Integrals and the Fundamental Theorem of Calculus | q005.html

(cod: P-68-40-3) Fundamental Theorem of Calculus: If \(f(x)\) is a continuous function on an interval \([a,b]\) and \(F\) is a primitive or antiderivative of \(f\), that is, \[\dfrac{d}{dx}F(x)=f(x),\] then \[\int_a^b f(x) \, dx = F(b) - F(a).\]

The objective of this exercise is to present a proof for the Fundamental Theorem of Calculus (F.T.C.). For each natural \(n > 3\), consider the regular partition \[\small{a = x_0 < x_1 < x_2 < \cdots < x_{n-1} < x_n = b}\] of \([a,b]\) into subintervals of length \(\Delta x = (b-a)/n\). Note that we can express the variation of \(F\) on \([a,b]\) as the sum of the variations of \(F\) on the subintervals. In fact, we have \begin{eqnarray*} & & F(b) - F(a) \\ & & \\ &=& F(x_n) - F(x_0) \\ & & \\ &=& \small{(F(x_n)- F(x_{n-1})) + (F(x_{n-1}) - F(x_{n-2}))} \\ & & \\ &+& \small{\cdots + (F(x_{2}) - F(x_{1})) + (F(x_1)-F(x_0))}. \end{eqnarray*} Indeed, observe that, in the long sum, we can cancel several terms, leaving only the first and the last (telescoping sum). Thus, we can conclude that \(\small{\begin{equation}\label{equa1}F(b) - F(a) = \sum_{i = 1}^n F(x_i) - F(x_{i-1}).\end{equation}}\) Now, for each \(i \in \{1,2,\dots,n\}\), there exists \(x_i^* \in (x_{i-1},x_i)\) such that \[\dfrac{F(x_i) - F(x_{i-1})}{x_i - x_{i-1}} = F'(x_i^*),\] or, equivalently, \[\small{F(x_i) - F(x_{i-1}) = F'(x_i^*) \Delta x = f(x_i^*) \Delta x}.\] Therefore, it follows from equation \((\ref{equa1})\) that, for any \(n > 3\), \[F(b) - F(a) = \sum_{i = 1}^n f(x_i^*) \Delta x.\] Taking \(n \to + \infty\) (always considering \(x_i^*\) as mentioned), we conclude that \[F(b) - F(a) = \int_a^b f(x) \, dx,\] as we wanted to demonstrate.

We have two questions:

(1) Which classical theorem guarantees that the statement in brown is true?

(2) Why is the statement in blue true?

The statement in brown follows from the Extreme Value Theorem. To justify the statement in blue, note that \[\sum_{i = 1}^n f(x_i^*) \Delta x\] is a Riemann sum (with regular partition) of \(f\). Since \(f\) is continuous, it is integrable on \([a,b]\), hence \[\lim\limits_{n \to + \infty }\sum_{i = 1}^n f(x_i^*) \Delta x = \int_a^b f(x) \, dx.\] However, on the other hand, all these Riemann sums are equal to \(F(b) - F(a)\) (for any \(n\)).

The statement in brown follows from the Chain Rule. To justify the statement in blue, note that \[\sum_{i = 1}^n f(x_i^*) \Delta x\] is a Riemann sum (with regular partition) of \(f\). Since \(f\) is continuous, it is integrable on \([a,b]\), hence \[\lim\limits_{n \to + \infty }\sum_{i = 1}^n f(x_i^*) \Delta x = \int_a^b f(x) \, dx.\] However, on the other hand, all these Riemann sums are equal to \(F(b) - F(a)\) (for any \(n\)).

The statement in brown follows from the Mean Value Theorem of Lagrange. To justify the statement in blue, note that \[\sum_{i = 1}^n f(x_i^*) \Delta x\] is a Riemann sum (with regular partition) of \(f\). Since \(f\) is continuous, it is integrable on \([a,b]\), hence \[\lim\limits_{n \to + \infty }\sum_{i = 1}^n f(x_i^*) \Delta x = \int_a^b f(x) \, dx.\] However, on the other hand, these Riemann sums are always less than \(F(b) - F(a)\) (for any \(n\)).

The statement in brown follows from the Intermediate Value Theorem. To justify the statement in blue, note that \[\sum_{i = 1}^n f(x_i^*) \Delta x\] is a Riemann sum (with regular partition) of \(f\). Since \(f\) is continuous, it is integrable on \([a,b]\), hence \[\lim\limits_{n \to + \infty }\sum_{i = 1}^n f(x_i^*) \Delta x = \int_a^b f(x) \, dx.\] However, on the other hand, these Riemann sums are always greater than \(F(b) - F(a)\) (for any \(n\)).

The statement in brown follows from the Mean Value Theorem of Lagrange. To justify the statement in blue, note that \[\sum_{i = 1}^n f(x_i^*) \Delta x\] is a Riemann sum (with regular partition) of \(f\). Since \(f\) is continuous, it is integrable on \([a,b]\), hence \[\lim\limits_{n \to + \infty }\sum_{i = 1}^n f(x_i^*) \Delta x = \int_a^b f(x) \, dx.\] However, on the other hand, all these Riemann sums are equal to \(F(b) - F(a)\) (for any \(n\)).