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Volume as an integral
Let's consider
f: Rn → R.
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.
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.