Proof of Fundamental Theorem of Calculus I and II

Proof of Fundamental Theorem of Calculus I and II

Proof of Fundamental Theorem of Calculus I and II

Previously we have studied the material on Fundamental Theorem of Calculus on Integrals along with several example problems. The Fundamental Theorem of Calculus (abbreviated as FTC) is a special topic in integrals that makes it easier for us to calculate definite integral forms. In this article we will discuss the Proof of Fundamental Theorem of Calculus I and II.

As logical learners, you certainly don't just believe a statement. The statement must be proven first before its truth can be trusted. Now, let's prove the truth of the Fundamental Theorem of Calculus I and II.

Concepts Needed for the Proof:

Several concepts we will need to prove FTC I and FTC II are:

Additivity of Definite Integrals Definition of Function Derivative Squeeze Theorem
1. Additivity of Definite Integrals:

If \( f \) is an integrable function on an interval containing \( a \), \( b \), and \( c \), then:

\[ \int_a^c f(x)dx = \int_a^b f(x)dx + \int_b^c f(x)dx \]
2. Definition of Function Derivative:

The first derivative of function \( F(x) \) can be written as:

\[ \lim_{h \to 0} \frac{F(x+h)-F(x)}{h} = F'(x) = \frac{d}{dx} F(x) \]
3. Squeeze Theorem:

If there is a form \( k \leq f(x) \leq k \), then \( f(x) = k \).

Proof of Fundamental Theorem of Calculus I

Fundamental Theorem of Calculus I (FTC I)

If \( f \) is continuous on \( (a, b) \) and \( x \) is any point in \( (a, b) \), then:

\[ \frac{d}{dx} \int_a^x f(t)dt = f(x) \]
Proof of FTC I:
1

Define \( F(x) = \int_a^x f(t)dt \). Then:

\[ F(x+h) = \int_a^{x+h} f(t)dt \]

Using the additivity property of integrals:

\[ F(x+h) = \int_a^x f(t)dt + \int_x^{x+h} f(t)dt \quad \text{(Equation 1)} \]

where \( h \) is a positive real number.

2

Subtract \( F(x) \) from \( F(x+h) \):

\[ F(x+h)-F(x) = \left( \int_a^x f(t)dt + \int_x^{x+h} f(t)dt \right) - \left( \int_a^x f(t)dt \right) \]
\[ F(x+h)-F(x) = \int_x^{x+h} f(t)dt \]
3

Define:

\[ m = \text{minimum value of } f(x) \text{ for } x \text{ in } [a, b] \]
\[ M = \text{maximum value of } f(x) \text{ for } x \text{ in } [a, b] \]
Visual Representation:

Consider three areas under the curve \( y = f(t) \):

Area A: Rectangle with height \( m \) and width \( h \) → Area = \( m \times h \)

Area B: Area under curve from \( x \) to \( x+h \) → Area = \( \int_x^{x+h} f(t)dt = F(x+h) - F(x) \)

Area C: Rectangle with height \( M \) and width \( h \) → Area = \( M \times h \)

4

Establish the relationship between these three areas:

\[ \text{Area A} \leq \text{Area B} \leq \text{Area C} \]
\[ m \times h \leq F(x+h) - F(x) \leq M \times h \]

Divide by \( h \):

\[ \frac{m \times h}{h} \leq \frac{F(x+h) - F(x)}{h} \leq \frac{M \times h}{h} \]
\[ m \leq \frac{F(x+h) - F(x)}{h} \leq M \]
5

Apply the limit as \( h \to 0 \):

\[ \lim_{h \to 0} m \leq \lim_{h \to 0} \frac{F(x+h) - F(x)}{h} \leq \lim_{h \to 0} M \]

As \( h \to 0 \), both \( m \) and \( M \) approach \( f(x) \):

\[ \lim_{h \to 0} m = \lim_{h \to 0} M = f(x) \]
6

Thus we have:

\[ f(x) \leq \lim_{h \to 0} \frac{F(x+h) - F(x)}{h} \leq f(x) \]

By the Squeeze Theorem:

\[ \lim_{h \to 0} \frac{F(x+h) - F(x)}{h} = f(x) \]
7

From the definition of derivative:

\[ \lim_{h \to 0} \frac{F(x+h) - F(x)}{h} = F'(x) = \frac{d}{dx} F(x) \]

And since \( F(x) = \int_a^x f(t)dt \):

\[ \frac{d}{dx} \int_a^x f(t)dt = f(x) \]
Therefore, it is proven that:
\[ \frac{d}{dx} \int_a^x f(t)dt = f(x) \]

Proof of Fundamental Theorem of Calculus II

Fundamental Theorem of Calculus II (FTC II)

If \( f \) is continuous on \( [a, b] \) and \( F \) is an antiderivative of \( f \) on \( [a, b] \), then:

\[ \int_a^b f(x) dx = F(b) - F(a) \]
Proof of FTC II:
1

We will use FTC I to prove FTC II. From FTC I:

\[ f(x) = \frac{d}{dx} \int_a^x f(t) dt \]
2

Let \( g(x) \) be an antiderivative of \( f \), defined as:

\[ g(x) = \int_a^x f(t) dt \]

Let \( F(x) \) be another antiderivative of \( f \). Since antiderivatives differ only by a constant:

\[ F(x) = g(x) + C \]

where \( C \) is a constant.

3

Evaluate \( F(a) \) and \( F(b) \):

\[ F(a) = g(a) + C \quad \text{and} \quad F(b) = g(b) + C \]

Note that:

\[ g(a) = \int_a^a f(t) dt = 0 \]
\[ g(b) = \int_a^b f(t) dt \]
4

Subtract \( F(a) \) from \( F(b) \):

\[ F(b) - F(a) = (g(b) + C) - (g(a) + C) \]
\[ = g(b) + C - g(a) - C \]
\[ = g(b) - g(a) \]
\[ = \int_a^b f(t) dt - 0 \]
\[ = \int_a^b f(t) dt \]
Therefore, it is proven that:
\[ \int_a^b f(x) dx = F(b) - F(a) \]
Educational Note:

From the proof of both theorems above, it can be seen that more analysis is needed to understand the proof of the Fundamental Theorem of Calculus I and II. For high school level, our most important suggestion is focusing on their usage in solving problems.

For the proofs, it's better not to delve too deeply into them, except for very important needs, such as for teachers or at the university level. Actually, the Fundamental Theorem of Calculus I and II were previously studied only at the university level, but in the 2013 curriculum, the theory has been introduced, which of course will be difficult for students to understand in terms of proof.

Key Insights from the Proofs:

1. FTC I shows that differentiation and integration are inverse operations.

2. FTC II provides a practical method for evaluating definite integrals without using Riemann sums.

3. The proofs elegantly connect several fundamental concepts: limits, derivatives, integrals, and the Squeeze Theorem.

4. Understanding these proofs deepens appreciation for the unity of calculus.