Calculus at Moravian University

The graph verifies that the upper boundary of the region is given by \(f\) and the lower bound is given by \(g\text{.}\) Therefore the area of the region is the value of the integral

A quick calculation shows that \(f=g\) at \(x=1, 2\) and 4. One can proceed thoughtlessly by computing \(\ds \int_1^4\big(f(x)-g(x)\big)\ dx\text{,}\) but this ignores the fact that on \([1,2]\text{,}\) \(g(x)>f(x)\text{.}\) (In fact, the thoughtless integration returns \(-9/4\text{,}\) hardly the expected value of an area.) Thus we compute the total area by breaking the interval \([1,4]\) into two subintervals, \([1,2]\) and \([2,4]\) and using the proper integrand in each.

We give two approaches to this problem. In the first approach, we notice that the region's “top” is defined by two different curves. On \([0,1]\text{,}\) the top function is \(y=\sqrt{x}+2\text{;}\) on \([1,2]\text{,}\) the top function is \(y=-(x-1)^2+3\text{.}\)

Thus we compute the area as the sum of two integrals:

The second approach is clever and very useful in certain situations. We are used to viewing curves as functions of \(x\text{;}\) we input an \(x\)-value and a \(y\)-value is returned. Some curves can also be described as functions of \(y\text{:}\) input a \(y\)-value and an \(x\)-value is returned. We can rewrite the equations describing the boundary by solving for \(x\text{:}\)

Figure 7.1.9 shows the region with the boundaries relabeled. A differential element, a horizontal rectangle, is also pictured. The width of the rectangle is a small change in \(y\text{:}\) \(\Delta y\text{.}\) The height of the rectangle is a difference in \(x\)-values. The “top” \(x\)-value is the largest value, i.e., the rightmost. The “bottom” \(x\)-value is the smaller, i.e., the leftmost. Therefore the height of the rectangle is

The area is found by integrating the above function with respect to \(y\) with the appropriate bounds. We determine these by considering the \(y\)-values the region occupies. It is bounded below by \(y=2\text{,}\) and bounded above by \(y=3\text{.}\) That is, both the “top” and “bottom” functions exist on the \(y\) interval \([2,3]\text{.}\) Thus

Recognize that there are two “top” functions to this region, causing us to use two definite integrals.

We can also approach this by converting each function into a function of \(y\text{.}\) This also requires 2 integrals, so there isn't really any advantage to doing so. We do it here for demonstration purposes.

The “top” function is always \(x=\frac{7-y}2\) while there are two “bottom” functions. Being mindful of the proper integration bounds, we have

Of course, the final answer is the same. (It is interesting to note that the area of all 4 subregions used is 3/4. This is coincidental.)

The measurements of length can be viewed as measuring “top minus bottom” of two functions. The exact answer is found by integrating \(\ds \int_0^{12} \big(f(x)-g(x)\big)\ dx\text{,}\) but of course we don't know the functions \(f\) and \(g\text{.}\) Our discrete measurements instead allow us to approximate.

We have the following data points:

We also have that \(\dx=\frac{b-a}{n} = 2\text{,}\) so Simpson's Rule gives

Since the measurements are in hundreds of feet, \(\text{units}^2 = (100\ \text{ ft } )^2 = 10,000\ \text{ ft } ^2\text{,}\) giving a total area of \(440,133\ \text{ ft } ^2\text{.}\) (Since we are approximating, we'd likely say the area was about \(440,000\ \text{ ft } ^2\text{,}\) which is a little more than 10 acres.)