Friday, February 3, 2023
Google search engine
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

2 COMMENTS

LEAVE A REPLY

Please enter your comment!
Please enter your name here

- Advertisment -
Google search engine

Most Popular

Recent Comments

Insert math as
Block
Inline
Additional settings
Formula color
Text color
#333333
Type math using LaTeX
Preview
\({}\)
Nothing to preview
Insert