Question

Using the methods suggested in the preceding problem, find the area of the trapezoid bounded by the line y= x+3, the ordinates x=1, x=3, and the x axis.

Answer 1

We can find the area of the trapezoid as the limit of a sum of inscribed rectangular areas:

Let's divide the interval [1, 3] into $n$ equal segments with length $\frac{2}{n}$ by the points x₀, x₁, ..., x_n,

where

$x_0=1, \: x_1=1+\frac{2}{n},\: x_2=1+2\frac{2}{n},\: ...,\:x_n=1+n\frac{2}{n}=1+2=3.$

The heights of the n approximating rectangles are:

$h_1=1+3=4,\: h_2=1+\frac{2}{n}+3=4+\frac{2}{n},\: ...,\: h_n=1+(n-1)\frac{2}{n}+3=4+(n-1)\frac{2}{n},$

and the sum of their areas is

$S_n=4\cdot\frac{2}{n}+(4+\frac{2}{n})\cdot \frac{2}{n}+...+(4+(n-1)\frac{2}{n})\cdot \frac{2}{n}=$

$=[4n+(\frac{2}{n}+2\frac{2}{n}+...+(n-1)\frac{2}{n})]\cdot\frac{2}{n}=$

$=[4n+\frac{2}{n}(1+2+...+(n-1))]\cdot\frac{2}{n}=$

$=[4n+\frac{2}{n}\frac{1+(n-1)}{2}(n-1)]\cdot\frac{2}{n}=[4n+\frac{2}{n}\cdot\frac{n}{2}(n-1)]\cdot\frac{2}{n}=$

$=[4n+n-1]\cdot \frac{2}{n}=[5n-1]\cdot \frac{2}{n}=10-\frac{2}{n}.$

It is easy to see that $S_n$ tends toward the value 10 as $n$ gets larger and larger.

For this reason, we take $S=10.$

Answer: 10.