Friday, February 3, 2023
HomeMathematicsCalculusFirst and Second fundamental theorem of calculus

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

### Dimension theorem of a quotient space: If W be a subspace of a finite dimensional vector space V over ​( mathbb{R} )​ then ​(...

$${}$$