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:[font = section; inactive; ]
The Fundamental Theorem
:[font = text; inactive; ]
The Fundamental Theorem of Calculus says, roughly, that the following processes "undo" one another:

      - Computing slopes of tangent lines (Differentiation)

      - Computing areas under curves (Integration)

I'll illustrate this by defining two functions:

      - DifferenceQuotient[] will approximate the slope of the tangent line to a curve.

      - RectangleSum[] will approximate the area under a curve.

Then I'll show that if you do one of these functions followed by the other, you get (approximately) what you started with.
:[font = text; inactive; ]
I'll use the following test function:
:[font = input; nowordwrap; ]
f[x_] := Sin[x^2]
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DifferenceQuotient[] computes the (approximate) slope of the tangent line to a curve. If you compare it to the definition of the derivative, you'll see that it's essentially the same thing, except that I don't compute a limit.
:[font = input; nowordwrap; ]
DifferenceQuotient[f_, dx_] :=
   Function[(f[# + dx] - f[#]) / dx]
:[font = text; inactive; ]
Taking the limit would amount to letting dx go to 0. So I can get an approximation to the derivative by taking a small value for dx --- I'll use dx = 0.01.

Here's the approximate derivative of my test function Sin[x^2]:
:[font = input; startGroup; nowordwrap; ]
dqf = DifferenceQuotient[f, 0.01]
:[font = output; inactive; formatted; output; endGroup; nowordwrap; ]
(f[#1 + 0.01] - f[#1])/0.01 & 


;[o]
f[#1 + 0.01] - f[#1]
-------------------- & 
        0.01
:[font = text; inactive; ]
Let's check that it works. Here's the approximate value of the derivative at x = 1.3 as given by dqf:
:[font = input; startGroup; nowordwrap; ]
dqf[1.3]
:[font = output; inactive; formatted; output; endGroup; nowordwrap; ]
-0.3441669325131014


;[o]
-0.344167
:[font = text; inactive; backColorRed = 65280; backColorGreen = 65280; backColorBlue = 65280; fontColorRed = 0; fontColorGreen = 0; fontColorBlue = 0; plain; fontName = "Arial"; fontSize = 12; ]
And here's the "exact" value of the derivative at x = 1.3:
:[font = input; startGroup; startGroup; nowordwrap; ]
f'[1.3]
:[font = output; inactive; formatted; output; endGroup; nowordwrap; ]
-0.3091960801711923


;[o]
-0.309196
:[font = text; inactive; endGroup; ]
You can see that the two values are pretty close. I could get a better approximation by making dx smaller.
:[font = text; inactive; ]
Here's a function that approximates the area under a curve using a rectangle sum:
:[font = input; nowordwrap; ]
RectangleSum[f_, start_, dx_] :=
   Function[Sum[f[start + i dx],
   {i, 0, Floor[(# - start)/dx]}] dx]
:[font = text; inactive; ]
In RectangleSum[], dx represents the width of a rectangle, and I'm using the left-hand endpoint of each subinterval to get the rectangle's height. It's essentially the same as the definition of the definite integral of a function, except that I'm not computing a limit. I can obtain a good approximation to the actual integral by making dx small. I'll take dx = 0.01.

I apply my RectangleSum[] function to Sin[x^2] to get an approximation to the integral of Sin[x^2]:
:[font = input; startGroup; nowordwrap; ]
sumf = RectangleSum[f, 0, 0.01]
:[font = output; inactive; formatted; output; endGroup; nowordwrap; ]
Sum[f[0 + i*0.01], {i, 0, Floor[(#1 - 0)/0.01]}]*0.01 & 


;[o]
                                #1 - 0
Sum[f[0 + i 0.01], {i, 0, Floor[------]}] 0.01 & 
                                 0.01
:[font = text; inactive; ]
Again, I'll check my work by comparing my approximate integral to the "exact" integral. Here is the approximate integral from 0 to Sqrt[Pi] using sumf[]:
:[font = input; startGroup; nowordwrap; ]
sumf[Sqrt[Pi]]
:[font = output; inactive; formatted; output; endGroup; nowordwrap; ]
0.894834765072355


;[o]
0.894835
:[font = text; inactive; ]
And here is the "exact" value of the integral, which I've computed using Mathematica's numerical integration routines:
:[font = input; startGroup; nowordwrap; ]
NIntegrate[f[x], {x, 0, Sqrt[Pi]}]
:[font = output; inactive; formatted; output; endGroup; nowordwrap; ]
0.894831469484145


;[o]
0.894831
:[font = text; inactive; ]
You can see that the results are pretty close.
:[font = text; inactive; ]
Now I'm ready for my experiment. First, I'll use DifferenceQuotient[] to build a function which approximate the derivative of f[x] = Sin[x^2]:
:[font = input; startGroup; nowordwrap; backColorRed = 65280; backColorGreen = 65280; backColorBlue = 65280; fontColorRed = 0; fontColorGreen = 0; fontColorBlue = 0; bold; fontName = "Courier New"; fontSize = 12; ]
stepone = DifferenceQuotient[f, 0.1]
:[font = output; inactive; formatted; output; endGroup; nowordwrap; ]
(f[#1 + 0.1] - f[#1])/0.1 & 


