__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, \( F(b)-F(a)\leq \int\limits_{a}^{\bar{b}}f \)

\( \therefore \int\limits_{\underline{a}}^{b}f\leq F(b)-F(a)\leq \int\limits_{a}^{\bar{b}}f \)

\( \therefore \int\limits_{a}^{b}f\leq F(b)-F(a)\leq \int\limits_{a}^{b}f \)

\( \implies\int\limits_{a}^{b}f=F(b)-F(a) \)

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