Approximating Areas: Learn It 2

Approximating Area Cont.

Right-Endpoint Approximation

The second method for approximating area under a curve is the right-endpoint approximation. It is almost the same as the left-endpoint approximation, but now the heights of the rectangles are determined by the function values at the right of each subinterval.

Right-Endpoint Approximation

In the right-endpoint approximation, we estimate the area under a curve by constructing rectangles whose heights are determined by the function values at the right endpoints of subintervals.

 

The approximation of the area [latex]A[/latex] using [latex]n[/latex] subintervals is given by the formula:

[latex]A \approx R_n = \displaystyle\sum_{i=1}^{n} f(x_i)\Delta x[/latex]

 

where [latex]\Delta x =\frac{b-a}{n}[/latex] is the width of each subinterval, and [latex]x_{i}[/latex] are the right endpoints of the subintervals.

A diagram showing the right-endpoint approximation of area under a curve. Under a parabola with vertex on the y axis and above the x axis, rectangles are drawn between a=x0 on the origin and b = xn. The rectangles have endpoints at a=x0, x1, x2…x(n-1), and b = xn, spaced equally. The height of each rectangle is determined by the value of the given function at the right endpoint of the rectangle.
Figure 3. In the right-endpoint approximation of area under a curve, the height of each rectangle is determined by the function value at the right of each subinterval. Note that the right-endpoint approximation differs from the left-endpoint approximation in (Figure).

Since we have already seen how to solve using left-endpoint approximation, the right-endpoint approximation follows a similar process. The key difference is that the heights of the rectangles are determined by the function values at the right endpoints of the subintervals, rather than the left endpoints. This means:

  • Left-Endpoint Approximation: Uses [latex]f(x_{i -1})[/latex] for each subinterval.
  • Right-Endpoint Approximation: Uses [latex]f(x_i)[/latex] for each subinterval.

By adjusting the endpoint used, we slightly alter the position and height of the rectangles, which can affect the accuracy of the approximation depending on the behavior of the function.

The graphs in Figure 4 represent the curve [latex]f(x)=\frac{x^2}{2}[/latex].

Diagrams side by side, showing the differences in approximating the area under a parabolic curve with vertex at the origin between the left endpoints method (the first diagram) and the right endpoints method (the second diagram). In the first diagram, rectangles are drawn at even intervals (delta x) under the curve with heights determined by the value of the function at the left endpoints. In the second diagram, the rectangles are drawn in the same fashion, but with heights determined by the value of the function at the right endpoints. The endpoints in both are spaced equally from the origin to (3, 0), labeled x0 to x6.
Figure 4. Methods of approximating the area under a curve by using (a) the left endpoints and (b) the right endpoints.

In graph (a) we divide the region represented by the interval [latex][0,3][/latex] into six subintervals, each of width [latex]0.5[/latex]. Thus, [latex]\Delta x=0.5[/latex].

We then form six rectangles by drawing vertical lines perpendicular to [latex]x_{i-1}[/latex], the left endpoint of each subinterval.

We determine the height of each rectangle by calculating [latex]f(x_{i-1})[/latex] for [latex]i=1,2,3,4,5,6[/latex].

The intervals are [latex][0,0.5], \, [0.5,1], \, [1,1.5], \, [1.5,2], \, [2,2.5], \, [2.5,3][/latex].

We find the area of each rectangle by multiplying the height by the width.

Then, the sum of the rectangular areas approximates the area between [latex]f(x)[/latex] and the [latex]x[/latex]-axis.

When the left endpoints are used to calculate height, we have a left-endpoint approximation. Thus,

[latex]\begin{array}{ll} A \approx L_6 & =\displaystyle\sum_{i=1}^{6} f(x_{i-1})\Delta x=f(x_0)\Delta x+f(x_1)\Delta x+f(x_2)\Delta x+f(x_3)\Delta x+f(x_4)\Delta x+f(x_5)\Delta x \\ & =f(0)0.5+f(0.5)0.5+f(1)0.5+f(1.5)0.5+f(2)0.5+f(2.5)0.5 \\ & =(0)0.5+(0.125)0.5+(0.5)0.5+(1.125)0.5+(2)0.5+(3.125)0.5 \\ & =0+0.0625+0.25+0.5625+1+1.5625 \\ & =3.4375 \end{array}[/latex] 

In Figure 4(b), we draw vertical lines perpendicular to [latex]x_i[/latex] such that [latex]x_i[/latex] is the right endpoint of each subinterval, and calculate [latex]f(x_i)[/latex] for [latex]i=1,2,3,4,5,6[/latex].