;[o]
f[#1 + 0.1] - f[#1]
------------------- & 
        0.1
:[font = text; inactive; ]
Next, I'll feed my approximate derivative stepone into RectangleSum[] to build its approximate integral:
:[font = input; startGroup; nowordwrap; backColorRed = 65280; backColorGreen = 65280; backColorBlue = 65280; fontColorRed = 0; fontColorGreen = 0; fontColorBlue = 0; bold; fontName = "Courier New"; fontSize = 12; ]
steptwo = RectangleSum[stepone, 0, 0.1]
:[font = output; inactive; formatted; output; endGroup; nowordwrap; ]
Sum[((f[#1 + 0.1] - f[#1])/0.1 & )[0 + i*0.1], 
    {i, 0, Floor[(#1 - 0)/0.1]}]*0.1 & 


;[o]
     f[#1 + 0.1] - f[#1]                              #1 - 0
Sum[(------------------- & )[0 + i 0.1], {i, 0, Floor[------]}] 0.1 & 
             0.1                                       0.1
:[font = text; inactive; ]
Thus, steptwo is the result of feeding my original function f[x] = Sin[x^2] through the derivative and integral builders, one after the other. Here's the graph of steptwo:
:[font = input; startGroup; nowordwrap; ]
Plot[steptwo[x], {x, 0, 2Sqrt[Pi]}]
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;[o]
-Graphics-
:[font = text; inactive; ]
And here's the graph of f[x] = Sin[x^2]:
:[font = input; startGroup; nowordwrap; ]
Plot[Sin[x^2], {x, 0, 2Sqrt[Pi]}]
:[font = postscript; inactive; output; BITMAP; PostScript; pictureLeft = 100; pictureTop = 0; pictureWidth = 300; pictureHeight = 185; nowordwrap; ]
%!
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MathPictureEnd
:[font = output; inactive; formatted; output; endGroup; nowordwrap; ]
No Input Form was saved for this expression.


;[o]
-Graphics-
:[font = text; inactive; ]
The resemblance is obvious. This is numerical evidence that computing the derivative followed by the integral gives you what you started with.
:[font = text; inactive; ]
Next, I'll try it in the opposite order. First, the integral:
:[font = input; startGroup; nowordwrap; backColorRed = 65280; backColorGreen = 65280; backColorBlue = 65280; fontColorRed = 0; fontColorGreen = 0; fontColorBlue = 0; bold; fontName = "Courier New"; fontSize = 12; ]
stepthree = RectangleSum[f, 0, 0.1]
:[font = output; inactive; formatted; output; endGroup; nowordwrap; ]
Sum[f[0 + i*0.1], {i, 0, Floor[(#1 - 0)/0.1]}]*0.1 & 


;[o]
                               #1 - 0
Sum[f[0 + i 0.1], {i, 0, Floor[------]}] 0.1 & 
                                0.1
:[font = text; inactive; ]
Next, I feed the integral into the derivative:
:[font = input; startGroup; nowordwrap; backColorRed = 65280; backColorGreen = 65280; backColorBlue = 65280; fontColorRed = 0; fontColorGreen = 0; fontColorBlue = 0; bold; fontName = "Courier New"; fontSize = 12; ]
stepfour = DifferenceQuotient[stepthree, 0.1]
:[font = output; inactive; formatted; output; endGroup; nowordwrap; ]
((Sum[f[0 + i*0.1], {i, 0, Floor[(#1 - 0)/0.1]}]*0.1 & )[#1 + 0.1] - 
     (Sum[f[0 + i*0.1], {i, 0, Floor[(#1 - 0)/0.1]}]*0.1 & )[#1])/0.1 & 


;[o]
                                 #1 - 0
((Sum[f[0 + i 0.1], {i, 0, Floor[------]}] 0.1 & )[#1 + 0.1] - 
                                  0.1
 
                                    #1 - 0
    (Sum[f[0 + i 0.1], {i, 0, Floor[------]}] 0.1 & )[#1]) / 0.1 & 
                                     0.1
:[font = text; inactive; ]
Finally, I'll plot the result:
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Plot[stepfour[x], {x, 0, 2Sqrt[Pi]}]
:[font = postscript; inactive; output; BITMAP; PostScript; pictureLeft = 100; pictureTop = 0; pictureWidth = 300; pictureHeight = 185; nowordwrap; ]
%!
%%Creator: Mathematica
%%AspectRatio: .61803 
MathPictureStart
%% Graphics
/Courier findfont 10  scalefont  setfont
% Scaling calculations
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Again, I get a graph which resembles the graph of the original function: Computing the integral followed by the derivative gives you what you started with.
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The Fundamental Theorem of Calculus is a precise statement of the results I've described. As a practical point, it allows you to compute definite integrals using antiderivatives --- and thus, is a tool you'll use frequently in applications of integral calculus.
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Copyright 1997 by Bruce Ikenaga
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