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This was the question I was given:

Use a Riemann sum with n = 5 rectangles to approximate the area of the region bounded by the lines $x = 1$, $x = 2$, $y = 0$ and the curve $y = 1/x$. Use the appropriate endpoint of each subinterval to compute a lower sum.

When I saw this problem, this is what I came up with: $$\sum^n_{i=0} \frac51\frac1x \Delta x$$

However, this did not result in the right answer. Where did I go wrong when finding the Riemann Sum? How can I rectify this?

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    $\begingroup$ Welcome to Math Stack Exchange. Do you know how $x$ depends on $i$? $\endgroup$ – J. W. Tanner Jul 22 at 4:13
  • $\begingroup$ What do you mean? $\endgroup$ – burt Jul 22 at 4:17
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    $\begingroup$ Note: I don't know why you wrote $\dfrac51$; that does not belong $\endgroup$ – J. W. Tanner Jul 22 at 4:20
  • $\begingroup$ What I mean is this: when you compute that sum you wrote, there is a different value of $x$ for each $i$; if I tell you a particular value of $i$, can you tell me what is the value of $x$ corresponding to that $i$? $\endgroup$ – J. W. Tanner Jul 22 at 4:20
  • $\begingroup$ Note: if $i$ goes from $\color{red}0$ to $n$, that’s $n\color{red}{+1}$ rectangles $\endgroup$ – J. W. Tanner Jul 22 at 5:06
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The lower Riemann sum would be $$\sum_{i=1}^5 \dfrac 1{ x} \Delta x$$ with $\Delta x=\dfrac15$ and $x=1+i\Delta x$.

In other words, $\dfrac15\left(\dfrac1{1.2}+\dfrac1{1.4}+\dfrac1{1.6}+\dfrac1{1.8}+\dfrac12\right)$.

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  • $\begingroup$ so to figure out $\Delta x$ you divide 5 by 1? $\endgroup$ – burt Jul 22 at 4:21
  • $\begingroup$ Or did I just pull that out of nowhere? $\endgroup$ – burt Jul 22 at 4:22
  • $\begingroup$ The width of each rectangle is $\dfrac15=\dfrac{2-1}5$; we're dividing the region between $x=1$ and $x=2$ into $5$ equal sub-intervals $\endgroup$ – J. W. Tanner Jul 22 at 4:23
  • $\begingroup$ So I just did it backwards $\endgroup$ – burt Jul 22 at 4:23
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    $\begingroup$ Oh, I see - its $\frac{b-a}n$ $\endgroup$ – burt Jul 22 at 13:42
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I prefer to discuss Riemann sums in pictures. Here's the region whose area we are supposed to approximate: enter image description here You need to use $5$ rectangles. This means that you should subdivide the domain into $5$ equal pieces: enter image description here Next, we draw perpendicular lines up to the graph, ready to become the sides of rectangles: enter image description here Now, we we need to choose the height of the rectangles. We want the lower sums, hence we want the height of the rectangles to be as small as possible. As this is a decreasing function, the height of the rectangle will therefore be the function value at the rightmost point of its base, giving us our final picture: enter image description here These are our final five rectangles. If we compute the area of these rectangles, and sum them up, this will be our Riemann sum. Our rectangles all have a width of $0.2$. Their respective heights are $\frac{1}{1.2}, \frac{1}{1.4}, \frac{1}{1.6}, \frac{1}{1.8},$ and $\frac{1}{2}$. Thus, the Riemann sum is:

$$0.2 \cdot \frac{1}{1.2} + 0.2 \cdot \frac{1}{1.4} + 0.2 \cdot \frac{1}{1.6} + 0.2 \cdot \frac{1}{1.8} + 0.2 \cdot \frac{1}{2} \approx 0.65.$$

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  • $\begingroup$ Pictures are very helpful for this (+1) $\endgroup$ – J. W. Tanner Jul 22 at 5:00
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The width is going to be the same for all of the five rectangles (it's basically the length of the interval, which is $1$, divided by $5$): $$\Delta x=\frac{b-a}{n}=\frac{2-1}{5}=\frac{1}{5}.$$

A sample point on the interval $[1,2]$: $$x_i=a+i\cdot \Delta x=1+\frac{i}{5}.$$

You get the area under the curve (in this case, it's going to be an approximation, of course) by multiplying the height of a rectangle, which is your function $f(x)=\frac{1}{x}$ evaluated at the sample point $x_i$, by the width $\Delta x=\frac{1}{5}$, which is the same for all of the five rectangles:

$$ \sum_{i=1}^{5}f(x_i)\Delta x= \sum_{i=1}^{5}\frac{1}{1+\frac{i}{5}}\frac{1}{5}=\\ \sum_{i=1}^{5}\frac{1}{5+i}=\frac16+\frac17+\frac18+\frac19+\frac{1}{10}\approx 0.65. $$

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