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{displaymath}\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}\end{displaymath}    

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
\includegraphics[scale=.6,angle=0,
clip= true]{q10-f0.eps}



Javier Fernandez 2004-04-14