Challenge – Proving the Fundamental Theorem (HSC SSCE Mathematics Advanced): Revision Notes

Challenge – Proving the Fundamental Theorem

Introduction

This section develops a rigorous proof of the fundamental theorem of calculus. The fundamental theorem is one of the most important results in mathematics, as it connects differentiation and integration. This material is challenging and requires careful thought, so take your time working through it.

The definite integral as a function of its upper limit

When we write a definite integral $\int_a^b f(x) \, dx$, the value changes as we change the upper limit $b$. This means the definite integral can be viewed as a function of its upper limit.

To emphasize this functional relationship, we make two important changes to our notation:

• Replace the upper limit $b$ with $x$ (the conventional variable for functions)
• Replace the integration variable $x$ with $t$ (to avoid using $x$ twice)

This gives us the integral $\int_a^x f(t) \, dt$, which clearly shows the dependence on the upper limit $x$.

The signed area function

The signed area function for $f(x)$ starting at $x = a$ is defined as:

$A(x) = \int_a^x f(t) \, dt$

The diagram above shows how the area under a curve (top graph) corresponds to the signed area function (bottom graph). As we move right from $a$ to any point $x$, we accumulate the shaded area, and this accumulated value is $A(x)$.

Understanding signed areas

The term "signed area" is crucial because areas below the $x$-axis contribute negatively to the integral, while areas above contribute positively.

Consider the parabola shown above with axis of symmetry at $x = b$. The four regions marked $B$ all have equal area. Let's analyze the signed area function $A(x) = \int_a^x f(t) \, dt$:

In the interval $[a, b]$:

• The function $f(t)$ is negative and decreasing
• As we integrate from $a$ towards $b$, we accumulate negative area
• Therefore $A(x)$ decreases at an increasing rate

In the interval $[b, c]$:

• The function $f(t)$ is still negative but now increasing
• We continue to accumulate negative area, but more slowly
• Therefore $A(x)$ decreases at a decreasing rate
• At $x = c$, we reach the minimum value $A(c) = -2B$

In the interval $[c, d]$:

• The function $f(t)$ becomes positive
• We start accumulating positive area, which increases $A(x)$
• The signed area function increases at an increasing rate
• At $x = d$, we have $A(d) = -B$ (we've regained one area unit $B$)

Worked example: constructing a signed area function

Let's work through a concrete example to see how signed areas work in practice.

The fundamental theorem — differential form

We can now state and prove the first form of the fundamental theorem.

Proof of the differential form

The proof begins with the definition of the derivative as a limit:

$A'(x) = \lim_{h \to 0} \frac{A(x+h) - A(x)}{h}$

Using the diagram, we can express the difference in areas:

$A(x+h) - A(x) = \int_x^{x+h} f(t) \, dt$

Therefore:

$A'(x) = \lim_{h \to 0} \frac{1}{h} \int_x^{x+h} f(t) \, dt$

The key step is to use a sandwiching technique. Assume $f(t)$ is increasing in the interval $[x, x+h]$ (as shown in the diagram).

Setting up the sandwich:

• The lower rectangle on $[x, x+h]$ has height $f(x)$ and width $h$
• The upper rectangle has height $f(x+h)$ and width $h$
• The integral $\int_x^{x+h} f(t) \, dt$ represents the area under the curve

Therefore, the area under the curve is sandwiched between the two rectangles:

$h \times f(x) \leq \int_x^{x+h} f(t) \, dt \leq h \times f(x+h)$

Dividing by $h$ (assuming $h > 0$):

$f(x) \leq \frac{1}{h} \int_x^{x+h} f(t) \, dt \leq f(x+h)$

Taking the limit:

Because $f(x)$ is continuous, we know that $f(x+h) \to f(x)$ as $h \to 0$.

The middle expression is sandwiched between $f(x)$ and $f(x+h)$, so:

$\lim_{h \to 0} \frac{1}{h} \int_x^{x+h} f(t) \, dt = f(x)$

This means $A'(x) = f(x)$, as required.

Using the differential form

The differential form is particularly useful when we want to differentiate an integral without actually evaluating it first.

The fundamental theorem — integral form

The integral form is the version of the fundamental theorem you've been using to evaluate definite integrals throughout this unit.

Proof of the integral form

We now know from the differential form that both $F(x)$ and $\int_a^x f(t) \, dt$ are primitives of $f(x)$.

Step 1: Because any two primitives differ only by a constant:

$\int_a^x f(t) \, dt = F(x) + C$

for some constant $C$.

Step 2: Find the constant by substituting $x = a$:

$\int_a^a f(t) \, dt = F(a) + C$

But $\int_a^a f(t) \, dt = 0$ (zero width), so:

$0 = F(a) + C$

$C = -F(a)$

Step 3: Substitute back:

$\int_a^x f(t) \, dt = F(x) - F(a)$

Step 4: Change letters from $x$ to $b$ and from $t$ to $x$:

$\int_a^b f(x) \, dx = F(b) - F(a)$

This completes the proof of the integral form.

Verifying consistency

We can check that the differential and integral forms are consistent by evaluating integrals and then differentiating the results.

Understanding continuous functions

The fundamental theorem requires that $f(x)$ be continuous. Let's clarify what this means.

This last definition can seem confusing. Consider the reciprocal function $y = \frac{1}{x}$:

• The function has a discontinuity at $x = 0$ (a vertical asymptote)
• However, $x = 0$ is not in the domain of the function
• Since the function is continuous at every point in its domain, it is technically called a "continuous function"

Remember!