We multiply each [latex]f(x_i)[/latex] by [latex]\Delta x[/latex] to find the rectangular areas, and then add them. This is a right-endpoint approximation of the area under [latex]f(x)[/latex]. Thus,

[latex]\begin{array}{ll} A \approx R_6 & =\displaystyle\sum_{i=1}^{6} f(x_i)\Delta x=f(x_1)\Delta x+f(x_2)\Delta x+f(x_3)\Delta x+f(x_4)\Delta x+f(x_5)\Delta x+f(x_6)\Delta x \\ & =f(0.5)0.5+f(1)0.5+f(1.5)0.5+f(2)0.5+f(2.5)0.5+f(3)0.5 \\ & =(0.125)0.5+(0.5)0.5+(1.125)0.5+(2)0.5+(3.125)0.5+(4.5)0.5 \\ & =0.0625+0.25+0.5625+1+1.5625+2.25 \\ & =5.6875 \end{array}[/latex]

Use both left-endpoint and right-endpoint approximations to approximate the area under the curve of [latex]f(x)=x^2[/latex] on the interval [latex][0,2][/latex]; use [latex]n=4[/latex].

Sketch left-endpoint and right-endpoint approximations for [latex]f(x)=\frac{1}{x}[/latex] on [latex][1,2][/latex]; use [latex]n=4[/latex]. Approximate the area using both methods.

Looking at Figure 4 and the graphs in the previous example, we can see that when we use a small number of intervals, neither the left-endpoint approximation nor the right-endpoint approximation is a particularly accurate estimate of the area under the curve.

However, it seems logical that if we increase the number of points in our partition, our estimate of [latex]A[/latex] will improve. We will have more rectangles, but each rectangle will be thinner, so we will be able to fit the rectangles to the curve more precisely.

We can demonstrate the improved approximation obtained through smaller intervals with an example.

Let’s explore the idea of increasing [latex]n[/latex], first in a left-endpoint approximation with four rectangles, then eight rectangles, and finally [latex]32[/latex] rectangles. Then, let’s do the same thing in a right-endpoint approximation, using the same sets of intervals, of the same curved region.

Figure 8 shows the area of the region under the curve [latex]f(x)=(x-1)^3+4[/latex] on the interval [latex][0,2][/latex] using a left-endpoint approximation where [latex]n=4[/latex].

A graph of the left-endpoint approximation of the area under the given curve from a = x0 to b=x4. The heights of the rectangles are determined by the values of the function at the left endpoints.
Figure 8. With a left-endpoint approximation and dividing the region from a to b into four equal intervals, the area under the curve is approximately equal to the sum of the areas of the rectangles.

The width of each rectangle is

[latex]\Delta x=\frac{2-0}{4}=\frac{1}{2}[/latex]

The area is approximated by the summed areas of the rectangles, or

[latex]\begin{array}{ll} L_4 & =f(0)(0.5)+f(0.5)(0.5)+f(1)(0.5)+f(1.5)0.5 \\ & =7.5 \end{array}[/latex]
 

Figure 9 shows the same curve divided into eight subintervals.

A graph showing the left-endpoint approximation for the area under the given curve from a=x0 to b = x8. The heights of the rectangles are determined by the values of the function at the left endpoints.
Figure 9. The region under the curve is divided into [latex]n=8[/latex] rectangular areas of equal width for a left-endpoint approximation.

Comparing the graph with four rectangles in Figure 8 with this graph with eight rectangles, we can see there appears to be less white space under the curve when [latex]n=8[/latex]. This white space is area under the curve we are unable to include using our approximation.

