1. Assume that $f(x), x \in[a, b]$ is continuous & differentiable on $(a, b)$ so that $f^{\prime}(x)$ is known.

Given $f^{\prime}(x)$, can we find the $f(x)$?

2. Assume that $f(x)$ is a continuous function on $[a, b]$,

Area A ?

$f(x)=x, [0,a] \quad a>0$

$A(x) \quad 0<x<a$

$A(x)=\frac{1}{2}$ $\times$$x$$\times$$x$

$A(x)=\frac{1}{2} x^2$

$A(a)=\frac{1}{2} a^2, A(0)=0$

$\frac{d}{d x}(A(x))=\frac{d}{d x}\left(\frac{x^2}{2}\right)=\frac{2 x}{2}=x$

$\frac{d}{d x} (A(x)) = x$

Suppose $f(x)$ be a continuous function $[a, b]$ $A(x)$ is the area function, then

$A^{\prime}(x)=f(x) \quad \forall x \in[a, b] \text {. }$

$\frac{d}{d x}(\sin x+1)=\frac{d}{d x} \sin x+\frac{d}{d x} 1$

$=\frac{d}{d x} \sin x=\cos x$

$\cos x$ $\quad$ has $\quad$ $(\sin x+1)$ also as antiderivative.

$\frac{d}{d x}(\sin x+c)=\cos x$

$\sin x+c$ is anti derivative of $\cos x$

$c$ - arbitrary constant, $c \in \mathbb{R}$

$\frac{d}{d x}(F(x))=f(x) \text { then }$

$\frac{d}{d x}(F(x)+c)=f(x)$

$\{F(x)+c: c \in \mathbb{R}\}$

Set of all the anti-derivative of $f(x)$ or family of one parameter curves.

Integral:

$\int f(x) d x=F(x)+c,$ $\quad$ $c \in \mathbb{R}$

$\int$ = integral

f$(x)$ = integrand

$x$ = variable of integration

F$(x)$ = integral/ anti-derivative

c = arbitrary constant.

Remark

$\int f(t) d t=\int f(x) d x$

1. $\int \cos x d x= \quad \sin x+c$

2. $\int e^{n x} d x=\frac{e^{n x}}{n}+c$

3. $\int \sec ^2 x d x=\tan x+c$

4. $\int x d x=\frac{x^2}{2}+c$

5. $\int \cos t d t=\sin t+c$

Find Integral (anti-derivative) of following functions by inspection:

i)

$f(x)=\sin 2 x$

$\frac{d}{d x}\left(-\frac{\cos 2 x}{2}\right)=\sin 2 x$

$-\frac{1}{2} \cos 2 x+c$

= Anti derivative of sin2x

$f(x)=e^{4 x}$

$\frac{d}{d x}\left(\frac{e^{4 x}}{4}\right)=e^{4 x}$

$\Rightarrow$ Anti derivative of $e^{4 x} = \frac{e^{4 x}}{4}+ c.$

3.

$f(x)=\sin 2 x-4 e^{3 x}$

$\frac{d}{d x}\left(-\frac{1}{2} \cos 2 x-4 \cdot \frac{e^{3 x}}{3}\right)=\sin 2 x-4 e^{3 x}$

$\Rightarrow$ Anti derivative of

$\sin 2 x-4 e^{3 x} = -\frac{1}{2} \cos 2 x-\frac{4}{3} e^{3 x}+c$

$f^{\prime}(x)=g^{\prime}(x) \forall x \in I$

then $f(x)-g(x)=$ constant

$h(x)=f(x)-g(x) \forall x$

$h^{\prime}(x)=f^{\prime}(x)-g^{\prime}(x)=0$ $\forall x$

$h^{\prime}(x)=0 \Rightarrow h(x)=\text { constant }$

$\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^2}}$

$\frac{d}{d x}\left(\cos ^{-1} x\right)=-\frac{1}{\sqrt{1-x^2}}$

$\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^2}}=\frac{d}{d x}\left(-\cos ^{-1} x\right)$

$\sin ^{-1} x+\cos ^{-1} x=\text { constant }$

$\sin ^{-1} x+\cos ^{-1} x=\pi / 2$

Let $f(x)=e^x$ and now we see tangent at the point $P_ 0, P_ 1, P_ 2..$ and the points $Q_ 0,Q_ 1,Q_2 ..$

${P_0} = (0,0)$

$\frac{d y}{d x}|_ {P_ 0} = \frac{d}{d x} (e^x {-1})|_ {P_ 0}= {e^x|}_ {P_0}=1,$

${Q_0} = (1,e-1)$

$\frac{d y}{d x}|_ {Q_ 0}=\frac{d}{d x}(e^x-1)|_ {Q_ 0} = e^xt|_ {Q_0}=e^1=e$

$\frac{d y}{d x}|_{Q_1}=e$

${P_1} = (0,1)$

$\frac{d y}{d x}|_ {P_ 1} =\frac{d}{d x} (e^x)|_ {P_ 1}= {e^x|}_ {P_ 1}=1$

${Q_ 1} = (1,e), ~ \frac{d y}{d x}|_ {Q_ 1} =\frac{d}{d x} (e^x)|_ {Q _ 1} ={e^x|}_{Q _1}= e$

$A(x)={\int_0^x x} d x=\frac{x^2}{2}$

Here, ${\int_0^x x}$ = definite Integral.

def ${ }^n$ of Integrals.

anti-derivative or Integrals.

graphical representation of these Integrals.