5.6 Computer Calculation of Area

In this section we will discuss a Maple program for calculating approximate values of $A_a^bf$ for monotonic functions $f$ on the interval $[a,b]$. The programs will be based on formulas discussed in theorem 5.40.

Let $f$ be a decreasing function from the interval $[a,b]$ to $\mbox{{\bf R}}_{\geq 0}$, and let $P=\{x_0,x_1,\cdots ,x_n\}$ be a partition of $[a,b]$. We know that

$$\alpha(I_a^b(f,P)) \leq A_a^b f \leq \alpha(O_a^b(f,P)),$$

where

$$\alpha(I_a^b(f,P)) = \sum_{i=1}^n (x_i - x_{i-1}) f(x_i),$$ (5.83)

$$\alpha(O_a^b(f,P)) = \sum_{i=1}^n (x_i - x_{i-1}) f(x_{i-1}).$$ (5.84)

Let $V_a^b(f,P)$ be the average of $\alpha(I_a^b(f,P))$ and $\alpha(O_a^b(f,P))$, so

$$V_a^b(f,P) = { \alpha(I_a^b(f,P)) + \alpha(O_a^b(f,P)) \over ... ...m_{i=1}^n (x_i - x_{i-1})\cdot{f(x_{i}) + f(x_{i-1}) \over 2}.$$

Now $${(x_i - x_{i-1})\cdot{f(x_{i}) + f(x_{x-1}) \over 2}}$$ represents the area of the trapezoid with vertices $(x_{i-1},0)$, $(x_{i-1}, f(x_{i-1}))$, $(x_i, f(x_i))$ and $(x_i,0)$, so $V_a^b(f,P)$ represents the area under the polygonal line obtained by joining the points $(x_{i-1}, f(x_{i-1}))$ and $(x_i, f(x_i))$ for $1\leq i\leq n$.

In the programs below, leftsum(f,a,b,n) calculates

$$\displaystyle {\sum_{j=1}^{n}f\Bigg(a+(j-1)\Big( {{b-a}\over ... ... \sum_{j=1}^{n}f\Bigg(a+(j-1)\Big( {{b-a}\over n}\Big)\Bigg),$$

which corresponds to (5.84) when $P$ is the regular partition of $[a,b]$ into $n$ equal subintervals, and rightsum(f,a,b,n) calculates

$$\displaystyle {\sum_{j=1}^n f\Bigg(a+j\Big({{b-a}\over n}\Big... ...n}\Big) \sum_{j=1}^n f\Bigg(a+j\Big({{b-a}\over n}\Big)\Bigg).$$

which similarly corresponds to (5.83). The command average(f,a,b,n) calculates the average of leftsum(f,a,b,n) and rightsum(f,a,b,n).

The equation of the unit circle is $x^2+y^2=1$, so the upper unit semicircle is the graph of $f$ where $f(x)=\sqrt{1-x^2}$. The area of the unit circle is 4 times the area of the portion of the circle in the first quadrant, so

$$\pi=4A_0^1[\sqrt{1-t^2}].$$

Also

$$\ln (2)=A_1^2\Big[{1\over t}\Big].$$

My routines and calculations are given below. Here leftsum, rightsum and average are all procedures with four arguments, f,a,b, and n.

f is a function.

a and b are the endpoints of an interval.

n is the number of subintervals in a partition of [a,b].

The functions F and G are defined by F$(x) = 1/x$ and G $(x) = \sqrt{1-x^2}$. The command

             average(F,1.,2.,10000);

estimates $\ln (2)$ by considering the regular partition of $[1,2]$ into $10000$ equal subintervals. and the command

             4*average(G,0.,1.,2000);

estimates $\pi$ by considering the regular partition of $[0,1]$ into $2000$ equal subintervals.

> leftsum :=
>       (f,a,b,n) -> (b-a)/n*sum(f( a +((j-1)*(b-a))/n),j=1..n);

$${\it leftsum} := ( {f}, {a}, {b}, {n} ) \rightarrow {\dis... ...)} ( {b} - {a} )}{{n}}} \! \right) \! \right) }{{n}}}$$

> rightsum := 
>       (f,a,b,n) -> (b-a)/n*sum(f( a +(j*(b-a))/n),j=1..n);

$${\it rightsum} := ( {f}, {a}, {b}, {n} ) \rightarrow {\di... ...j} ( {b} - {a} )}{{n}}} \! \right) \! \right) }{{n}}}$$

> average :=
>       (f,a,b,n) -> (leftsum(f,a,b,n) + rightsum(f,a,b,n))/2;

$${\it average} := ( {f}, {a}, {b}, {n} ) \rightarrow {\dis... ...aystyle \frac {1}{2}} {\rm rightsum}( {f}, {a}, {b }, {n} )$$

> F := t -> 1/t;

$${F} := {t} \rightarrow {\displaystyle \frac {1}{{t}}}$$

> leftsum(F,1.,2.,10000);

$$.6931721810$$

> rightsum(F,1.,2.,10000);

$$.6931221810$$

> average(F,1.,2.,10000);

$$.6931471810$$

> ln(2.);

$$.6931471806$$

> G := t -> sqrt(1-t^2);

$${G} := {t} \rightarrow {\rm sqrt}( 1 - {t}^{2} )$$

> 4*leftsum(G,0.,1.,2000);

$$3.142579520$$

> 4*rightsum(G,0.,1.,2000);

$$3.140579522$$

> 4*average(G,0.,1.,2000);

$$3.141579521$$

> evalf(Pi);

$$3.141592654$$

Observe that in these examples, average yields much more accurate approximations than either leftsum or rightsum.