area under curve. No?? – most venerable sir Jan 12 '16 at 02:23
  • But the problem is the area is NEVER a square! Because one side of it is missing the endpoint 0. – most venerable sir Jan 12 '16 at 02:24
  • @Doeser The approximating rectangles can have area $1$ - just don't choose $x_1^*$ as a representative of any of the intervals. Then you're treating it exactly like a square. You can see that the area under the graph graph is definitely contained in a square (with area $1$) and contains a rectangle with height $1$ built on the interval $[\varepsilon,1]$ (with area $1-\varepsilon$) - so the area must be no more than $1$ and more than $1-\varepsilon$ for any $\varepsilon$. Thus, it must be $1$. This is basically what the proof's saying. – Milo Brandt Jan 12 '16 at 02:29
  • But you forgot that for x=0, which point should included in the width/base of the rectangle to make it a square, f is 0. Should the base be 0.99999999999999.... – most venerable sir Jan 12 '16 at 02:36
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    @Doeser I agree that the base is $.999999\ldots$. It's just that that number equals $1$. This might be where you misunderstanding is - if one doesn't accept that $.999\ldots$, they're not going to accept that this integral is $1$. – Milo Brandt Jan 12 '16 at 02:40
  • Ohhhh alright, thanks for dat – most venerable sir Jan 12 '16 at 02:41
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If the partition is $\{0,\ 1/6,\ 3/6,\ 4/6,\ 5/6,\ 1\}$ then $\Delta x_1= \dfrac 1 6 - 0 = \dfrac 1 6 \ne 0$.

In your first case, $x_1^* \ne 0$ and in your second case $x_1^*=0$. But in either case $\Delta x_1 = x_1 - x_0$ need not be $0$. However, if it is $0$, then the statement that it approaches $0$ would still be true.

Michael Hardy
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  • No, delta x_1 is formed between x_0 and x_1, because it's on [0,1], x_0 is 0. But then x_1 is also zero for the second case of this problem... – most venerable sir Jan 10 '16 at 02:43