~THE DEFINITE INTEGRAL~
~Question: "what is a definite
integral?"
Answer: "A number" (that's it!)
~Question:
"what does it represent?"
Answer: "Possibly an area, volume,
length, or some other numerical value of some quantity."
~Question: "can
it be a negative number?"
Answer:"sure can"
~To
understand how this can happen, we go back to the basic area
problem.
~Some simple areas can be found by inscribing or circumscribing
rectangles between the curve and the x axis and then letting the number increase
to infinity. If the sum is formulated properly, the limit of the sum will
equal the area in question. (see link, Area under a curve) Be careful,
since sums below the x axis are negative.
~These sums are called Riemann
Sums.
~The tie between the limit of a Reimann Sum and the area in
question can be shown in a very involved proof. It is lengthy and requires much
background in calculus. It is omitted here.
~It also can be proven, in a
less involved way, that the area under a curve and between x=a and x=b can be
found by integration. (uses one of the fundamental theorems, namely, the tie
between differentialion and integration) All you have to do is to set up the
definite integral from a to b and evaluate it.
~The definite integral
from x=a to x=b is denoted as ∫abf(x)dx and is evaluated by
integrating f(x), then substituing the upper limit in the expression, then
subtracting the lower limit substituted in the expression. If we denote F(x) as
the integral (without the constant), it would look this way: F(b) - F(a).
~Examples involving this technique have been given in the link, Area
under a curve, and many more will be done in class.
Properties of the Definite Integral
1) ∫aaf(x)dx =
0
2) ∫abf(x)dx = ∫baf(x)dx
3) If
a < c < b, then ∫acf(x)dx + ∫cbf(x)dx = ∫abf(x)dx
4) If
f(x)≥ g(x) on [a;b], then ∫abf(x)dx ≥
∫abg(x)dx
5) If
M=Max{f(x)} and m=Min{f(x)} over [a;b], then m(b-a) ≤
∫abf(x)dx ≤
M(b-a)
(says the definte integral will be between the numerical value
of one circumscribed & one inscribed rectangle)
~Well, since the
limit of a Riemann Sum equals the area and a definite integral equals the area,
we define the definite integral as the limit of a Riemann Sum. Therefore,
many problems are solved by setting up Riemann Sums.
~Consequently,
defining the definite integral as this limit, we have a direct tie to the area
problem and other problems that can be expressed and formulated as limits of
these types of Reimann Sums. This will give us a way to solve all kinds of
applied problems (more complicated area problems, volume problems, surface area
problems, work problems, arc length problems, fluid pressure problems, just to
name a few).
~Therefore, if a problem is subject to analysis by using a
limit of a Reimann Sum, it can (theoretically) be solved by a definite integral.
~Basically, Here is how the problem must be
formulated
lim ∑f(xk)Dxk = ∫abf(x)dx
Dx → 0
~So, informally, if a problem involves a
solution from x=a to x=b, we partition the interval into k subintervals, each of
length Dx, form the Riemann sum, then set up it's
limit. In actuality, only one representative term of the sum is needed since
they are all formed the same way (this is known as a "specimen" piece).(i.e., to
find the volume of a solid, we can think of the solid as the sum of an infinite
number of smooth disks of different sizes. However, the volume of each one is
represented the same way, so all we need is a specimen disk in the formulation
that represents each one. This would be called our specimen volume piece)
However, we don't need to evaluate the limit, just set it up so that we can
recognize the funcion f. Then all we have to do is integrate f(x) from a to b
for our answer. Solid disks may not be appropiate for all volumes. Depending on
their cross-section and how they were generated, a different specimen piece must
be chosen (i.e., washers, shells, or a special one with a given cross-sectional
area to fit our problem).
~Informally, one can think of the symbol ∫ab as the
lim∑ over [a;b] and the Dx in the sum
becoming dx in the integral.
~It turns out that many applied
problems can be "set up" by using Riemann Sums of these types, therefore, can be
solved by a definite integral.
~For those of you who go on to calculus
II, you will see many.
~The above mention problems are at the heart of
calculus II.