The area of the rectangles is

[latex]\begin{array}{ll} L_8 & =f(0)(0.25)+f(0.25)(0.25)+f(0.5)(0.25)+f(0.75)(0.25) \\ & +f(1)(0.25)+f(1.25)(0.25)+f(1.5)(0.25)+f(1.75)(0.25) \\ & =7.75 \end{array}[/latex]
 

The graph in Figure 10 shows the same function with 32 rectangles inscribed under the curve.

A graph of the left-endpoint approximation of the area under the given curve from a = x0 to b = x32. The heights of the rectangles are determined by the values of the function at the left endpoints.
Figure 10. Here, 32 rectangles are inscribed under the curve for a left-endpoint approximation.

There appears to be little white space left. The area occupied by the rectangles is

[latex]\begin{array}{ll} L_{32} & =f(0)(0.0625)+f(0.0625)(0.0625)+f(0.125)(0.0625)+\cdots+f(1.9375)(0.0625) \\ & =7.9375 \end{array}[/latex]
 

We can carry out a similar process for the right-endpoint approximation method. A right-endpoint approximation of the same curve, using four rectangles (Figure 11), yields an area

[latex]\begin{array}{ll} R_4 & =f(0.5)(0.5)+f(1)(0.5)+f(1.5)(0.5)+f(2)(0.5) \\ & =8.5 \end{array}[/latex]

 

A graph of the right-endpoint approximation for the area under the given curve from x0 to x4. The heights of the rectangles are determined by the values of the function at the right endpoints.
Figure 11. Now we divide the area under the curve into four equal subintervals for a right-endpoint approximation.

Dividing the region over the interval [latex][0,2][/latex] into eight rectangles results in [latex]\Delta x=\frac{2-0}{8}=0.25[/latex]. The graph is shown in Figure 12. The area is

[latex]\begin{array}{ll} R_8 & =f(0.25)(0.25)+f(0.5)(0.25)+f(0.75)(0.25)+f(1)(0.25) \\ & +f(1.25)(0.25)+f(1.5)(0.25)+f(1.75)(0.25)+f(2)(0.25) \\ & =8.25 \end{array}[/latex]

 

A graph of the right-endpoint approximation for the area under the given curve from a=x0 to b=x8.The heights of the rectangles are determined by the values of the function at the right endpoints.
Figure 12. Here we use right-endpoint approximation for a region divided into eight equal subintervals.

Last, the right-endpoint approximation with [latex]n=32[/latex] is close to the actual area (Figure 13). The area is approximately

[latex]\begin{array}{ll} R_{32} & =f(0.0625)(0.0625)+f(0.125)(0.0625)+f(0.1875)(0.0625)+\cdots+f(2)(0.0625) \\ & =8.0625 \end{array}[/latex]

 

A graph of the right-endpoint approximation for the area under the given curve from a=x0 to b=x32. The heights of the rectangles are determined by the values of the function at the right endpoints.
Figure 13. The region is divided into 32 equal subintervals for a right-endpoint approximation.

Based on these figures and calculations, it appears we are on the right track; the rectangles appear to approximate the area under the curve better as [latex]n[/latex] gets larger.

Furthermore, as [latex]n[/latex] increases, both the left-endpoint and right-endpoint approximations appear to approach an area of [latex]8[/latex] square units.

The table below shows a numerical comparison of the left- and right-endpoint methods. 

Converging Values of Left- and Right-Endpoint Approximations as [latex]n[/latex] Increases
Values of [latex]n[/latex] Approximate Area [latex]L_n[/latex] Approximate Area [latex]R_n[/latex]
[latex]n=4[/latex] [latex]7.5[/latex] [latex]8.5[/latex]
[latex]n=8[/latex] [latex]7.75[/latex] [latex]8.25[/latex]
[latex]n=32[/latex] [latex]7.94[/latex] [latex]8.06[/latex]

The idea that the approximations of the area under the curve get better and better as [latex]n[/latex] gets larger and larger is very important, and we now explore this idea in more detail.