Question

prove part 1 and 2 of the fundamental theory of calculus

Answer 1

Proof of the first part

For a given f(t), define the function F(x) as

F(x) = \int_a^x f(t) \,dt\,.

For any two numbers x1 and x1 + Δx in [a, b], we have

F(x_1) = \int_{a}^{x_1} f(t) \,dt

and

F(x_1 + \Delta x) = \int_a^{x_1 + \Delta x} f(t) \,dt\,.

Subtracting the two equations gives

F(x_1 + \Delta x) - F(x_1) = \int_a^{x_1 + \Delta x} f(t) \,dt - \int_a^{x_1} f(t) \,dt. \qquad (1)

It can be shown that

\int_{a}^{x_1} f(t) \,dt + \int_{x_1}^{x_1 + \Delta x} f(t) \,dt = \int_a^{x_1 + \Delta x} f(t) \,dt.
(The sum of the areas of two adjacent regions is equal to the area of both regions combined.)

Manipulating this equation gives

\int_{a}^{x_1 + \Delta x} f(t) \,dt - \int_{a}^{x_1} f(t) \,dt = \int_{x_1}^{x_1 + \Delta x} f(t) \,dt.

Substituting the above into (1) results in

F(x_1 + \Delta x) - F(x_1) = \int_{x_1}^{x_1 + \Delta x} f(t) \,dt. \qquad (2)

According to the mean value theorem for integration, there exists a c in [x1, x1 + Δx] such that

\int_{x_1}^{x_1 + \Delta x} f(t) \,dt = f(c) \Delta x \,.

Substituting the above into (2) we get

F(x_1 + \Delta x) - F(x_1) = f(c) \Delta x \,.

Dividing both sides by Δx gives

\frac{F(x_1 + \Delta x) - F(x_1)}{\Delta x} = f(c).
Notice that the expression on the left side of the equation is Newton's difference quotient for F at x1.

Take the limit as Δx → 0 on both sides of the equation.

\lim_{\Delta x \to 0} \frac{F(x_1 + \Delta x) - F(x_1)}{\Delta x} = \lim_{\Delta x \to 0} f(c).

The expression on the left side of the equation is the definition of the derivative of F at x1.

F'(x_1) = \lim_{\Delta x \to 0} f(c). \qquad (3)

To find the other limit, we will use the squeeze theorem. The number c is in the interval [x1, x1 + Δx], so x1 ≤ c ≤ x1 + Δx.

Also, \lim_{\Delta x \to 0} x_1 = x_1 and \lim_{\Delta x \to 0} x_1 + \Delta x = x_1\,.

Therefore, according to the squeeze theorem,

\lim_{\Delta x \to 0} c = x_1\,.

Substituting into (3), we get

F'(x_1) = \lim_{c \to x_1} f(c)\,.

The function f is continuous at c, so the limit can be taken inside the function. Therefore, we get

F'(x_1) = f(x_1) \,.

which completes the proof.
Proof of the second part
This is a limit proof by Riemann sums. Let ƒ be (Riemann) integrable on the interval [a, b], and let ƒ admit an antiderivative F on [a, b]. Begin with the quantity F(b) − F(a). Let there be numbers x1, ..., xn such that

a = x_0 < x_1 < x_2 < \cdots < x_{n-1} < x_n = b. \,

It follows that

F(b) - F(a) = F(x_n) - F(x_0). \,

Now, we add each F(xi) along with its additive inverse, so that the resulting quantity is equal:

\begin{matrix} F(b) - F(a) & = & F(x_n)\,+\,[-F(x_{n-1})\,+\,F(x_{n-1})]\,+ \cdots +\,[-F(x_1) + F(x_1)]\,-\,F(x_0) \\ & = & [F(x_n)\,-\,F(x_{n-1})]\,+\,[F(x_{n-1})\,+ \cdots -\,F(x_1)]\,+\,[F(x_1)\,-\,F(x_0)]. \end{matrix}

The above quantity can be written as the following sum:

F(b) - F(a) = \sum_{i=1}^n \,[F(x_i) - F(x_{i-1})]. \qquad (1)

Next we will employ the mean value theorem. Stated briefly,

Let F be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then there exists some c in (a, b) such that

F'(c) = \frac{F(b) - F(a)}{b - a}.

It follows that

F'(c)(b - a) = F(b) - F(a). \,

The function F is differentiable on the interval [a, b]; therefore, it is also differentiable and continuous on each interval [xi −1, xi ]. According to the mean value theorem (above),

F(x_i) - F(x_{i-1}) = F'(c_i)(x_i - x_{i-1}) \,.

Substituting the above into (1), we get

F(b) - F(a) = \sum_{i=1}^n \,[F'(c_i)(x_i - x_{i-1})].

The assumption implies F'(ci) = f(ci). Also, xi − xi − 1 can be expressed as Δx of partition i.

F(b) - F(a) = \sum_{i=1}^n \,[f(c_i)(\Delta x_i)]. \qquad (2)

A converging sequence of Riemann sums. The numbers in the upper right are the areas of the grey rectangles. They converge to the integral of the function.

Notice that we are describing the area of a rectangle, with the width times the height, and we are adding the areas together. Each rectangle, by virtue of the Mean Value Theorem, describes an approximation of the curve section it is drawn over. Also notice that Δxi need not be the same for all values of i, or in other words that the width of the rectangles can differ. What we have to do is approximate the curve with n rectangles. Now, as the size of the partitions get smaller and n increases, resulting in more partitions to cover the space, we will get closer and closer to the actual area of the curve.

By taking the limit of the expression as the norm of the partitions approaches zero, we arrive at the Riemann integral. We know that this limit exists because ƒ was assumed to be integrable. That is, we take the limit as the largest of the partitions approaches zero in size, so that all other partitions are smaller and the number of partitions approaches infinity.

So, we take the limit on both sides of (2). This gives us

\lim_{\| \Delta \| \to 0} F(b) - F(a) = \lim_{\| \Delta \| \to 0} \sum_{i=1}^n \,[f(c_i)(\Delta x_i)].

Neither F(b) nor F(a) is dependent on ||Δ||, so the limit on the left side remains F(b) − F(a).

F(b) - F(a) = \lim_{\| \Delta \| \to 0} \sum_{i=1}^n \,[f(c_i)(\Delta x_i)].

The expression on the right side of the equation defines the integral over ƒ from a to b. Therefore, we obtain

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

which completes the proof.

It almost looks like the first part of the theorem follows directly from the second, because the equation g(x) - g(a) = \int_a^x f(t) \, dt,  where g is an antiderivative of ƒ, implies that F(x) = \int_a^x f(t)\, dt\,  has the same derivative as g, and therefore F ′ = ƒ. This argument only works if we already know that ƒ has an antiderivative, and the only way we know that all continuous functions have antiderivatives is by the first part of the Fundamental Theorem.[8] For example if ƒ(x) = e−x2, then ƒ has an antiderivative, namely

g(x) = \int_0^x f(t) \, dt\,

and there is no simpler expression for this function. It is therefore important not to interpret the second part of the theorem as the definition of the integral. Indeed, there are many functions that are integrable but lack antiderivatives that can be written as an elementary function. Conversely, many functions that have antiderivatives are not Riemann integrable (see Volterra's function).