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?
On
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:
$$ \sum_{i=1}^{5}f(x_i)\Delta x= \sum_{i=1}^{5}\frac{1}{1+\frac{i}{5}}\frac{1}{5}=\\ \sum_{i=1}^{5}\frac{1}{5+i}=\frac16+\frac17+\frac18+\frac19+\frac{1}{10}\approx 0.65. $$
On
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:
$$0.2 \cdot \frac{1}{1.2} + 0.2 \cdot \frac{1}{1.4} + 0.2 \cdot \frac{1}{1.6} + 0.2 \cdot \frac{1}{1.8} + 0.2 \cdot \frac{1}{2} \approx 0.65.$$
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, $\dfrac15\left(\dfrac1{1.2}+\dfrac1{1.4}+\dfrac1{1.6}+\dfrac1{1.8}+\dfrac12\right)$.