MA005: Calculus I

The basic shape we will use is the rectangle; the area of a rectangle is $\text { (base) } \bullet \text { (height). }$ If the units for each side of the rectangle are "meters," then the area will have the units $\text { ("meters") •("meters") = "square meters" } = \mathrm{m}^{2}$. The only other area formulas needed for this section are for triangles, $\text { area }=\mathrm{b} \cdot \mathrm{h} / 2$, and for circles, $\text { area }=\pi \cdot \mathrm{r}^{2}$. Three other familiar properties of area are assumed and will be used:

Addition Property: The total area of a region is the sum of the areas of the non–overlapping pieces which comprise the region. (Fig. 1)

Inclusion Property: If region B is on or inside region A, then the area of region B is less than or equal to the area of region A. (Fig. 2)

Location–Independence Property: The area of a region does not depend on its location. (Fig. 3)

Example 1: Determine the area of the region in Fig. 4a

Solution: The region can easily be broken into two rectangles, Fig. 4b, with areas 35 square inches and 3 square inches respectively, so the area of the original region is 38 square inches.

Practice 1: Determine the area of the region in Fig. 5 by cutting it in two ways: (a) into a rectangle and triangle and (b) into two triangles.

We can use the three properties of area to get information about areas that are difficult to calculate exactly. Let $A$ be the region bounded by the graph of $f(x)=1 / x$, the x–axis, and vertical lines at $\mathrm{x}=1$ and $\mathrm{x}=3$. Since the two rectangles in Fig. 6 are inside the region A and do not overlap each, the area of the rectangles, $1 / 2+1 / 3=5 / 6$, is less than the area of region A.

Practice 2: Build two rectangles, each with base 1 unit, outside the shaded region in Fig. 6 and use their areas to make a VALID statement about the area of region A.

Practice 3: What can be said about the area of region A in Fig. 6 if we use both inside and outside rectangles with base 1/2 unit?

Example 2: In Fig. 7, there are 32 dark squares, 1 centimeter on a side, and 31 lighter squares of the same size. We can be sure that the area of the leaf is smaller than what number?

Solution: The area of the leaf is smaller than $32+31=63 \mathrm{~cm}^{2}$.

Practice 4: We can be sure that the area of the leaf is at least how large?

Functions can be defined in terms of areas. For the constant function $\mathrm{f}(t)=2$, define $\mathrm{A}(x)$ to be the area of the rectangular region bounded by the graph of $f$, the t–axis, and the vertical lines at $t=1$ and $t=x$ (Fig. 8a). $\mathrm{A}(2)$ is the area of the shaded region in Fig. 8b, and $\mathrm{A}(2)=2$. Similarly, $\mathrm{A}(3)=4$ and $\mathrm{A}(4)=6$. In general, $\mathrm{A}(x)=(\text { base })(\text { height })= (x-1)(2)=2 x-2$  for any $x \geq 1$. The graph of $\mathrm{y}=\mathrm{A}(x)$ is shown in Fig. 8c, and $\mathrm{A}^{\prime}(x)=2$ for every value of $x>1$.

Practice 5: For $\mathrm{f}(t)=2$, define $\mathrm{B}(x)$ to be the area of the region bounded by the graph of $f$, the t–axis, and vertical lines at $t=0 \text { and } t=x$. Fill in the table in Fig. 9 with the values of $B$. How are the graphs of $y=\mathrm{A}(x) \text { and } y=\mathrm{B}(x)$ related?

Sometimes it is useful to move regions around. The area of a parallelogram is obvious if we move the triangular region from one side of the parallelogram to fill the region on the other side and ending up with a rectangle (Fig. 10). At first glance, it is difficult to estimate the total area of the shaded regions in Fig. 11a . However, if we slide all of them into a single column (Fig. 11b), then it is easy to determine that the shaded area is less than the area of the enclosing $\text { rectangle }=(\text { base })(\text { height })=(1)(2)=2$.

Practice 6: The total area of the shaded regions in Fig. 12 is less than what number?