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286 INTRODUCTION TO CALCULUS
Figure A.7 Lower and upper sums over a partition and its refinement. The lower sum has increased and the upper sum has decreased in the refinement. The area under the curve is always between the lower and upper sums.
and the upper sum
Mx0,...,xn =Yj mi X (xi — xi-i)
We can refine the partition by adding one more x value to it. Let xi,..., x'n+1 be a refinement of the partition x1,..., xn. Then x0 = x0, xn+1 = xn, xi = xi for all
i < k, and xi+i = xi for all i > k. xk is the new value added to the partition. In the lower and upper sum, all the bars except for the kth are unchanged. The kth bar has been replaced by two bars in the refinement. Clearly,
Mx x < MXo x
X0’...,Xn + 1 — x0?...?xn
Lx0,...,xn+1 — LXo,...,x„ .
The lower and upper sums for a partition and its refinement are shown in Figure A.7. We see that refining a partition must make tighter bounds on the area under the curve.
Next we will show that for any continuous function defined on a closed interval [a, b],we can find a partition x0,...,xn for some n that will make the difference between the upper sum and the lower sum as close to zero as we wish. Suppose б > 0 is the number we want the difference to be less than. We draw lines S =
apart parallel to the horizontal (x) axis. (Since the function is defined on the closed interval, its maximum and minimum are both finite.) Thus a finite number of the horizontal lines will intercept the curve y = f (x) over the interval [a, b]. Where one of the lines intercepts the curve, draw a vertical line down to the horizontal axis. The x values where these vertical lines hit the horizontal axis are the points for our partition. For example, the function f (x) = 1 + л/4 — x2 is defined on the interval [0, 2]. The difference between the upper sum and the lower sum for the partition for that б is given by
Mx0,...,xn LX0,...,xn S X [(x1 x0) + (x2 x1) + ... + (xn xn-1)]
= S X [b — a]
INTRODUCTION TO CALCULUS 287
Figure A.8 The partition induced for the function f (x) = 1 + \/4 — x2 where e1 = 1 and its refinement where e2 = 2.
We can make this difference as small as we want to by choosing e > 0 small enough.
Let ek = k for k = 1,..., ro. This gives us a sequence of partitions such that limk—TO ek = 0. Hence
lim Mx0 x — Lxn x =0.
The partitions for ei and e2 are shown in Figure A.8. Note that Sk = 2k.
That means that the area under the curve is the least upper bound for the lower sum, and the greatest lower bound for the upper sum. We call it the definite integral and denote it
/ f(x)dx .
Note the variable x in the formula above is a dummy variable:
p b p b
/ f (x)dx = f (y)dy.
Basic Properties of Definite Integrals
Theorem 7 Let f (x) and g(x) be functions defined on the interval [a, 6], and let c be a constant. Then the following properties hold.
1. The definite integral of a constant times a function is the constant times the definite integral of the function:
p b p b
/ cf(x)dx = c / f(x)dx.
2. The definite integral of a sum of two functions is a sum of the definite integrals of the two functions:
p b p b p b
/ (f (x) + g(x))dx = f (x)dx + / g(x)dx .
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288 INTRODUCTION TO CALCULUS
Fundamental Theorem of Calculus
The methods of finding extreme values by differentiation and finding area under a curve by integration were known before the time of Newton and Liebniz. Newton and Liebniz independently discovered the fundamental theorem of calculus that connects differentiation and integration. Because each was unaware of the others work, they are both credited with the discovery of the calculus.
Theorem 8 Fundamental theorem of calculus. Let f (x) be a continuous function defined on a closed interval. Then:
1. The function has antiderivative in the interval.
2. If a and b are two numbers in the closed interval such that a < b, and F (x) is any antiderivative function of f (x), then
f f (x)dx = F(b) — F(a).
This function shows the area under the curve y = f (x) between a and x. Note that the area under the curve is additive over an extended region from a to x + h:
In other words, I(x) is an antiderivative of f (x). Suppose F(x) is any other antiderivative of f (x). Then
For x e (a, b), define the function
f (x)dx = f (x)dx + f (x)dx .
By definition, the derivative of the function I (x) is
for all values x' e [x, x + h). Thus
F (x) = I (x) + c
INTRODUCTION TO CALCULUS 289
Figure A.9 The function f (x) = x 1/2.
for some constant c. Thus F (b) — F (a) = I (b) — I (a) = /ab f (x)dx, and the theorem is proved.
For example, suppose f (x) = e-2x for x > 0. Then F(x) = — 1 x e-2x is an antiderivative of f (x). The area under the curve between 1 and 4 is given by
r4 1 1
J f (x)dx = F(4) — F(1) = — ^ x e-2x4 + - x e-2x1.
Definite Integral of a Function f (x) Defined on an Open Interval
Let f (x) be a function defined on the open interval (a, b). In this case, the antiderivative F(x) is not defined at a and b. We define