THE FUNDAMental THEOREM OF CALCULUs AN INTRODUCTION TO THE FUNDAMENTAL THEOREM In the previous section, we went through all four problems in the discovery project, and you’ll agree that each problem yielded unprecedented but hugely significant results. In this section, we’ll discuss each problem and the implications of each result obtained. These results form the basis of one of the most important theorems in calculus: the Fundamental Theorem of Calculus. As we all know, calculus is categorized into two branches: Differential Calculus and Integral Calculus. Integral calculus arose from the area problem, while differential calculus arose from an apparently unrelated problem, the tangent problem. The fundamental theorem of calculus not only establishes a precise connection between these two branches of calculus, it also goes on to show that these two branches are actually inverse processes; that is, integration is the inverse of differentiation, and vice versa. The discovery of this relationship was made by Isaac Barrow (1630 – 1677), who also happened to be Sir Isaac Newton’s teacher in Cambridge. Years later, it was Leibniz and Newton who actually exploited this relationship and developed calculus into the systematic mathematical method it has become today. More importantly, this theorem enabled these mathematicians to compute integrals easily without having to compute limits of sums. In the late 16th century, there were intense disputes between Newton’s followers and those of Leibniz as to who invented calculus first. Newton arrived at his version of calculus first, but for fears of disparagement and controversy, he did not publish it immediately. It was in 1684 that Leibniz’s version of calculus was first published. The truth of the matter is that both men invented calculus autonomously. You can read an article about this controversy at wikipedia.org. You can also read more about Sir Isaac Newton and Gottfried Wilhelm Leibniz at wikipedia.org. The Fundamental Theorem is divided into two parts. So, for the rest of this section, we’ll discuss the results of the problems we dealt with in the discovery project, from which we’ll derive the two parts of the fundamental theorem. First, let’s analyze problem 1. We found the area under the line y = 2t + 1 using geometry and from the geometrical method used, an expression was derived for the area (in terms of the limit), which was:
A(x) = t2 + t – 2 So far, we have simply performed the process of integration. Now here’s where the whole process takes an interesting twist: We differentiate the area function above and we actually got back the original equation y = 2t + 1. The problem simply teaches us that integration and differentiation are inverse processes: If we obtain the integral of a function on a given interval and then differentiate the result, we end up with the original equation. Thus, it’s basically an issue of one process reversing the other, and vice versa. A similar situation occurs in problem 2: We are given an integral from which we derived an explicit expression for the area:
A(x)
=
4/3
+
x
+
After that, we differentiate A(x) and obtain
A’(x)
=
1 + x2
Now, let’s go back to the integral in question:
A(x)
=
∫
x 1
(1 + t2) dt
(x3/3)
Note that the area function A depends only on x (which happens to be the upper limit of the integral. If x is a fixed number, then the integral A(x) is a definite area. But, if x varies, however, then the integral also varies. Let’s look at the integral once again. We find that the integrand
y = 1 + t2 is positive under the region from t = 1 and t = x (that is, f(t) ≥ 0). Since f(x) ≥ 0, then the integral
A(x)
∫
=
x
(1 + t2) dt
1
can be interpreted as the area under the graph of y = 1 + t 2 (where x can be any number between a and b). In order to compute A’(x) from the definition of a derivative, we first realize that, for a small, positive number h, then
A(x + h) – A(x) Is obtained by subtracting areas. This area is the small orange rectangle drawn in the graph of problem 2 in the discovery project. For small h, the area is A(x + h) – A(x) approximately equal to the orange rectangle with height f(x) and breadth h. So,
A(x + h) – A(x)
=
hf(x)
Therefore,
A' (x)
A(x + h ) – A(x)
=
≈
h
f(x)
This would mean that as h → 0, A’(x) → f(x). Alternatively, we are saying that
A' (x)
=
lim
A(x + h ) – A(x)
h →0
≈
h
f(x)
Equation 1
This is the technique we used to solve problem 3 in the Discovery Project. We found that, if
cos (x2)
f(x) = and
g(x)
∫
=
x 0
(cos t2) dt
Then, like equation 1, we have
g' (x)
=
lim
g(x + h ) –g(x)
h →0
≈
h
f(x)
This leads us to Part 1 of the Fundamental Theorem:
If f is a continous function on the interval [a, b], then a new function g defined by
g(x) =
∫
x a
f( t) dt
is also continuous on [a, b], and is such that
g(x) =
f(x)
In other words, this theorem simply says that the derivative of a definite integral with respect to its upper limit x is the integrand evaluated at the upper limit. For easy reference, this theorem is abbreviated FTC 1. We have derived part 1 the fundamental theorem of calculus from the discovery project, now let’s actually prove this theorem. If the values x and (x + h) exist in the interval [a, b], then
A(x+h) – A(x)
∫
=
x+h a
∫ xa
–
f(t) dt
f(t) dt
Since A(x + h) is the sum of two adjacent areas, that is,
A(x+h)
∫ xa
=
f(t) dt
∫
+
x+h a
f(t) dt
Then, following property 5, we get
A(x+h) – A(x)
∫
=
x
+
f(t) dt
a
∫
x+h a
f(t) dt
–
∫
x a
f(t) dt
which results in
A(x+h) – A(x)
∫
=
x+h a
Equation 2
f(t) dt
Thus, if h > 0 and we want to express equation 2 like equation 1, then we have
A(x + h ) – A(x)
=
h
∫
1/h
x+h a
Equation 3
f(t) dt
We can also prove FTC 1 using the Extreme Value Theorem. Recall this theorem says that
For a continuous function on [a, b], there are two numbers u and v in [a, b] such that f (u) = m and f (v) = M, where m and M are the absolute minimum and absolute values of f on [a, b].
