The Area Problem

Sigma Notation

Definition : = + + + . . . + + . Where each is a Real number. Similarly, = f(m) + f(m+1) + . . . +f(n) where f is a function and m and n are integers and .

> Sum(a[i],i=1..7)=sum(a[i],i=1..7);

If f(i) = i, then we get the very important sum of the first n integers:

> Sum(i,i=1..n)= `1 + 2 + 3 + . . . + n`;

If then we can write the sum of the first 5 suared integers as (and you should check):

> Sum(i^2,i=1..5)=sum(i^2,i=1..5);

problem: Show that = -

> LHS:=Sum(i^2-3*i,i = 2 .. 5);

> value(%);

> RHS:=Sum(i^2,i=2..5)-3*Sum(i,i=2..5);

> value(%);

=> They are equal since LHS = RHS.

Properties and Special Sums:

Prove the following very important properties:

1)

2) +

3) Same as 2) except with a minus sign!

Memorize the following special sums :

1) = n 2) 3) 4) 5)

problem: Use the above properties and sums to confirm the following:

1)

> Sum(i^2-3*i+4,i=1..9)=sum(i^2-3*i+4,i=1..9);

2)

> Sum(i^2-3*i+4,i=1..n);

> simplify(value(%));

3)

> Sum((i+2)^2,i=1..n);

> simplify(value(%));

4) Find

> S:=sum(1*(i/n)^2/n,i = 1 .. n);

> simplify(%);

> limit(S,n=infinity);

Area under f and more

Big Problem: Suppose f(x) > 0 for all x [a,b]. Find the area between f and the x-axis from x = a to x = b.

Notation: If f is a positive function on [a,b] then the area above is denoted .

e.g. Find the area under y = 2x + 1 and above [0,4].

1. The Easy Way.

Below is a picture of the area (shaded) to be measured (nevermind the specific commands used for the picture). You can see that we can calculate the area using formulas from geometry i.e. we can split the area into the area of a rectangle + the area of a triangle. You should get:

Area of Rectangle + Area of Triangle = (4)(1) + (1/2)(4)(9 - 1) = 4 + 16 = 20 .

> with(plots):inequal({y<=2*x+1,x<4,x>0},x=0..6,y=0..10,optionsexcluded=(color=white));

2. The Hard way.

> with(student):f:=x->2*x+1;

> rightbox(f(x),x=0..4,4,shading=white,colour=black);

If you calculate the area of the 4 rectangles you will get:

> area4:=rightsum(f(x),x=0..4,4);value(%);

That not even close to the real area so we try more rectangles:

> rightbox(f(x),x=0..4,8,shading=white,colour=black);

> area8:=rightsum(f(x),x=0..4,8);value(%);

This is closer to the real answer but still not great. Let's try more rectangles:

> rightbox(f(x),x=0..4,16,shading=white,colour=black);

> area16:=rightsum(f(x),x=0..4,16);value(%);evalf(%);

Better, but still not exact. Let's get a formula for the area of n such rectangles:

> area_n:=simplify(sum(f(0+4*i/n)*4/n,i=1..n));

Let's verify the formula for the previous n's:

> subs(n=4,area_n);subs(n=8,area_n);subs(n=16,area_n);

Define the area to be the limit as n -> :

> Limit(area_n,n=infinity)=limit(area_n,n=infinity);

And this is what we got when we did it the easy way!!!

It follows that = 20 square units.

e.g. Find

> restart:with(student):f:=x->x^2+2;

We do this the same (hard) way: by considering the area of n "approximating" rectangles. Here are some pictures (and the approximations) for increasing numbers of approximating rectangles,

> rightbox(f(x),x=1..2,4,shading=white,colour=black,title=`Four Approximating Rectangles`);rightsum(f(x),x=1..2,4);value(%);evalf(%);

> rightbox(f(x),x=1..2,20,shading=white,colour=black,title=`Twenty Approximating Rectangles`);rightsum(f(x),x=1..2,20);value(%);evalf(%);

> rightbox(f(x),x=1..2,60,shading=white,colour=black,title=`Sixty Approximating Rectangles`);rightsum(f(x),x=1..2,60);value(%);evalf(%);

Here's the formula for the area of n of these rectangles:

> area_n:=Sum(f(1+i/n)*1/n,i=1..n)=sum(f(1+i/n)*1/n,i=1..n);

> area_N:=rhs(area_n);

> simplify(area_N);

> area:=limit(area_N,n=infinity);

> evalf(%);

More general:

If f is a continuous function on [a,b] then = where [a,b] has been divided into n subintervals of equal width and is a sample point from the subinterval. The sum is a special case of what is called a Reimann Sum.

If f is continuous and positive on [a,b] then gives the area under f and above [a,b]. If each is the right end-point of the base of the approximating rectangle then and .

If f is any continuous function on [a,b] then can be thought of as the sum of areas under f and above the x-axis plus the negative of the areas above f and under the x-axis. Consider the picture below for such a function:

> f:=x->4*x-x^3;

> rightbox(f(x),x=0..3,24,shading=white,colour=black);

> Sum(f(3*i/n)*3/n,i=1..n);

> S:=simplify(value(%));

> Limit(S,n=infinity)=limit(S,n=infinity);

Properties of the Definite Integral

Most of the following important properties can be easily proved by the Reimann Sum definition of and the properties of summation notation. When f is a positive function most can also be seen to have a geometric interpretation in terms of areas.

Basic Must Know Properties:

1) 2)

3) = 4) =

5) Replace "+" by "-" in number (2) 6) where b is any number (not necessarily between a and c)

Comparison Properties:

7) If for x in [a,b] then

8) If for x in [a,b] then

9) If f(x) [m,M] for all x [a,b] then [m(b - a), M(b - a)]

Comparison of 3 different Area Approximations

Consider the area under between x = 1 and x = 3 and approximate it using 8 rectangles:

i) using right endpoints for the heights

ii) using midpoints

iii) using left endpoints

> restart:with(student):f:=x->x^2+2;

> rightbox(f(x),x=1..3,8,shading=white,colour=black,title="Using right endpoints");

> REP:=rightsum(f(x),x=1..3,8);evalf(%);

> middlebox(f(x),x=1..3,8,shading=white,title="Using midpoints",colour=black);

> MP:=middlesum(f(x),x=1..3,8);evalf(%);

> leftbox(f(x),x=1..3,8,shading=white,title="Using left endpoints",colour=black);

> LEP:=leftsum(f(x),x=1..3);evalf(%);

The actual area is:

> Area:=Int(f(x),x=1..3);evalf(Area);

Notice that LEP < MP < REP and that MP is the best estimate of the area (see below). This is usually true.

> evalf(REP-Area);evalf(MP-Area);evalf(LEP-Area);