| www.whyslopes.com Volume 3, Why Slopes and More Math |
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-/[]\- Logic
Mastery
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Online Volume 2, Three Skills for Algebra, Chapters 1 to 25 - skip 18., verbalizes and explains key skills and concepts, those needed in calculus, again to make the hard easier. A visual understanding of complex numbers may serve as back ground info for partial fraction decomposition. |
Area Calculation Problem
The selection of labels x and y for the horizontal and
vertical axes for most of graphs met so far is
arbitrary. It can be changed. The letters s and q
could have been used instead in all the previous graphs.
You should imagine this replacement, and the effect, if
any, it has on your knowledge or opinion of mathematics.
Problem: Suppose G(x) and h(x) are continuous at each point x in the interval [a,b]. Further suppose that the slope G¢(x) of G(x) satisfies G¢(x) = h(x) for a £ x £ b. Find a formula for the area A under the curve q = h(s) between s = a and s = b.The solution to this problem follows in three steps.
Step 1. Define an Area Function
First, introduce a function F(x) as follows. For each x between a and b inclusive let
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Here F(a) = 0 and F(b) = A is the area to be computed. Note that the area computation will be based on (a) finding the slope or derivative F¢(x) and then (b) observing how to obtain F(x) from a knowledge of F¢(x). The value of A = F(b) is required. It will be given by a formula involving the function G(x).
Step 2. Area Function Slope Calculation
Second, the following diagram leads to a formula for F¢(x).
In this figure, the area under q = h(s) from s = x to s = x+Dx is given by
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Since h(s) is assumed to be continuous on the interval [a,b], it is continuous at the point s1 = x. Therefore, for every whole number k, there is a number n such thatwhen |s-x| £ [1/2]·10-n. Now given k and such an n, if x £ s £ x+Dx and 0 £ Dx £ [1/2]·10-n then |s-x| £ |Dx| £ [1/2]·10-n as well. For such numbers s, it follows that |h(s)-h(x)| £ [1/2]·10-k. The latter in turn (see diagram) implies that the region B [between q = h(x) and q = h(s) above the s-interval from s = x to s = x+Dx] has an area
|h(s)-h(x)| £ 1
2·10-k This is equivalent to
|Area B| = |F(x+Dx) -F(x) -h(x)Dx| £ Dx · 1
2·10-k Therefore
ê
ê
êArea B
Dxê
ê
ê= ê
ê
ê|F(x+Dx) -F(x) -h(x)Dx
Dxê
ê
ê£ 1
2·10-k when 0 £ Dx £ [1/2]·10-n. Since the foregoing argument holds for every whole number k > 0, it implies that the limiting value of [(F(x+Dx) -F(x))/(Dx)] = h(x) when Dx > 0 approaches zero.
ê
ê
êArea B
Dxê
ê
ê= ê
ê
êF(x+Dx) -F(x)
Dx- h(x) ê
ê
ê£ 1
2·10-k
Step 3. Difference of Two Functions
Third, the previous discussion of vertical motions (and earthquakes) implies or suggests that for x in the interval [a,b], the difference F(x)-G(x) = C for some constant number C which does not depend on x. Recall F(a) = 0. This implies C = F(a)-G(a) = 0-G(a) = -G(a). Therefore, the constant value is given by C = -G(a) because F(a) = 0. Now F(x)-G(x) = C implies
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Remark. The latter formula is correct whenever G(x) is a function whose slope or derivative is h(x) for every x in the interval [a,b]. Given a formula for the function h(s) or h(x), the area calculation problem can be solved easily if methods of anti-differentiation can provide a G(x). For some h(x), this is possible. In particular, methods for anti-differentiation can be employed (sometimes) to find several functions G(x) whose derivative or slope on the interval x = a to x = b coincides everywhere with the slope h¢(x) of F(x). The problem statement above assumed one such function G(x) was available.
Methods for anti-differentiation or reversing slope calculations
say how to find possible
formulas for a function f(x) from a single formula for
its derivative (slope) m = f¢(x). These methods are ad hoc.
They do not work in all examples, but they do work in a large
number. Methods for finding or obtaining a
function from its derivative or slope lead to formulas for
the calculation of areas, volumes, weights, masses, forces,
totals etc met in geometrical, physical and some business
computations.
Remark. The computation of many geometric, physical and business quantities can be related to the computation of the area under the graph of some function q = h(s). The unit of area in these graphs is given by a product of the units of the horizontal and vertical coordinates q and s.
| www.whyslopes.com Volume 3, Why Slopes and More Math |
Volumes
Foreword, One Calculus preview and Online Chapters:
(V) signals video (RealPlayer Format) to
watch
Units in Calculations:
Enriched material: The Appendices of Volume
3 are located in the Real
Analysis Area.
Integration
& Lipschitz
Continuity
These appendices continue the
decimal viewpoint of limits, error
control and continuity begun
in Chapter 14. The One Sided
Range Theorem is a postscript,
not in printed version.