Conference rusure::math

    Oops!  make that 
    
           In this quadrature, you need to evaluate the sum of
           the *squares* first n integers ...
    
    Chuck 

    Chuck,
    
         I thought about the problem for a while, and the only result I got
    is that your daughter is in a very good school :-).  I am wodering how 
    the kids were shown that the area of a circle is pi*r^2?  If I still 
    remember, it was done with the more or less same limiting process.
    
    Eugene

    I don't think the school is all that great.  The teacher throws
    out these problems all of the time.  They seem to be more of a test
    for the parents than the kids.  I solve them all but some of the
    solutions get me blank stares from my daughter.
    
    Chuck

    I seem to recall that there IS a way to get the area under a parabola
    without invoking either calculus or limits.
    
    Hmmm...

    Nuts, I don't remember the geometric "trick" that one can use to get the
    area under a parabola offhand; it was known by the Greeks.

    But is this "without calculus" yet another kind of "new math"?

    To me, it seems that dividing the area under the curve into small
    rectangles is the most intuitively obvious thing there is - you can
    actually "see" how the error comitted at the boundaries of the curve
    gets smaller and smaller as the subdivision gets finer.  That's the
    historical route - and probably still the best way to introduce such
    an important idea...  there's only a little algebra to support it.

    I recall laboriously plotting a parabolic graph as a kid and convincing
    myself that this worked.

    Of course this neglects such things as "nowhere integrable functions"
    that analysts love to think about :-)

    - Jim

The area under a parabola y=x^2 is the same as the volume of a pyramid
with 4 sides of side x and height x.  Consider the volume as the "area"
under the area curve. The volume of a pyramid of height .5 and side 1 is
1/6 since six pyramids fill a cube. This is easy to see as a vertical
cross section of the pyramid that is parallel to the side is a right
triangle so that the pyramids form fit the cube. Then if we streatch the
pyramid so its height is 1, then the area is 2*1/6=1/3. Now streatch all 
dimensions by x, then the volume is X^3/3.



        TOM

    Re. .6
    
    Neat!  I like it.
    
    Re. .5
    
    Like I say, the teacher throws out hard problems once in a while.
    It seems to be just that teacher.  Frustrating for the kids but
    I guess harmless.
    
    Thanks very much.
    Chuck

The standard three-point formula for approximate integration on even 
intervals is exact for parabolas (and pretty close for other functions that
are reasonably smooth). One point - (0,0) - drops out immediately, and the
rest should be easy to memorize. Is this what the teacher was driving at? 

    re .6
    
    I don't get it.  Could you explain further?
    
    Eugene

    re .9
    
    The way I took it, ther area of the base of the pyramid is the
    function y = x^2.  Now the area under the curve is the sum of these
    areas times (God forgive me) the infinitesimal height - you
    can use better language here - which is the volume.  That is, the
    volume of the pyramid is the area under the parabola.  Slick!
    
    Anyway, I hope I can put it in a nice way for my daughter - it has
    the intuitive appeal I was looking for (I think).
    
    Chuck

    re -1
    
    Now I get it.  Thanks.
    
    Eugene

>< Note 1054.0 by HIBOB::SIMMONS "Tristram Shandy as an equestrian" >
>                        -< Quadrature without calculus >-
>
>    My daughter (7th grade) was given the problem of finding the area
>    under the curve y = x^2 on an interval starting at the origin without
>    calculus.  I showed her the obvious quadrature dividing the interval
>    into n equal subintervals and summing the areas of the small
>    rectangles.  In this quadrature, you need to evaluate the sum of
>    the first n integers and observe that for large n, this sum is very
>    "close" to (n^3)/3.  The exact answer follows.
>    
>    My question.  Does anyone know a more intuitive or more easily
>    explained quadrature (without calculus) for this problem?  The
>    quadrature above has some reasoning which seems to give even a bright
>    7th grader trouble.

If you are allowed any technique to solve the problem, I can think of several
of the 'barometer off the Empire State building' variety that will appeal
to engineers and other seekers-of-the-other-way-around-the-establishment.

way 1:

Someone elsewhere in this notesfile has mentioned using a graph on paper.
Circumscribe the parabolic area with a rectangle, carefully cut out the
parabolic and rectangular regions, and weigh with an analytical balance.
The weights will be proportional to the corresponding areas.

way 2:

Use s = (1/2)*a*t^2, the formula for the distance dropped in a uniform
gravitational field from rest as a function of time since the release of
the object.

To do this you need a meter stick and a stopwatch (an accurate regular watch
with .1 second precision might suffice).

Use a moderately heavy compact object (low drag to weight ratio at reasonable
speeds) like a ball bearing (not the Ming vase or a younger brother).  Drop
the object (hereafter referred to as a 'bearing', even if it isn't) from
different heights, timing it.  Getting the release time right is the hardest
part, and is left to the reader as an exercise; additional apparatus can
be devised.  Use a linoleum or other semi-soft surface that you can
hear a satisfying impact sound from, but not one that dents or chips.
Best to perform this experiment when Mom is not around anyway.

Adjust the height until the fall time corresponds to the abscissa value for
which you want the integral.  Several tries at each height will probably be
necessary to get your reflexes synchronised with when to stop the watch.
The drop height is then the area under the curve.  Correct for the finite
speed of sound in air from the floor to your ear after estimating if this
is worthwhile.  Provide only significant figures.

This will teach about uniform acceleration, parabolas, air drag, ball bearings,
stopwatches, feedback, acoustics, guessing at the relevant factors,
significant figures, linoleum, and the psychology of mothers.

The instructor should be overjoyed.

    Re .12 - mathemeticians don't like anything without an abstract
    axiomatic basis, so it probably wouldn't do...  the earlier idea
    of the 6 pyramids fitting together to make the cube is really neat though.

    There is an archimedian theorem about the area bounded by a parabola
    and a chord - it's 2/3 of the area bounded by the triangle made of the
    chord and the tangent lines where the chord meets the parabola.

    The parabola can be in general position in the plane.

    The proof consists in showing that a line drawn between the midpoints
    of the tangential sides is tangent to the parabola (!), thus you
    have the same situation as before in the two pieces of the parabola,
    and the process can be recursively repeated and the series telescopes.

    Unfortunately tangency is a calculus concept when you get down to it -
    but the geometry very nice.  You can show that the construction works
    by relying on the geometric definition of a parabola (directrix and
    focal point...)

    Also since a parabola is still a parabola under general affine
    transformations of the plane (x' = a*x + b*y + c, y' = d*x + e*y + f),
    that important idea could be introduced.  Simple computer graphics
    would be helpful in showing the shearing, translation, and rotation
    involved.  You could show that ratios of areas are invariant under
    affine transformations, all kinds of stuff, all with pictures to help
    the intuition.

    - Jim

    Just to complete things,  the pyramid solution came too late for
    my daughter to use.  There were two kids in the class who had
    solutions, my daughter and one other.  Both solutions were the direct
    summation of the squares of the first n integers which comes from
    adding up all of the little rectangles.  Jessica did tell the guy
    about the pyramid solution and she said he liked it.  Anyway, this
    problem was a test of the parents.
    
    Maybe the guy hopes to find a second Gauss in his class someday!
    
    Chuck