The fundamental theorem of calculus

The fundamental theorem of calculus states, in part, that

Theorem 8061

 If $f:D\rightarrow(-\infty,\infty)$ is continuous on [a,b] and if there is some function $F:[a,b]\rightarrow(-\infty,\infty)$ with F'(x) = f(x) for $x\in(a,b)$ then

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

To this point, we don't have a definition of what the left-hand side of this equation is, although if f is non-negative on [a,b] we have agreed to let it stand for the area of the region bounded below by the x-axis, above by y = f(x), to the left by x = b and to the right by x=a. In fact, with this interpretation we showed on the first day that if p is a positive integer and b > 0 then

$$\int_0^b x^p\;dx = \frac{b^{p+1}}{p+1}$$

We shall now show why Theorem 5 is plausible. Our tool will be the Mean Value Theorem.

First two examples. Consider $f:(-\infty,\infty)\rightarrow(-\infty,\infty)$where f(u) = u³. We know that if F(u) = u⁴/4 then F'(u) = u³. We can use this to show that the area, A, of the region between the u-axis, y=u³, u=0 and u = 2 must be 4. In order to keep things simple, suppose we divide [0,2] into four subintervals, [0,2/3], [2/3,1], [1,3/2] and [3/2,2]. You will see that the number and location of the subintervals is totally irrelevant to our argument. You can see from the diagram below that the area in question is less than or equal to the sum of the areas of the four rectangles on the right and greater than or equal to the sum of the areas of the four rectangles on the left.

Now apply the Mean Value Theorem to $F:[0,2/3]\rightarrow(-\infty,\infty)$,F(u) = u⁴/4. We get that for some $c_1\in (0, 2/3)$ we have

F(2/3) - F(0) = F'(c₁)(2/3 - 0) = (c₁)³(2/3).

Since

$$0^3(2/3 - 0) \leq (c_1)^3(2/3 - 0) \leq (2/3)^3(2/3-0)$$

we have  

$$0^3(2/3 - 0) \leq F(2/3) - F(0) \leq (2/3)^3(2/3-0).$$ (4)

Next apply the Mean Value Theorem to $F:[2/3,1]\rightarrow(-\infty,\infty)$,F(u) = u⁴/4. We get that for some $c_2 \in (2/3,1)$ we have

F(1) - F(2/3) = F'(c₂)(1 - 2/3 ) = (c₂)³(1 - 2/3)

Since

$$(2/3)^3(1 - 2/3) \leq (c_2)^3(1 - 2/3) \leq (1)^3(1- 2/3)$$

we have  

$$(2/3)^3(1 - 2/3) \leq F(1) - F(2/3) \leq (1)^3(1- 2/3)$$ (5)

Next apply the Mean Value Theorem to $F:[1, 3/2]\rightarrow(-\infty,\infty)$,F(u) = u⁴/4. We get that for some $c_3\in (1,3/2)$ we have

F(3/2) - F(1) = F'(c₃)(3/2 - 1) = (c₃)³(3/2 - 1).

Since

$$(1)^3(3/2 - 1) \leq (c_3)^3(3/2 - 1) \leq (3/2)^3(3/2 - 1)$$

we have  

$$(1)^3(3/2 - 1) \leq F(3/2) - F(1) \leq (3/2)^3(3/2 - 1)$$ (6)

Finally, apply the Mean Value Theorem to $F:[3/2, 2]\rightarrow(-\infty,\infty)$, F(u) = u⁴/4. We get that for some $c_4\in (3/2,2)$ we have

F(2) - F(3/2) = F'(c₄)(2 - 3/2) = (c₄)³(2 - 3/2).

Since

$$(3/2)^3(2- 3/2) \leq (c_4)^3(2 - 3/2) \leq (2)^3(2 - 3/2)$$

we have  

$$(3/2)^3(2- 3/2) \leq F(2) - F(3/2) \leq (2)^3(2 - 3/2).$$ (7)

Now notice that if we add the inequalities (4) through (7) together we see that the total area of the rectangles lying under the graph of y = u³ is less than or equal to F(2) - F(0), while the total area of the rectangles lying over the graph of y = u³ is greater than equal to F(2) - F(0). If we could show that there is only one such number, we would have shown that the area in question is given by F(2) - F(0) = 2⁴/4 - 0⁴/4 = 4.

