Calculus Of One Real Variable – By Pheng Kim Ving Chapter 9: The Integral – Section 9.4: The Fundamental Theorem Of Calculus 9.4 The Fundamental Theorem Of Calculus

 1. Integrals And Antiderivatives

In Section 9.3 Example 3.1 we calculated definite integrals directly from the definition by using summation formulas. This
is often hard and time-consuming, and isn't always possible for many other functions. In this section we're going to
present a way to calculate definite integrals of a function
f without directly using their definition. This way utilizes an
antiderivative of
f. In the process we're also going to discuss the relationship between derivatives and integrals.

Integral Functions

{1.1} See Section 9.3 Remarks 3.1 iii.

 Fig. 1.1   Definite Integral.

 Fig. 1.2   Integral Function

 Fig. 1.3   Integral Function.

 Fig. 1.4   Integral Function.

All Integral Functions Are Antiderivatives

It follows that the difference between any 2 integral functions of a function is a constant.

This reminds us of the property that the difference between any 2 antiderivatives of a function is a constant; see
Section 5.7 Remarks 2.1 v. Possibly an integral function of a function
f is an antiderivative of f. In Section 7.1 Parts 2 And 3 we
saw that the derivative of the area function under 1/
x over x > 0 defined to be ln x is 1/x itself, or in integral

 Fig. 1.5   Difference between any 2 integral functions is a constant.

terms the derivative of an integral function of 1/x is 1/x, ie an integral function of 1/x is an antiderivative of 1/x. Wow
probably an integral function of a function
f is an antiderivative of f.

Any integral function of the function (1/3)x is an antiderivative of (1/3)x. Wow very probably any integral function of a
function
f is an antiderivative of f. OK. It turns out that certainly any integral function of a function f is an antiderivative of

# Any integral function of function (1/3)x is an antiderivative of (1/3)x.

f. This fact is stated and proved in Theorem 2.1 below.

{1.2}

# Area A = 21 square units.

Equations Of Integral Functions

Its regular (non-integral) equation is the equation of an antiderivative. Its graph has x-intercept a. As integral functions

# Integral Functions Of f(x) = x2 On [1, 4].

are antiderivatives, the vertical distance between the graphs of any 2 of them is a constant; see Section 5.7 Graphs Of
Antiderivatives
.

On Other Closed Bounded Intervals And On The Entire Real Line

# Integral Functions And Antiderivatives Of f(x) = x2 on [1, 4].

Calculating Definite Integrals By Using Any Antiderivative

Or more formally for the general case:

 Fig. 1.11

This formula establishes the calculating of definite integrals by using any antiderivative. It'll be formally proved in
Theorem 2.1 below. Note that it means that the definite integral over [
a, b] is equal to the common change of all the
antiderivatives over [
a, b].

 Fig. 12

 Fig. 1.13

 Fig. 1.14

 2. The Fundamental Theorem Of Calculus

As seen above any integral function of a function f is an antiderivative of f and the definite integral of f over [a, b] is the
common change of all antiderivatives of
f over [a, b]. As any integral function of f is an antiderivative of f the derivative
of any integral function of
f is f itself. See Fig. 2.1.

 Fig. 2.1   Derivative of integral of f is f itself:

#### Theorem 2.1 – The Fundamental Theorem Of Calculus

 Suppose that the function f is continuous on a closed interval I and let a and b be any 2 points in I.   1.  Define the function G on I by:

Proof
1.  As
f(x) is continuous on any closed sub-interval of I, it's integrable there (See Section 9.3. Theorem 5.1.). So:

Suppose h > 0. By continuity, f attains a maximum and a minimum on [x, x + h]. Let M and m be in [x, x + h] such that

f( M ) is that maximum and f(m) that minimum. Thus:

where the last equality is obtained by the fact that f is continuous at x. For h < 0, a similar argument leads to the
same conclusion.

2.  By part 1 the function G defined on I by:

EOP

# Fig. 2.3

Abbreviations

For the remaining of this section we use the abbreviations “FTC”, “FTC1”, and “FTC2” for “Fundamental Theorem Of
Calculus”, “Fundamental Theorem Of Calculus Part 1”, and “Fundamental Theorem Of Calculus Part 2” respectively.

## Remarks 2.1 – On The FTC (Theorem 2.1)

1. For the function f in Fig. 2.1, from the above proof we have:

Thus for infinitesimally small h we can think of the light-blue strip as becoming the rectangle of base h and height f(x).
As a consequence we can think of the infinitesimally short piece of the graph of
f over an interval of infinitesimally small
width
h as a horizontal straight line forming the top side of the rectangle. Note that any rectangle of height f(x) and any
base
h, not just infinitesimally small h or small h, is also such that f(x) = (area of rectangle)/(its base h).

2.  The hypothesis of the FTC doesn't require that f be non-negative-valued on its domain. Clearly the proof doesn't need
that condition. The theorem applies to any continuous function.

3. The proof of part 1 doesn’t require that a be some particular point in I. So the point a is arbitrary in I, and thus the
derivative of any integral function of
f on I is f itself. Intuitively, in Fig. 2.1, the value f(x) of f at x is of course the same
wherever in
I a is located. The derivative of an integral function of f on I is f itself wherever in I a is located.

The FTC1 asserts that any integral function of f is an antiderivative of f, or, in other words, that f is the derivative of any
of its integral functions.

4. The FTC2 provides us a with a technique to calculate the exact  values of definite integrals of functions whose
antiderivatives are known or can be found.

