5.1 Approximating areas  (Page 4/17)

In [link] (b), we draw vertical lines perpendicular to $x_i$ such that $x_i$ is the right endpoint of each subinterval, and calculate $f\left(x_i\right)$ for $i=1,2,3,4,5,6.$ We multiply each $f\left(x_i\right)$ by Δ x to find the rectangular areas, and then add them. This is a right-endpoint approximation of the area under $f\left(x\right).$ Thus,

Approximating the area under a curve

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

First, divide the interval $\left[0,2\right]$ into n equal subintervals. Using $n=4,\text{Δ}x=\frac{\left(2-0\right)}{4}=0.5.$ This is the width of each rectangle. The intervals $\left[0,0.5\right],\left[0.5,1\right],\left[1,1.5\right],\left[1.5,2\right]$ are shown in [link] . Using a left-endpoint approximation, the heights are $f\left(0\right)=0,f\left(0.5\right)=0.25,f\left(1\right)=1,f\left(1.5\right)=2.25.$ Then,

$\begin{array}{cc}{L}_4\hfill & =f\left(x_0\right)\text{Δ}x+f\left(x_1\right)\text{Δ}x+f\left(x_2\right)\text{Δ}x+f\left(x_3\right)\text{Δ}x\hfill \\ & =0\left(0.5\right)+0.25\left(0.5\right)+1\left(0.5\right)+2.25\left(0.5\right)\hfill \\ & =\mathrm{1.75.}\hfill \end{array}$

The right-endpoint approximation is shown in [link] . The intervals are the same, $\text{Δ}x=0.5,$ but now use the right endpoint to calculate the height of the rectangles. We have

$\begin{array}{cc}{R}_4\hfill & =f\left(x_1\right)\text{Δ}x+f\left(x_2\right)\text{Δ}x+f\left(x_3\right)\text{Δ}x+f\left(x_4\right)\text{Δ}x\hfill \\ & =0.25\left(0.5\right)+1\left(0.5\right)+2.25\left(0.5\right)+4\left(0.5\right)\hfill \\ & =\mathrm{3.75.}\hfill \end{array}$

The left-endpoint approximation is 1.75; the right-endpoint approximation is 3.75.

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Looking at [link] and the graphs in [link] , 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 A 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 n , first in a left-endpoint approximation with four rectangles, then eight rectangles, and finally 32 rectangles. Then, let’s do the same thing in a right-endpoint approximation, using the same sets of intervals, of the same curved region. [link] shows the area of the region under the curve $f\left(x\right)={\left(x-1\right)}^3+4$ on the interval $\left[0,2\right]$ using a left-endpoint approximation where $n=4.$ The width of each rectangle is

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