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Summing Area by Rectangles

April 6th, 2009 |

Author: Merrin

Calculating the area between a function, f(x), and the x axis over a certain interval is called integration. Integration is a general problem leading to the second main branch of calculus. Integration is closely linked to differentiation. Integration is facilitated by using antiderivatives, functions whose derivatives give a target function. As an example of an antiderivative, the derivative of

$\displaystyle\cfrac{1}{2}x^2$

is x. It is said that g(x) is the antiderivative of x. It turns out that the general formula for the area under the curve y=x is related to the antiderivative depending over which domain is chosen to generate the area. The entire link between integrals, derivatives, and antiderivatives will be made clear once we reach the fundamental theorem of calculus.

To start off integration, we will present a roundabout method to calculate the approximate area beneath a line, yet is general enough for most continuous curves. We use the following formulas from geometry to calculate areas. The area of a triangle is

$A=\cfrac{1}{2} bh$

and the area of a rectangle is bh. Notice for a simple function like y=x, generating a triangular area, the base is just from zero to x and the height is just zero to y=x so the area is

$\cfrac{1}{2}x^2$

like we alluded to before. We will get to these relations in due time.

Example 1. Estimate the area of a triangle which has vertices at the coordinates (0, 0), (1, 0), and (1, 1) by breaking the area into ten rectangles whose tops intersect the triangle at their midpoints.

Solution 1. The exact area of this triangle is 1/2, but we will try to approximate the area by drawing a series of 10 rectangles and adding their areas as if we didn’t know the total area. We take the width of each rectangle to be Delta x=0.1 such that n=10 fit between x=0 and x=1. The top midpoints of the rectangles are drawn so as to intersect the line y=x. Note from the figure of this example that the left parts of the rectangles overhanging y=x have areas that exactly cancel the holes under the sloped line to the right of the tops. For the following calculations in this section, we will make heavy use of this formula

$\displaystyle S_{n} =\sum _{k=1}^{n}k =1+2+...+n=\cfrac{n(n+1)}{2}$

This formula was derived in the introductory material where we explained sigma notation as well as the general calculation of sums of powers. The formula for our appoximate midpoint area is

$\displaystyle A_{M} =\sum_{i=1}^{10}(\Delta x) h(x_i) = \cfrac{1}{2} \sum _{k=0}^{9}(2k+1)(\Delta x)^{2}$

$\displaystyle A_M=\cfrac{1}{2} \left(2\cdot \cfrac{9\cdot 10}{2} +10\right)(\Delta x)^{2}$

$\displaystyle A_M =\cfrac{1}{2} 10^{2} (\Delta x)^{2} =\cfrac{1}{2}$

The generalization to n divisions where n=1/Delta x is

$\displaystyle A_{M} =\cfrac{1}{2} \sum _{k=0}^{n-1}(2k+1)(\Delta x)^{2} =\cfrac{1}{2} \left(2\cdot \cfrac{(n-1)n}{2} +n\right)(\Delta x)^{2} =\cfrac{1}{2} n^{2} (\Delta x)^{2} =\cfrac{1}{2}$

When the midpoints of the tops of the rectangles intersect the line then no matter how finely subdivided the intervals are, the area is still correct area. This is basically a fluke of where we chose the rectangle to intersect the top of the triangle. Choosing the intersections to the left or right result in overestimates or underestimates of the area.

rectm

For this example, when the rectangles intersect through the midpoints then the exact area is calculated

Example 2. Estimate the area of the same triangle using ten rectangles that intersect with their upper right hand corners.

Solution 2. The upper right hand corner intersections give an overestimate of the area.

$\displaystyle A_{R} =\sum _{k=1}^{10}k(\Delta x)^{2} =(\Delta x)^{2} \cfrac{10\cdot 11}{2} =\cfrac{1}{2} \cdot \cfrac{11}{10} = \cfrac{1}{2}\left( 1+\cfrac{1}{10} \right)$

For an arbitrary number of divisions of width

$\Delta x=\cfrac{1}{n}$

the area is

$\displaystyle A_{R} =\sum _{k=1}^{n}k (\Delta x)^{2} =\cfrac{1}{2} (\Delta x)^{2} n(n+1)=\cfrac{1}{2} \left(1+\cfrac{1}{n} \right)$

For an infinite number of divisions, the calculated area agree with the area calculated from the midpoint intersections. The limit as n goes to infinity, says the area is 1/2 which is correct.

rectr

For this example, when the rectangles intersect through the upper right corners the area is overestimated

Example 3. Estimate the area of the same triangle using ten rectangles whose upper left corners coincide with y=x.

Solution 3. The upper left hand rectangles coinciding at the upper left hand corners will underestimate of the area.

$\displaystyle A_{L} =\sum _{k=0}^{9}k(\Delta x)^{2} =\cfrac{9\cdot 10}{2} (\Delta x)^{2} =\cfrac{1}{2}\cdot\cfrac{9}{10}=\cfrac{1}{2}\left(1-\cfrac{1}{10}\right)$

$\displaystyle A_{L} =\sum _{k=0}^{n-1}k (\Delta x)^{2} =\cfrac{1}{2} (\Delta x)^{2} (n-1)n=\cfrac{1}{2} \left(1-\cfrac{1}{n} \right)$

In the limit as n goes to infinity, the error shrinks to zero.

rectl

For this example, when the rectangles intersect through the upper left corners the area is underestimated

I want to emphasize two points of these calculations. If we had chosen a function other than a line y=x to calculate the area under the curve then the exact calculations with summations would have been more difficult but qualitatively similar. We know how to sum squares or nth powers so we could have done the functions f(x)=x^n just as easily as for f(x)=x. In the limit of infinite rectangles, the area from upper left hand rectangles, midpoint rectangles, and upper right hand rectangles are all equal.

$\displaystyle \mathop{\lim }\limits_{n\to \infty } A_{L} =\mathop{\lim }\limits_{n\to \infty } A_{M} =\mathop{\lim }\limits_{n\to \infty } A_{R} =\cfrac{1}{2}$

This implies there is a more general way to write down the area under a curve in terms of limits which will be formally introduced in the next section.

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