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Volume as an integral

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Let's consider

 f: RnR.

The volume of the box

 B = ∏i=1n [ ai, bi ]

is

 V = ∏i=1n ( bi - ai ),

i.e. the volume of the product of n intervals is the product of their lengths.

In dim = 2, the area is

 ∆A = ( b1 - a1 ) ⋅ ( b2 - a2 ),

and in dim = 3, the volume is

 ∆V = ( b1 - a1 ) ⋅ ( b2 - a2 ) ⋅ ( b3 - a3 ).

In general, the volume of each little box is

 ∆V = ∏i=1n ( bi - ai ) / kn 
    → 0 as k → ∞.

The Riemann sum of f: B → ℝ for this partition:

 Each [ ai, bi ] is divided into k intervals, and each interval has a point in it, call it cii.

Riemann sum dim 2 and 3.jpg

In dim = 2:

i,j f( cij ) ∆A, where 
 the 3-volume of each box is f( cij ) ∆A, and 
 the height is f( cij ), and 
 the base is ∆A.

In dim = 3:

i,j,k f( cijk ) ∆V, where 
 the 4-volume of each box is f( cijk ) ∆V, and 
 the height is f( cijk ), 
 and the 3-volume (of the base) is ∆V.

Back to dim = 1. Consider

 limk→∞,i=1k f( ci ) ∆x
 (or lim∆x→0 ...),

where ci denotes the midpoint of the respective intervals. If this limit exists, it is called the Riemann integral of f from a to b. We write

ab f(x) dx,

and call f an integrable function on [ a, b ].

For dim = 2 it's all very similar

 limk→∞i=1k f( cij ) ∆A = ∫∫B f(x) dA,

where

 B = [ a1, b1 ] × [ a2, b2 ].

Example. Let's compute something. Let

 f(x) = x, a = 0, b = 1, and 
 xi = 1 / k i  the endpoints for i = 1, ..., k (k intervals).
 ci = 1 / k i chosen points for i = 0, ..., k-1. 

Then

 limk→∞i=1k f( ci ) ∆x
   = limk→∞i=1k ci ∆x
   = limk→∞ ( c1 + ... + ck ) 1 / k
   = limk→∞ ( 0 + (1 / k) + (2 / k ) + ... + (( k - 1 ) / k ) k
   = limk→∞ ( (1 / k2) + (2 / k2) + ... + (( k - 1 ) / k2) )     arithmetic progression
   = ( k ( k - 1 ) / 2 ) / k2
   = 1 / 2

So,

[0,1] x dx 
            = ∫01 x dx 
            = ( x2 / 2 )|01 
            = 1 / 2.