Area Problem Revisited
We already talked about area problem and presented there one approach to solve this problem. Here we will present another approach.
So, suppose we are given a function `f(x)` that is positive on `[a,b]` and we want to find an area `S` under the curve. It is not so easy to find the area of a region with curved sides. So, we will start from rectangle approximation.
To get a better understanding, let's start from example.
Example. Use rectangles to estimate the area under the parabola `f(x)=x^2` from 0 to 1.
Let's divide area `S` into five strips. We can approximate area of each strip by area of rectangle whose base is same as base of strip and whose height is same as the right edge of a strip (value of function at the rightest point of the strip).
Since we divided interval `[0,1]` on five strips then length of base of each strip is `(1-0)/5=0.2`.
So, we have following five subintervals: `[0,0.2]`, `[0.2,0.4]`, `[0.4,0.6]`, `[0.6,0.8]`, `[0.8,1]`.
As was already stated above length of base of each strip is 0.2. This is also base of each rectangle.
Now let's see what is the area of each rectangle. Recall that height is value of function at right endpoint.
First rectangle: height is `0.2^2`, so area is `S_1=0.2*0.2^2.`
Second rectangle: height is `0.4^2`, so area is `S_2=0.2*0.4^2`.
Third rectangle: height is `0.6^2`, so area is `S_3=0.2*0.6^2`.
Fourth rectangle: height is `0.8^2`, so area is `S_4=0.2*0.8^2`.
Fifth rectangle: height is `1^2`, so area is `S_5=0.2*1^2`.
Sum of areas of these approximating rectangles is `R_5=0.2*0.2^2+0.2*0.4^2+0.2*0.6^2+0.2*0.8^2+0.2*1^2=11/25=0.44`
As can be seen `R_5` is greater than area `S`, so `S<0.44`.
Instead of using the right endpoints we could use the smaller rectangles whose heights are the values of at the left-hand endpoints of the subintervals (the leftmost rectangle has collapsed because its height is 0).
Let's see what is the area of each rectangle. Recall that height is value of function at left endpoint.
First rectangle: height is `0^2`, so area is `S_1=0.2*0^2`.
Second rectangle: height is `0.2^2`, so area is `S_2=0.2*0.2^2`.
Third rectangle: height is `0.4^2`, so area is `S_3=0.2*0.4^2`.
Fourth rectangle: height is `0.6^2`, so area is `S_4=0.2*0.6^2`.
Fifth rectangle: height is `0.8^2`, so area is `S_5=0.2*0.8^2`.
Sum of areas of these approximating rectangles is `L_5=0.2*0^2+0.2*0.2^2+0.2*0.4^2+0.2*0.6^2+0.2*0.8^2=6/25=0.24`
As can be seen area `S` is larger than `L_5` so `0.24<S<0.44`.
Finally we can take height of rectangles to be value of function at midpoint of interval.
Corresponding approximation is `M_5=0.2*0.1^2+0.2*0.3^2+0.2*0.5^2+0.2*0.7^2+0.2*0.9^2=33/100=0.33`.
Since it is not clear what is greater: `S` or `M_5`, then we just write `S~~0.33`.
What will be if we increase number of strips? Clearly the greater number of strips, the better approximation we will obtain.
So, let's see what will be if we approximate area by the very large number of rectangles.
Suppose we divided area into `n` strips. Then length of base of each strip is `(1-0)/n=1/n`.
Let's use right-endpoint approximation. In this case height of i-th rectangle is `(1/i)^2` and area of i-th rectangle is `A_i=1/n*(i/n)^2`.
Summing areas of rectangles will give approximate area:
`S_n=1/n*(1/n)^2+1/n*(2/n)^2+...+1/n*(n/n)^2=sum_(i=1)^n 1/n (i/n)^2=1/n^3 sum_(i=1)^ni^2`.
To rewrite this formula we need the following formula: `sum_(i=1)^ni^2=(n(n+1)(2n+1))/6`.
So, we can write that `S_n=1/n^3 (n(n+1)(2n+1))/6=((n+1)(2n+1))/(6n^2)=(2n^2+3n+1)/(6n^2)=1/3+1/(2n)+1/(6n^2).`
If we now make `n` very large we will find required area `S`: `S=lim_(n->oo)S_n=lim_(n->oo)(1/3+1/(2n)+1/(6n^2))=1/3` (this is just limit of the sequence).
Let's apply this idea to our initial problem - problem of finding area under continuous curve `y=f(x)` on interval `[a,b]`.
We start by dividing `S` into `n` strips `S_1` , `S_2` ,...,`S_n` of equal width.
The widht of the interval `[a,b]` is `b-a`, so width of each of the `n` strips is `Delta x=(b-a)/n`.
These strips divide interval `[a,b]` into `n` strips:
`[x_0,x_1],[x_1,x_2],...,[x_(n-1),x_n]` where `x_0=a` and `x_n=b`.
In fact if we let `n->oo` then it doesn't matter what height of rectangle to choose: `h_i=a+iDelta x` (right endpoint), `h_i=a+(i-1)Delta x` (left endpoint), `h_i=a+iDelta x-Delta x/2` (midpoint). We can in fact choose any sample point `x_i^**` from interval `[x_(i-1),x_i]`.
In this case area of i-th rectangle is `A_i=f(x_i^**) Delta x`.
Now area `S` can be approximated by summing areas of `n` rectangles: `S_n=f(x_1^**)Delta x+f(x_2^**) Delta x+...+f(x_n^**) Deltax=sum_(i=1)^nf(x_i^**) Delta x`.
This approximation becomes better and better as `n->oo`.
Fact. The area `S` of the region that lies under the graph of the continuous function `f` is the limit of the sum of the areas of approximating rectangles: `S=lim_(n->oo)S_n=lim_(n->oo)sum_(i=1)^nf(x_i^**) Delta x`.
It can be proved that above limit always exists, since we are assuming that `f` is continuous.