# Using Riemann sums to approximate area under the curve $y = 1/x$

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?

• Welcome to Math Stack Exchange. Do you know how $x$ depends on $i$? – J. W. Tanner Jul 22 at 4:13
• What do you mean? – burt Jul 22 at 4:17
• Note: I don't know why you wrote $\dfrac51$; that does not belong – J. W. Tanner Jul 22 at 4:20
• 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 $i$? – J. W. Tanner Jul 22 at 4:20
• Note: if $i$ goes from $\color{red}0$ to $n$, that’s $n\color{red}{+1}$ rectangles – J. W. Tanner Jul 22 at 5:06

## 3 Answers

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

• so to figure out $\Delta x$ you divide 5 by 1? – burt Jul 22 at 4:21
• Or did I just pull that out of nowhere? – burt Jul 22 at 4:22
• 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 – J. W. Tanner Jul 22 at 4:23
• So I just did it backwards – burt Jul 22 at 4:23
• Oh, I see - its $\frac{b-a}n$ – burt Jul 22 at 13:42

I prefer to discuss Riemann sums in pictures. Here's the region whose area we are supposed to approximate: You need to use $$5$$ rectangles. This means that you should subdivide the domain into $$5$$ equal pieces: Next, we draw perpendicular lines up to the graph, ready to become the sides of rectangles: 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: 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:

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

• Pictures are very helpful for this (+1) – J. W. Tanner Jul 22 at 5:00

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:

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