In this case, let’s assume we’re dealing with a function which is continuous on the interval [ x, x+h]. So, by property 10 of integrals, we have
∫
≤
mh
x+h a
≤
f(t) dt
Mh
Since f (u) = m and, f (v) = M, then we have
f(u)h
∫
≤
x+h a
f(t) dt
≤
f(v)h
Since h > 0, dividing the inequality will give
f(u)
≤
1/h
∫
x+h a
f(t) dt
≤
f(v)
Now, we use equation 3 to replace the middle expression of this inequality:
f(u)
≤
A(x + h ) – A(x) h
≤
f(v)
Since u and v lie between x and x+h, if we let h → 0, then u → x and v → x. Thus,
=
lim f(u)
h→0
lim f(u) u→x
=
f(x)
=
f(x)
AND
=
lim f(v)
h→0
lim f(v) v→x
Because f is continuous, then the two sided limits above are equal.
f(u)
=
lim f(u) u→x
A(x + h ) – A(x)
≤
AND
f(x)
≤
h
lim f(v) v→x
=
f(v)
f(x)
Using the squeeze theorem, we have
lim f(u) h →0
lim
≤
A(x + h ) – A(x)
h →0
≤
h
lim f(v)
h →0
which means
g' (x)
g(x + h ) –g(x)
lim
=
h →0
h
≈
f(x)
Using Leibniz notation for derivatives, we can write FTC 1 as
d/dx
∫
x+h a
f(t) dt
=
f(x)
The second part of the Fundamental Theorem which we abbreviate as FTC 2, can be derived from FTC 1 and a corollary from the mean value theorem. This theorem is stated like this:
If f’(x) = g’(x) for all x and v in an (a, b), then f – g is a constant on (a, b); that is, f(x) = g(x) + c Where c is a constant PROOF: If
F(x) = f(x) – g(x), then F’(x) = f’(x) – g’(x) = 0
Now, let
g(x)
=
∫ xa
f(t) dt
Equation 4
We know from FTC 1 that g’(x) = f (x); that is, g is an antiderivative of f. If F is any other antiderivative of f on [a, b], then we know from the theorem above that the antiderivatives differ by a constant c:
F(x) = g(x) + c
(for a ≤ x ≤ b)
Equation 5
Note that F and g are continuous functions. By taking the limits of both sides as x→ a+, we have (from equation 5):
lim
lim F(x)
=
x→a+
lim F(x)
=
x→a+
x→a+
x→a+
F(a)
=
[g(x)
+ c]
lim g(x) +
g(a)
+
lim
x→a+
c
C
If we also take the limits of both sides as x→b, we have
lim
lim F(x)
=
x→b
lim F(x)
=
x→b
x→b
x→b
F(b)
=
[g(x)
+ c]
lim g(x) +
g(b)
+
lim
x→b
c
C
We also find that these limits exist, even when x = a and x = b. If we put x = a in equation 4, we have
∫
g(a) =
a a
=
f( t) dt
0
Similarly, if we put x = b in equation 4, we have
g(b)
∫
=
b a
f( t) dt
So, using equation 5 along with x = a and x = b, we have
F(b) – F(a)
=
[ g(b) + c ] – [ g(a) + c ]
F(b) – F(a)
=
g(b) + c – g(a) + c
F(b) – F(a)
=
g(b) – g(a) + c – c
F(b) – F(a)
=
g(b) –
F(b) – F(a)
=
∫
b a
g(a)
f( t) dt
This leads us directly to part 2 of the Fundamental Theorem:
If f is continuous on [a, b] then
∫
b a
f( x) dx
=
F(b) – F(a)
Where F is ANY antiderivative of f; that is, a function such that
F’ = f
All this while, we had believed that the definite integral could only be evaluated by a complicated procedure (the Riemann Method) involving all the values of f(x) on the interval [a, b]. But now, with the Fundamental theorem, we see that a definite integral can be computed easily. All we need to know are the two endpoints of the interval [a, b]: a and b. Note that, in applying part 2 of the fundamental theorem, we use a particular antiderivative of f, not necessarily the most general antiderivative. Note that
F(b) – F(a) can also be expressed as
F(x) a
F(x)] a
[F(x)]
b
b
b a
Thus, alternative notations for FTC 2 are
∫ ∫ ∫
]
b a
f( x) dx
=
F(x)
b a
f( x) dx
=
F(x)
b a
f( x) dx
= [F(x)] ab

b a b a
So now, let’s bring both parts of the Fundamental Theorem together:
Suppose f is continuous on [a, b]. Then 1.
