The fundamental theorem of calculus
(In words, as much as possible)
David C, 2014
The fundamental theorem of calculus says that the process of finding the area
under a curved graph line is mathematically ...
Suppose I use a straight line to approximate the slope between points x1 and x2.
12
12
xx
yy
m



y1
y2
x1 x2
y1
y2
x1 x2
I could rearrange this as shown, for mysterious reasons that I’ll keep to myself.
  1212 yyxxm 
12
12
xx...
Now any time you multiply two numbers together, like m and (x2-x1),
this is arithmetically the same as finding the area of...
m
What could this possibly mean in real terms?
x1 x2
Now any time you multiply two numbers together, like m and (x2-x1),
t...
m
Well this could be the area under a part of a curved graph line.
x1 x2
  1212 yyxxm 
But what could this graph poss...
m
m(x).
x1 x2
  1212 yyxxm 
The height of the rectangle is m,
so it suggests that the vertical axis is m(x).
m
Now remember that m(x) is a slope function,
so the area of the rectangle is kind of like
the area under the graph of a s...
m
m(x).
Therefore the area of a very skinny rectangle
under the graph m(x) is the same as y2 - y1.
x1 x2
  1212 yyxxm 
m
m(x).
And what is y2 - y1?
Well, think about it:
If m is the derivative of y,
then y is the anti-derivative of m.
x1 x2
...
m
m(x).
x1 x2
So to get the area under part of a curvy line, you pretend that the area is a
rectangle and anti-differentia...
m(x).
x1 x2
You can make it work, though, by putting a whole lot of rectangles
together, side-by-side.
The area of each re...
m(x).
x2
You can make it work, though, by putting a whole lot of rectangles
together, side-by-side.
The area of each recta...
m(x).
x2
So the area under any graphline
is equal to the antiderivative of the left and right edges
of the region you’re l...
m(x).
9
An example of this theorem in action would be
finding the area under a line that curves according to, say, a cubic...
m(x).
The traditional notation for this process is a little bit more ornate
but it means exactly the same thing.

9
1
3
...
[END]
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Dc32343

  1. 1. The fundamental theorem of calculus (In words, as much as possible) David C, 2014
  2. 2. The fundamental theorem of calculus says that the process of finding the area under a curved graph line is mathematically equivalent to un-doing a differentiation. What does that mean? I’ll try to show you in an informal, hand-waving sort of way.
  3. 3. Suppose I use a straight line to approximate the slope between points x1 and x2. 12 12 xx yy m    y1 y2 x1 x2
  4. 4. y1 y2 x1 x2 I could rearrange this as shown, for mysterious reasons that I’ll keep to myself.   1212 yyxxm  12 12 xx yy m   
  5. 5. Now any time you multiply two numbers together, like m and (x2-x1), this is arithmetically the same as finding the area of a rectangle. m x1 x2   1212 yyxxm 
  6. 6. m What could this possibly mean in real terms? x1 x2 Now any time you multiply two numbers together, like m and (x2-x1), this is arithmetically the same as finding the area of a rectangle.   1212 yyxxm 
  7. 7. m Well this could be the area under a part of a curved graph line. x1 x2   1212 yyxxm  But what could this graph possibly be?
  8. 8. m m(x). x1 x2   1212 yyxxm  The height of the rectangle is m, so it suggests that the vertical axis is m(x).
  9. 9. m Now remember that m(x) is a slope function, so the area of the rectangle is kind of like the area under the graph of a slope function, especially if x2 and x1 are very close together. m(x). x1 x2   1212 yyxxm 
  10. 10. m m(x). Therefore the area of a very skinny rectangle under the graph m(x) is the same as y2 - y1. x1 x2   1212 yyxxm 
  11. 11. m m(x). And what is y2 - y1? Well, think about it: If m is the derivative of y, then y is the anti-derivative of m. x1 x2   1212 yyxxm  Therefore the area of a very skinny rectangle under the graph m(x) is the same as y2 - y1.
  12. 12. m m(x). x1 x2 So to get the area under part of a curvy line, you pretend that the area is a rectangle and anti-differentiate the formula at the left and right edges of the rectangle. The problem, however, is that the rectangle has to be very skinny if this approximation is to be any good.
  13. 13. m(x). x1 x2 You can make it work, though, by putting a whole lot of rectangles together, side-by-side. The area of each rectangle is equal to antiderivative( right edge ) - antiderivative( left edge )
  14. 14. m(x). x2 You can make it work, though, by putting a whole lot of rectangles together, side-by-side. The area of each rectangle is equal to antiderivative( right edge ) - antiderivative( left edge ) When you add all the rectangles together, all those antiderivatives cancel out except for the ones at the far edges, left and right. x1
  15. 15. m(x). x2 So the area under any graphline is equal to the antiderivative of the left and right edges of the region you’re looking at. (This, in words, is the fundamental theorem.) x1
  16. 16. m(x). 9 An example of this theorem in action would be finding the area under a line that curves according to, say, a cubic formula. 1 3 2xm  4 4 2 xy  Area between x=1 and x=9 is    4 4 24 4 2 19  3280
  17. 17. m(x). The traditional notation for this process is a little bit more ornate but it means exactly the same thing.  9 1 3 2 dxxarea  9 1 4 4 2 x ...etc 91
  18. 18. [END]
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