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. ​ $\displaystyle\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}=f(x)$ ​
​ $\implies F'(x)=f(x)$ ​
 Also Read: Every continuous function f defined on [a, b] is Riemann integrable

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, ​ $F(b)-F(a)\leq \int\limits_{a}^{\bar{b}}f$ ​ 
​ $\therefore \int\limits_{\underline{a}}^{b}f\leq F(b)-F(a)\leq$ $\int\limits_{a}^{\bar{b}}f$​
​ $\therefore \int\limits_{a}^{b}f\leq F(b)-F(a)\leq \int\limits_{a}^{b}f$ ​
​ $\implies\int\limits_{a}^{b}f=F(b)-F(a)$ ​