g(x)
If
∫
2.
b a
∫
=
f(x) dx
=
x
f( t) dt
a
F(b) – F(a)
then g' (x) = f(x) where F' = f
Verbally, these two parts can be interpreted like so: If f is integrated and the result is differentiated, we arrive back at the original function f. If we know an antiderivative F of f, then
∫
b a
f(x) dx
can be evaluated by subtracting the values of F at endpoints a and b. These two equations
d/dx AND
∫
b a
∫
x a
f(t) dt
f(x) dx
=
=
f(x)
F(b) – F(a)
Taken together make up what we call the Fundamental Theorem of Calculus and the overall message it sends is that differentiation and integration are inverse processes; one undoes the other, and vice versa In the days of Eudoxus, Archimedes, Fermat and Galileo, area and volume problems, lengths of curves e.t.c were very challenging problems. Well, not anymore!!!
With the use of the systematic method of calculus, we’ll find that these problems have become child's play! Let’s study these two examples which illustrate the Parts 1 and 2 of the Fundamental Theorem.
Example 1 Sketch the area represented by g(x). Then find g’(x) in two ways: a)
By Using FTC 1
b)
Evaluate the integral using FTC 2 and then differentiate:
g(x)
=
∫
x 1
(t2) dt
SOLUTION (a). Before we begin to solve this problem, we have to first ensure that the function in question is continuous on the given interval [1, x]. In this case the integrand y = t2 is continuous on any interval [a, b]. Its graph clearly shows that it is continuous; the area g(x) is also shown:
Since the function is continuous, part 1 of the Fundamental Theorem gives
g’(x) = x2
(b). To evaluate the integral using Part 2 of the Fundamental Theorem, we need an antiderivative of the integrand. So, if
f (t) = t2 Then, using the general antiderivative formula we have
F (t) = t3/3
For this integral, the endpoints are a = 1 and b = x. Therefore, Part 2 of the FTC gives
∫ ∫
b a
g(x)
=
g(x)
=
g(x)
=
(x3/3) – (13/3)
g(x)
=
(x3/3) – (1/3)
g(x)
=
(x3/3) – (1/3)
x 1
f(x) dx (t2) dt
=
F(b) – F(a)
=
F(x) – F(1)
This gives
Therefore,
If we differentiate g, we get
g’(x)
=
d/dx (x3/3)
g’(x)
=
x2 – 0
g’(x)
=
x2
–
d/dx (1/3)
See how both parts of the Fundamental Theorem work? In the end, we arrived at the same result.
Example 2 Sketch the area represented by g(x). Then find g’(x) in two ways: a)
By Using FTC 1
b)
Evaluate the integral using FTC 2 and then differentiate:
∫
x π
(2 + cos t) dt
SOLUTION
(a).
Since the function
f(t) = 2 + cos t is continuous, part 1 of the FTC gives the derivative of g as
g’(x) = 2 + cos x The area represented by the area function g is shown above:
(b).
For the integral g, the integrand is
f (t) = 2 + cos t Therefore, the antiderivative F of f is
F (t) = 2t + sin t Here, we have a = π and b = x. So, Part 2 of the Fundamental Theorem gives
g(x)
=
g(x)
=
(Remember that is
π
∫ ∫
x π
b a
f(x) dx
=
(2 + cos t) dt
F(b) – F(a) =
F( 2x + sin x) – F( 2π + sin π)
in radians)
g(x)
=
2x + sin x – 2π – sin π
g(x)
=
2x + sin x – 2π – 0
g(x)
=
2x + sin x – 2π
g’(x)
=
d/dx (2x) + d/dx (sin x) – d/dx (2π)
g’(x)
=
2 + cos x – 0
g’(x)
=
2 + cos x
Differentiating g gives
Examples 1 and 2 have shown that definite integrals can be evaluated using both parts of the Fundamental Theorem. Note how each part works differently. In the following section, we treat more examples.
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