Here is the general statement of what is shown in our example.

Theorem 8126

 Suppose that

• $f:D\rightarrow(-\infty,\infty)$ is continuous on [a,b];
• There is some function $F:[a,b]\rightarrow(-\infty,\infty)$ with F'(u) = f(u) for all $u\in (a,b)$;
• Suppose that u₀ = a, u_N = b, and $u_0 < u_1 < u_2 \cdots < u_{N-1} < u_N$;
• The range of $f:[u_{j-1}, u_j]\rightarrow(-\infty,\infty)$ is [m_j, M_j].

Then

$$\begin{array} {rcl} m_1(u_1 - u_0) + m_2(u_2 - u_1) + \cdots... ... + \cdots + M_N(u_N - u_{N-1}) & \geq & F(b) - F(a)\end{array}$$

A more compact way to write the conclusion of this theorem is  

$$\sum_{j=1}^N m_j(u_j - u_{j-1}) \leq F(b) - F(a) \leq \sum_{j=1}^N M_j(u_j - u_{j-1}).$$ (8)

An important theorem from real analysis says that for a function f which is continous on [a,b] there is only one number which satisfies the double inequality (8), and this number is called the integral of f on [a,b], and is denoted by

$$\int_a^b f(u)\;du.$$

You should think of the $\int_a^b$ as telling you to sum. The du tell us that the variable is u, and that we should divide [a,b] into subintervals. The du becomes important if the expression for f has several variables.

$$\int_a^b x^u du$$

and

$$\int_a^b x^u dx$$

have quite different meanings.

The proof of our last theorem goes just the way of the example. First, apply the Mean Value Theorem to $F:[u_0,u_1]]\rightarrow(-\infty,\infty)$. We get that for some $c_1\in (u_0,u_1)$ we have

F(u₁) - F(u₀) = F'(c₁)(u₁ - u₀) = f(c₁)(u₁-u₀).

Since

$$m_1 \leq f(c_1) \leq M_1$$

and (u₁ - u₀) > 0 we have

$$m_1(u_1 - u_0) \leq f(c_1)(u_1 - u_0) \leq M_1(u_1 - u_0).$$

Therefore

$$m_1(u_1 - u_0) \leq F(u_1) - F(u_0) \leq M_1(u_1 - u_0).$$

Next, apply the Mean Value Theorem to $F:[u_1,u_2]]\rightarrow(-\infty,\infty)$. We get that for some $c_2\in (u_1,u_2)$ we have

F(u₂) - F(u₁) = F'(c₂)(u₂ - u₁) = f(c₂)(u₂-u₁).

Since

$$m_2 \leq f(c_2) \leq M_2$$

and (u₁ - u₀) > 0 we have

$$m_2(u_2 - u_1) \leq f(c_2)(u_2 - u_1) \leq M_2(u_2 - u_1).$$

Therefore

$$m_2(u_2 - u_1) \leq F(u_2) - F(u_1) \leq M_2(u_2 - u_1).$$

We continue in this fashion. For the interval [u_j-1, u_j] we have: Apply the Mean Value Theorem to $F:[u_{j-1},u_j]]\rightarrow(-\infty,\infty)$. We get that for some $c_j\in (u_{j-1},u_j)$ we have

F(u_j) - F(u_j-1) = F'(c_j)(u_j - u_j-1) = f(c_j)(u_j-u_j-1).

Since

$$m_j \leq f(c_j) \leq M_j$$

and (u_j - u_j-1) > 0 we have

$$m_j(u_j - u_{j-1}) \leq f(c_j)(u_j - u_{j-1}) \leq M_j(u_j - u_{j-1}).$$

Therefore  

$$m_j(u_j - u_{j-1}) \leq F(u_j) - F(u_{j-1}) \leq M_j(u_j - u_{j-1}).$$ (9)

If we now add up the N double inequalities indicated by $(\ref{gencase})$we obtain the conclusion of the theorem.

QED