The fundamental theorem of calculus can be broken down into 2 different parts:

• 1st part: 1st fundamental theorem of calculus states that if \(f\) is continuous on the closed interval \([a, b]\) and \(F\) is the indefinite integral of \(f\) on $[a, b],$ then \(\int_{a}^{b} f(x) dx = F(b) - F(a)\). In other words the definite integration of a function is related to its antiderivative, and can be reversed by differentiation. This guarantees the existence of antiderivatives for continuous functions.

• Proof:

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

  \(\forall\; x_{1}, x_{1}+\Delta x \in [a, b]\),

  we have

  \(F(x_{1}) = \int_{a}^{x_{1}} f(t) d(t)\) &

  \(F(x_{1}+\Delta x) = \int_{a}^{x_{1}+\Delta x} f(t) dt\)

  \begin{equation} \label{eqn1} F(x_{1} + \Delta x) - F(x_{1}) = \int_{a}^{x_{1} + \Delta x} f(t) dt \;-\; \int_{a}^{x_{1} f(t) dt} \end{equation}

  \(\int_{a}^{x_{1}} f(t) dt\; +\; \int_{x_{1}}^{x_{1} + \Delta x} f(t) dt \;=\; \int_{a}^{x_{1} + \Delta x} f(t) dt\)

  Rearranging the above equation we have:

  \(\int_{a}^{x_{1} + \Delta x} f(t) dt \;-\; \int_{a}^{x_{1} \Delta x} f(t) dt \;=\; \int_{x_{1}}^{x_{1} + \Delta x} f(t) dt\)

  substituting into equation \ref{eqn1}

  \begin{equation} \label{eqn2} F(x_{1} + \Delta x) - F(x_{1}) \;=\; \int_{x_{1}}^{x_{1} + \Delta x} f(t) dt \end{equation}

  The Mean Value Theorem of integration tells us that there exists a number \(c(\Delta x) \in \mathbb{R}, \in [x_{1}, x_{1} + \Delta x]\) such that

  \(\int_{x_{1}}^{x_{1} + \Delta x} f(t) dt \;=\; f(c(\Delta x)) \Delta x\)

  Substituting into equation \ref{eqn2}

  \(F(x_{1} + \Delta x) - F(x_{1}) = f(c(\Delta x)) \Delta x\)

  dividing both side by \(\Delta x\)

  \(\frac{F(x_{1} + \Delta x) \;-\; F(x_{1})}{\Delta x} \;=\; f(c(\Delta x))\)

  The left side is Newton's difference quotient for \(F\) at \(x_{1}\)

  \(\lim_{\Delta x\rightarrow 0}\frac{F(x_{1} + \Delta x) \;-\; F(x_{1})}{\Delta x} \;=\; \lim_{\Delta x\rightarrow 0}f(c(\Delta x))\)

  Expression on the left side is the definition of the derivative of \(F\) at \(x_{1}\)

  \begin{equation} \label{eqn3} F'(x_{1}) \;=\; \lim_{\Delta x\rightarrow 0} f(c(\Delta x)) \end{equation}

  We use Squeeze Theorem to find the other limit. We know that \(c \in [x_{1}, x_{1}+\Delta x]\), so \(x_{1}\le c(\Delta x)\le x_{1}+\Delta x\)

  \(\lim_{\Delta x\rightarrow 0} x_{1} = x_{1}\) and \(\lim_{\Delta x\rightarrow 0} x_{1} + \Delta x = x_{1}\)

  According to the Squeeze Theorem

  \(\lim_{\Delta x\rightarrow 0} c(\Delta x) = x_{1}\)

  substituting into equation \ref{eqn3}

  \(F'(x_{1}) = \lim_{c(\Delta x)\rightarrow x_{1}} f(c(\Delta x))\)

  Since \(f\) is continuous at \(c(\Delta x)\), therefore, the limit can be taken inside the function

  \(F'(x_{1}) = f(x_{1})\).

