The Fundamental Theorem of Calculus in Analysis 📂Analysis

The Fundamental Theorem of Calculus in Analysis

Theorem¹

Given that function $f$ is Riemann integrable on the interval $[a,b]$ , and there exists a function $F$ that is differentiable on $[a,b]$ , satisfying $F^{\prime}=f$ . Then, the following holds true.

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

Explanation

This theorem is famously known as the Fundamental Theorem of Calculus Part 2, often abbreviated as FTC2[^Funcamental Theorem of Calculus1]. It implies that the definite integral of $f$ is represented by the difference of the values of the antiderivative $F$ at the endpoints.

Proof

Assuming $\varepsilon >0$ is given. Since $f$ is integrable on $[a,b]$ , by the necessary and sufficient condition, there exists a partition $P=\left\{a= x_{0}, \cdots, x_{n}=b \right\}$ of interval $[a,b]$ that satisfies the following.

$U(P,f)-L(P,f) < \varepsilon$

Since $F$ is assumed to be differentiable, hence continuous, by the Mean Value Theorem, there exists a $t_{i}\in [x_{i-1},x_{i}]$ that satisfies the following.

$F(x_{i})-F(x_{i-1})=f(t_{i})\Delta x_{i},\quad (i=1,\dots,n)$

Adding the above equation for all $i$ , we get:

$\begin{align*} \sum \limits _{i=1} ^{n} f(t_{i})\Delta x_{i}&=\left( F(b)-F(x_{n-1}) \right)+\cdots+\left( F(x_{1})-F(a) \right) \\ &= F(b) -F(a) \end{align*}$

Auxiliary Lemma
$\left| \sum \limits_{i=1} ^{n} f(t_{i})\Delta \alpha_{i} - \int _{a} ^{b}f (x)d\alpha (x) \right| < \varepsilon$

By the above auxiliary lemma, the following is true.

$\begin{align*} \left| \sum \limits _{i=1} ^{n} f(t_{i})\Delta \alpha_{i} - \int _{a} ^{b}f (x)d\alpha (x) \right| &= \left|\big( F(b)-F(a) \big) - \int _{a} ^{b}f (x)d\alpha (x) \right| \\ &< \varepsilon \end{align*}$

Here, since $\varepsilon$ is arbitrary positive number, we obtain:

$\int _{a} ^{b}f (x)d\alpha (x)=F(b)-F(a)$

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