[Math] Using Riemann sums to approximate area under the curve $y = 1/x$

calculusdefinite integralsriemann sum

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?

Best Answer

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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