How can rectangles approximate the area under a curve?

Think of covering the region between the curve and the horizontal axis with a row of narrow rectangles. For each rectangle, its width is the length of one piece of the horizontal interval. Its height comes from the curve’s value at a chosen sample point in that piece. Multiply width by height to get that rectangle’s area, then add the areas.

In symbols, one rectangle contributes f(xᵢ*)Δx: f(xᵢ*) is the chosen height and Δx is the subinterval’s width. Adding these contributions gives a Riemann sum, which estimates the area. A finite sum is an approximation, not necessarily the exact area; the definite integral is defined through a limit of such sums. [S1]

The curve may rise or fall within a rectangle’s span, so the rectangle’s flat top will not generally match the curve throughout that span. The estimate is built from the chosen widths and sample heights.

Suppose the curve is f(x)=x² on the interval [0,2]. For a simple finite estimate, divide the interval into four equal pieces and use the left endpoint of each piece as the sample point.

1. Width of each rectangle: Δx=(2−0)/4=0.5.
2. Left-endpoint sample points: 0, 0.5, 1, and 1.5.
3. Heights from f(x)=x²: 0, 0.25, 1, and 2.25.
4. Rectangle areas: 0×0.5=0; 0.25×0.5=0.125; 1×0.5=0.5; 2.25×0.5=1.125.
5. Add them: 0+0.125+0.5+1.125=1.75.

So the labeled finite estimate is: interval [0,2]; four rectangles; width 0.5 each; left-endpoint heights 0, 0.25, 1, 2.25; total rectangle area 1.75 square units. This is a Riemann-sum approximation to the area under the curve, not a claim that the integral equals 1.75. The number comes from these particular choices of width and sample points.

The sample-point rule changes the heights, so it changes the finite sum. In the example, choosing right endpoints gives sample points 0.5, 1, 1.5, and 2. Their heights are 0.25, 1, 2.25, and 4. Multiplying each by 0.5 and adding gives 3.75. That is another estimate, produced by a different choice—not the same calculation as the left-endpoint estimate of 1.75.

Using more, narrower rectangles can also produce a different finite sum. For the specific curve and interval above, you can compare estimates made with the same endpoint rule and successively narrower widths. Don’t assume a particular finite result is exact: the general definition of the definite integral uses a limit of Riemann sums, rather than identifying one finite sum as the integral. [S1]

A careful description could be: “Using four equal rectangles with left-endpoint heights, the estimated area under f(x)=x² on [0,2] is 1.75 square units. This is a finite approximation, not the exact integral.”

Use this fill-in template:

“On [a,b], I divide the interval into n equal pieces, so each width is Δx=(b−a)/n=___. I use [left endpoints/right endpoints/midpoints] as sample points: ___. The corresponding function heights are ___. My rectangle areas are height × width: ___. Adding them gives a finite estimate of ___ square units. This is a Riemann-sum approximation, not necessarily the exact area or integral.”

Before calculating, check that you have named both the interval and the sample-point rule. Those choices determine the rectangle widths and heights. Show the individual contributions before adding them, so a reader can see how the total was formed. If a task asks for an exact definite integral rather than a finite approximation, don’t present a rectangle sum as though it completed that different task.