# First and Second fundamental theorem of calculus

## Primitive or Antiderivative :

If $F'(x)=f(x)$ then F is called primitive or antiderivative of f.
e.g. $F(x)=x^2sin(\frac{1}{x})$
$\therefore F'(x)=2xsin(\frac{1}{x})+x^2.cos(\frac{1}{x}).(\frac{-1}{x^2})$
$\therefore F'(x)=2xsin(\frac{1}{x})-cos(\frac{1}{x})$
Here, $F'(0)=\displaystyle\lim_{x\to0}\frac{F(x)-F(0)}{x-0}$
$=\displaystyle\lim_{x\to0}\frac{x^2sin(\frac{1}{x})}{x}$
$=\displaystyle\lim_{x\to0}xsin(\frac{1}{x})$
$=0$

## Indefinite R-integral :

Let $f:[a, b]\to \mathbb{R}$ be R-integrable then function $F:[a, b]\to \mathbb{R}$ defined by $F(x)=\int\limits_{a}^{x} f(t)dt, x\in[a, b]$ is called indefinite R-integral of f.

## First Fundamental Theorem of Calculus :

Let $f:[a, b]\to \mathbb{R}$ be continuous and $F(x)=\int\limits_{a}^{x} f(t)dt, x\in[a, b]$ then F is differentiable and $F'(x)=f(x), \forall x\in[a, b]$.

Proof : Let $\epsilon>0$ be arbitrary.
As, $f:[a, b]\to \mathbb{R}$ be continuous, f is uniformly continuous, so that, $u, v\in[a, b]$, $|u-v|<\delta\implies|f(u)-f(v)|<\epsilon$
Choose, h small enough such that, $x\in(a, b)\implies x+h\in(a, b)$ & $|h|<\delta$.
$|\frac{F(x+h)-F(x)}{h}-f(x)|$
$=\frac{1}{|h|}|F(x+h)-F(x)-hf(x)|$
$=\frac{1}{|h|}|\int\limits_{a}^{x+h}f(t)dt-\int\limits_{a}^{x}f(t)dt-hf(x)|$
$=\frac{1}{|h|}|\int\limits_{x}^{x+h}f(t)dt-\int\limits_{x}^{x+h}f(x)dt|$
$=\frac{1}{|h|}|\int\limits_{x}^{x+h}[f(t)-f(x)]dt|$ ..... (1)
Case (1) : $h>0\implies|h|=h$
Therefore, (1) gives
$=\frac{1}{h}|\int\limits_{x}^{x+h}[f(t)-f(x)]dt|$
$\leq\frac{1}{h}\int\limits_{x}^{x+h}|f(t)-f(x)||dt|$
$<\frac{1}{h}\int\limits_{x}^{x+h}\epsilon dt$
$=\frac{1}{h}\times \epsilon\times h$
$=\epsilon$
Thus, $\forall \epsilon>0, \exists \delta>0$, such that,
$|\frac{F(x+h)-F(x)}{h}-f(x)|<\epsilon$
Case (2) : $h<0\implies|h|=-h$
$=\frac{1}{-h}|(-1)\int\limits_{x}^{x+h}[f(t)-f(x)]dt|$
$\leq\frac{1}{-h}\int\limits_{x+h}^{x}|f(t)-f(x)||dt|$
$<\frac{1}{-h}\int\limits_{x+h}^{x}\epsilon dt$
$=\frac{1}{-h}\times \epsilon\times (-h)$
$=\epsilon$
Thus, $\forall \epsilon>0, \exists \delta>0$, such that,
$|\frac{F(x+h)-F(x)}{h}-f(x)|<\epsilon$
Thus, in either case,
$\forall \epsilon>0, \exists \delta>0$, such that,
$|\frac{F(x+h)-F(x)}{h}-f(x)|<\epsilon$
i.e. $\displaystyle\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}=f(x)$
$\implies F'(x)=f(x)$

## Second Fundamental Theorem of Calculus :

If $f:[a, b]\to \mathbb{R}$ be Riemann integrable and F is primitive of f then $\int\limits_{a}^{b} f(x)dx=F(b)-F(a)$ or $\int\limits_{a}^{b} F'(x)dx=F(b)-F(a)$.

Proof : Let $f:[a, b]\to \mathbb{R}$ be Riemann integrable and F is primitive of f.
$\int\limits_{\underline{a}}^{b}f=\int\limits_{a}^{b}f=\int\limits_{a}^{\bar{b}}f$ ..... (1)
Let $P=\{a=x_0, x_1, x_2, ... , x_n=b\}$ be a partition of [a, b].
Let $m_k$ be infimum of f on $[a, b]$ and $M_k$ be supremum of f on $[x_{k-1}, x_k]$.
Since, F is primitive of f, $F'(x)=f(x)$, $\forall x\in(a, b)$.
$\implies$ F is differentiable function on (a, b) and since every differentiable function is continuous, F is continuous on [a, b].
$\therefore$ by L.M.V.T.,
$\exists t_k\in(x_{k-1}, x_k)$ such that
$F'(t_k)=\frac{F(x_k)-F(x_{k-1})}{x_k-x_{k-1}}$
$F'(t_k)(x_k-x_{k-1})=F(x_k)-F(x_{k-1})$
Since, $F'(x)=f(x)$, $\forall x\in(a, b)$
$F'(t_k)=f(t_k)$, $k=1, 2, 3, ... , n$
$\therefore f(t_k)(x_k-x_{k-1})=F(x_k)-F(x_{k-1})$ ..... (2)
For $[x_{k-1}, x_k]$,
$m_k\leq f(x)\leq M_k$, $\forall x\in[x_{k-1}, x_k]$
$\therefore m_k(x_k-x_{k-1})\leq f(x)(x_k-x_{k-1})\leq M_k(x_k-x_{k-1})$, $\forall k=1,2, ... , n$
$\therefore\displaystyle\sum_{k=1}^{n}m_k(x_k-x_{k-1})\leq\displaystyle\sum_{k=1}^{n}f(x)(x_k-x_{k-1})\leq \displaystyle\sum_{k=1}^{n}M_k(x_k-x_{k-1})$
$\therefore L(f, P)\leq\displaystyle\sum_{k=1}^{n}[F(x_k)-F(x_{k-1})]\leq U(f, P)$ ..... (2)
$\therefore L(f, P)\leq F(b)-F(a)\leq U(f, P)$
Since, P is arbitrary partition of [a, b],
$\therefore sup\{L(f, P) | P\in \mathscr{P}[a, b]\}\leq F(b)-F(a)$
$\therefore \int\limits_{\underline{a}}^{b}f\leq F(b)-F(a)$
Also, $F(b)-F(a)\leq \int\limits_{a}^{\bar{b}}f$
$\therefore \int\limits_{\underline{a}}^{b}f\leq F(b)-F(a)\leq \int\limits_{a}^{\bar{b}}f$
$\therefore \int\limits_{a}^{b}f\leq F(b)-F(a)\leq \int\limits_{a}^{b}f$
$\implies\int\limits_{a}^{b}f=F(b)-F(a)$