Real Analysis/Fundamental Theorem of Calculus

 Real Analysis Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is said to be the central theorem of elementary calculus. It states effectively that "Differentiation" and "Integration" are inverse operations.

Conventionally, the theorem is presented in two parts

First Form of the Fundamental Theorem

Theorem

Let $f,F:[a,b]\to\mathbb{R}$

Let $F$ be differentiable on $[a,b]$ and let $F'(x)=f(x)$ for all $x\in [a,b]$

Let $f$ be Riemann integrable on $[a,b]$

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

Proof

Let $\int_a^b f=L$ and let $\varepsilon>0$ be given.

Then, there exists $\delta>0$ such that for a partition $\|\mathcal{\dot{P}}\|<\delta$ implies that $|S(f,\mathcal{\dot{P}})-L|<\varepsilon$

Consider a partition $\mathcal{P}$ and let $x_{i-1},x_i\in\mathcal{P}$. By Lagrange's Mean Value Theorem, we have that there exists $c_i\in (x_{i-1},x_i)$ that satisfies $\frac{F(x_i)-F(x_{i-1})}{x_i-x_{i-1}}=F'(c_i)=f(c_i)$

Let the tagged partition $\mathcal{\dot{P}}$ be the partition $\mathcal{P}$ along with the tags $c_i$

Thus, $S(f,\mathcal{\dot{P}})=\sum_{i=1}^n f(c_i)(x_i-x_{i-1})=\sum_{i=1}^n \tfrac{F(x_i)-F(x_{i-1})}{x_i-x_{i-1}}(x_i-x_{i-1})=\sum_{i=1}^n (F(x_i)-F(x_{i-1}))=F(b)-F(a)$

But we know that $|S(f,\mathcal{\dot{P}})-L|<\varepsilon$ and hence, $|F(b)-F(a)-L|<\varepsilon$.

As $\varepsilon>0$ is arbitrary, $|F(b)-F(a)-L|=0$ that is, $\int_a^b f(x)dx=F(b)-F(a)$

Second Form of the Fundamental Theorem

We first define what is known as the "Indefinite Integral"

Definition

Let $f:[a,b]\to\mathbb{R}$ be Riemann integrable on $[a,b]$.

We define the Indefinite Integral of $f$ to be the function $F:[a,b]\to\mathbb{R}$ given by

$F(x)=\int_a^x f$ for all $x\in [a,b]$

Theorem

Let $f:[a,b]\to\mathbb{R}$ be continuous at $c\in [a,b]$

Let $F:[a,b]\to\mathbb{R}$ be the indefinite integral of $f$

Then, $F$ is differentiable at $c$ and $F'(c)=f(c)$

Proof

Let $\varepsilon>0$ be given, and let $x\in [a,b]$ but $x\neq c$

Observe that $F(x)-F(c)=\int_c^x f=L$(say).

There exists $\delta>0$ such that if a partition $\|\mathcal{\dot{P}}\|<\delta$ then, $|S(f,\mathcal{\dot{P}})-L|<\varepsilon |x-c|$ (note that in this proof, all the Riemann sums are over the interval $[c,x]$).

As $f$ is integrable over $[c,x]$, it is bounded over that interval. Hence, let $M=\sup f([c,x])$. Thus, $S(f,\mathcal{\dot{P}})\leq \sum_{i=1}^n M(x_i-x_{i-1})=M(x-c)$

As $f$ is continuous at $c$, there exists $\delta>0$ such that $M(x-c)-S(f,\mathcal{\dot{P}})<\varepsilon |x-c|$ whenever $|x-c|<\delta$.

Now consider $x\in V_{\delta}(c)$

Then, $\left| \tfrac{F(x)-F(c)}{x-c}-f(c)\right| =\left| \tfrac{L}{x-c}-f(c)\right| <\left| \tfrac{S(f,\mathcal{\dot{P}})+\varepsilon|x-c|}{x-c}-f(c)\right|<\left| \tfrac{M(x-c)+2\varepsilon|x-c|}{x-c}-f(c)\right|$

$<\left| \tfrac{f(c)(x-c)+3\varepsilon|x-c|}{x-c}-f(c)\right|<3\varepsilon$

That is, $\lim_{x\to c}\frac{F(x)-F(c)}{x-c}=f(c)$, or $F'(c)=f(c)$