5.  The FTC establishes the relationship between the derivative and the integral (part 1: derivative of integral function of f is

f, or integral function is antiderivative; part 2: definite integral equals common change of all antiderivatives). Hence the

#### Evaluation

We have for example:

When Antiderivative Is A Constant Multiple Of A Function

Suppose the antiderivative in the evaluation symbol is a constant multiple of a function, ie of the form kF(x), where k is a constant. Then:

The proof is a piece of cake:

the same answer as obtained above.

## Antidifferentiability Of Continuous Functions

A function is antidifferentiable if it has an antiderivative. Part i of the fundamental theorem clearly asserts the existence of
an antiderivative of a continuous function: since
F '(x) = f(x), F(x) is an antiderivative of f(x). We state this existence as
a corollary.

## Corollary 2.1

 If a function is continuous on a closed interval then it's antidifferentiable there.

Section 9.3 Theorem 5.1 states that every function continuous on a closed interval is integrable there. We can combine
that theorem with the fundamental theorem in that order to produce the following chain of properties:

So if a function is differentiable then it's antidifferentiable (if differentiable then continuous then antidifferentiable). Recall
that not every continuous function is differentiable.

### Example 2.1

Use the FTC1 then the FTC2 to differentiate each of the following functions:

Evaluate:

### Example 2.3

Prove this limit:

##### Solution

Then:

EOS

The expression of the limit looks like a Riemann sum of some function. So we do some algebraic manipulation to render
this look more obvious. We recognize that it indeed is a Riemann sum of the function
f(x) = 1/(1 + x) on [0, 1] with the
regular partition {
x0 = 0, x1, x2, ..., xn = 1} of order n of the interval [0, 1]. The right endpoints x1, x2, ..., xn are utilized
in our Riemann sum. Now,
f is continuous on [0, 1]. Thus f is integrable there; see Section 9.3 Theorem 5.1. The given
limit is the limit at infinity of a Riemann sum of
f, and hence, by definition, it's the definite integral of f over [0, 1].

We have f '(x) = – 1/(1 + x)2 < 0 for all x in [0, 1]. So f is decreasing there. Consequently, on each sub-interval [xi–1, xi],
f(xi) is the minimum of f on that sub-interval. Thus, (1/n)( f(x1) + f(x2) + f(x3) + ... + f(xn)) is actually a lower Riemann
sum. But we don't have to worry about it, because
f is continuous. It's enough to refer simply to the general Riemann
sum.

 3. Indefinite Integrals

The general antiderivative represents all antiderivatives. As seen above under the heading Are All Antiderivatives Integral
Functions?
, some antiderivatives are integral functions on a closed bounded interval; antiderivatives that aren't integral
functions on an interval may be integral functions on other intervals; and all antiderivatives on the entire real line are
integral functions there. As seen above in Theorem 2.1, any antiderivative, whether or not it's an integral function, can be
utilized to calculate definite integrals. So if the general antiderivative of a function is known or found, definite integrals of
that function can readily be computed. For these reasons the general antiderivative is considered as an integral function.
It's an integral function. It's a function and thus not a definite integral (which is a number if both upper and lower limits of
integration are numbers or constants), as a consequence in integral vocabulary it's called the indefinite integral.

Definition 3.1 – The Indefinite Integral

 The general antiderivative of a function f(x) is also an integral function of f(x) and is called the indefinite integral of f(x), and thus is denoted as:     which is read “integral of f(x) dx” or “sum of f(x) dx”. Note that there are no limits of integration in this notation.

We see that if F(x) is an antiderivative of f(x) then:

where C is an arbitrary constant.

Note that an indefinite integral is a function while a definite integral is a number.

## Relationship Between The Definite And Indefinite Integrals

If F(x) is an antiderivative of f(x) then:

The relationship between the definite and indefinite integrals is:

The definite integral from a to b is the change of the indefinite integral over [a, b]. This is easy to remember if we keep in
mind that the function in the evaluation symbol is any antiderivative which can be the general antiderivative and the indefinite
integral is the general antiderivative. When we employ the indefinite integral to compute a definite integral, we usually omit the
constant of integration
C for the same reason as discussed above. For example:

There are situations where it's simpler to find the general antiderivative in the form of indefinite integral first and then use it in
the calculation of the definite integral.

## Distinction Between The Integrals

Distinction Between The Integrals

 4. Integration

Definition 4.1 – Integration

 Integration is the process of finding a definite integral, sometimes also called definite integration to be more specific, or the process of finding the indefinite integral, sometimes also called indefinite integration to be more specific. Whether integration is definite or indefinite will be clear from the context.

Example 2.2 above is an example of definite integration.

## Integration Is An Inverse Operation Of Differentiation

Definite integration involves the finding of an antiderivative. Indefinite integration is the finding of the general
antiderivative. So integration in general is ant differentiation. Since ant differentiation is the inverse operation of
differentiation, integration is an inverse operation of differentiation.

Why does integration involve ant differentiation? Differentiation involves division and integration involves multiplication, as
the derivative is the limit of a quotient and the integral is the limit of a product, as seen in Section 9.3 Part 7. Now
multiplication is the inverse operation of division (for example (
x x 2)/2 = x and (x/2) x 2 = x for any number x). Thus
integration involves the inverse operation of differentiation, which is ant differentiation.

 Problems & Solutions

1. Calculate the following derivatives.

Solution

2. Compute the following definite integrals.

Solution

Solution

4. Consider the function y = 1/x2, whose graph is shown in the figure below.

If we calculate the area A of the shaded region using the fundamental theorem then we get:

Clearly A cannot be –2 or any other negative value. It even appears to be substantially greater than 2. What's wrong?

Solution

The function y = 1/x2 is discontinuous at x = 0. So it's discontinuous on any interval containing x = –1 and x = 1. Thus
the fundamental theorem doesn't apply to it.

5. Prove this limit:

Solution

Let:

So: