Quiz 10 (Solution)

Date: MATH 1210-4 - Spring 2004

Note: Unless stated otherwise, answers without justification receive no credit. Please, show your work!

Find the area of the region under the curve $y=2x+2$ over the interval $[0,3]$. To do this, divide the interval $[0,3]$ into $n$ equal subintervals, calculate the area of the corresponding circumscribed polygon -the area determined by the $n$ rectangles- and, then, let $n\rightarrow \infty$.

The base of the region has length $3$, so if we subdivide into $n$ intervals of equal length, each interval has length $\Delta x = \frac{3}{n}$. For this subdivision, the partition points are $x_j = j \Delta x$, for $j=1,\ldots,n$ (these are the right end points of the corresponding intervals).

We construct the rectangles by taking the ``sample points'' $\bar{x}_j = x_j = j\Delta x$. We have to use these sample points (right end points) because the condition that we use a ``circumscribed polygon'' says that the rectangles must contain the given region.

Then, the Riemann sums give the area for the $n$ rectangles:

$$\begin{split}R_n &= \sum_{j=1}^n f(\bar{x}_j) \Delta x = \sum... ...(n+1)}{2} + 2 \frac{3}{n} n = 9 (1+\frac{1}{n}) + 6 \end{split}$$

To complete the computation of the area we let $n\rightarrow \infty$ and obtain that the area is $9+6=15$.

Remark: We can use simple geometry to check this result. Figure 1 shows the region whose area we are computing. Clearly the region decomposes as a rectangle (with base $3$ and height $2$) and a right triangle (with sides $3$ and $6$). Then, the total area is $6+\frac{18}{2} = 15$.

Figure 1: Region