• 2nd part: 2nd fundamental theorem of calculus states that if \(F\) is defined by \(F(x) = \int_{a}^{x}f(t) dt\) then \(F'(x) = f(x)\) at each point on the open interval \(I\) iff \(f\) is a continuous function on that open interval \(I\) and letting \(a\) be any point in \(I\). In other words the definite integral of a function can be computed by using any one of its infinitely-many antiderivatives.

• Proof:

  Limit proof by Riemann sums.

  Let \(f\) be Riemann integrable on the interval \([a, b]\)

  Let \(f\) produce an antiderviative on \(F\) on \([a, b]\)

  We begin with the quantity \(F(b) - F(a)\).

  Lets denote the numbers \(x_{1}, \dots, x_{n}\) such that \(a = x_{0} < x_{1} < x_{2} < \dots < x_{n-1} < x_{n} = b\). Thus, \(F(b) - F(a) = F(x_{n}) - F(x_{0})\). Adding each \(F(x_{n})\) along with its additive inverse we get:

  \(F(b) - F(a) = F(x_{n}) + [-F(x_{n-1}) + F(x_{n-1})] + \dots + [-F(x_{1}) + F(x_{1})] - F(x_{0}) = [F(x_{n}) - F(x_{n-1})] + [F(x_{n-1}) - F(x_{n-2})] + \dots + [F(x_{2}) - F(x_{1})] + [F(x_{1}) - F(x_{0})]\). Written as a sum looks like this:

  \begin{equation} \label{eqn_fdl_cal} F(b) - F(a) = \sum_{n=1}^{N} [F(x_{n}) - F(x_{n-1})] \end{equation}

  Using the mean value theorem we have:

  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 \([a, b]\), terefore, it is also differentiable and continuous on each interval \([x_{n-1}, x_{n}]\). According to the mean value theorem we have \(F(x_{n}) - F(x_{n-1}) = F'(c_{n})(x_{n} - x_{n-1})\). Substituting into \ref{eqn_fdl_cal}, we get \(F(b)\) \(- F(a) = \sum_{n=1}^{N}[F'(c_{n})(x_{n} - x_{n-1})]\). The assumption implies that \(F'(c_{n}) = f(x_{n})\). Also, \(x_{n} - x_{n-1}\) can be expressed as \(\Delta_{x}\) of partition \(n\).

  \begin{equation} \label{eqn_2nd} F(b) - F(a) = \sum_{n=1}^{N}[f(c_{n})(\Delta x_{n})]. \end{equation}

  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 \(\Delta x_{n}\) need not be the same for all values of n, 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 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 \(f\) 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.

  Taking the limit of both sides of \ref{eqn_2nd}, gives us

  \[\lim_{\Vert\Delta x_{n}\Vert\rightarrow 0} F(b) - F(a) = \lim_{\Vert\Delta x_{n}\Vert\rightarrow 0} \sum_{n=1}^{N}[f(c_{n})(\Delta x_{n})].\]

  Both \(F(b)\) and \(F(a)\) are not dependent on \(\Vert\Delta x_{n}\Vert\), so the limit on the left side remains \(F(b) - F(a)\).

  \[F(b) - F(a) = \lim_{\Vert\Delta x_{n}\Vert\rightarrow 0} \sum_{n=1}^{N}[f(c_{n})(\Delta x_{n})].\]

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

  \[F(b) - F(a) = \int_{a}^{b}f(x)dx\]

  Suppose \(G\) is an antiderivative of \(f\). Then by the second theorem, we have \(G(x) - G(a) = \int_{a}^{x} f(t) dt\). Now, suppose \(F(x)=\int_{a}^{x} f(t) dt = G(x) - G(a)\). Then \(F\) has the same derivative as \(G\), and therefore \(F' = f\). This argument only works, if we already know that \(f\) has an antiderivative, and the only way we know that all continuous functions have antiderivatives first part of the Fundamental Theorem. For example, if \(f(x) = e^{-x^{2}}\) then \(f\) 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 elementary antiderivatives, and discontinuous functions can be integrable but lack any antiderivatives at all. Conversely, many functions that have antiderivatives are not Riemann integrable (Volterra's function).