The derivative of a function \(f(x)\) is defined as the limit of the difference quotient:
\begin{equation*}
f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
\end{equation*}
The numerator involves subtraction: \(f(x + h) - f(x)\text{,}\) which is the difference in function values. The denominator is \(h\text{,}\) and the whole thing is a division (quotient) by \(h\text{.}\) So, the differentiation fundamentally relies on subtraction (differences) and division. This captures the instantaneous rate of change by "subtracting" values and "dividing" by the tiny change in input.
A integral
\begin{equation*}
\int_a^b f(x) \, dx
\end{equation*}
is defined as the limit of a Riemann sum:
\begin{equation*}
\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x_i
\end{equation*}
Each term is \(f(x_i^*) \cdot \Delta x_i\text{,}\) which is a multiplication (function value times small width). Then you add (sum) all those terms together. So, integration is built on repeated multiplication (scaling areas/volumes) followed by addition (summing them up). This is why we describe integration as "continuous addition" enhanced by multiplication for varying heights.
If \(F(x)\) is an antiderivative of \(f(x)\text{,}\) then the most general form of the antiderivative is \(F(x) + C\text{,}\) where \(C\) is an arbitrary constant. This is because when you differentiate a constant, it becomes zero, so any constant added to the antiderivative will not affect the derivative. If
\begin{equation*}
\frac{d}{dx}F(x) =f(x)
\end{equation*}
then we write
\begin{equation*}
\int f(x)\, dx = F(x) + C
\end{equation*}
to represent the family of all antiderivatives of \(f(x)\text{.}\) Here, \(f(x)\) is called integrand, \(dx\) is the differential of the variable of integration, and \(C\) is the constant of integration. Integral is called "antiderivative" because it reverses the process of differentiation. Just as: Subtraction is the "anti" operation to addition (it reverses it), Division is the "anti" to multiplication, antidifferentiation (finding the antiderivative) is the inverse operation to differentiation. Integration is inverse process of differentiation. For example: the derivative of \(x^3\) is
\begin{equation*}
\frac{d}{dx}(x^3) =3x^{3-1} = 3x^2
\end{equation*}
the antiderivative of \(3x^2\) is
\begin{equation*}
\int 3x^2\, dx = 3\frac{x^{2+1}}{2+1}+C = 3\frac{x^3}{3}+C = x^3 +C.
\end{equation*}
Here \(C\) is constant term that may have vanished during differentiation of the function. It can be determined using initial condtion or geometric condition. See examples below:
Example 4.0.1.
\begin{equation*}
\frac{d}{dx}(3x^2+5) = 3\cdot 2 x+0 = 6x
\end{equation*}
and
\begin{equation*}
\int 6x\, dx = 6\frac{x^2}{2}+C = 3x^2+C
\end{equation*}
Here \(C\) represents missing \(5\) in the function.
If \(\frac{d}{dx}f(x) = 6x^2\) such that \(F(1)=5\text{.}\) Then find the function \(f(x).\) Solution:
\begin{align*}
\frac{d}{dx}f(x) \amp = 6x^2\\
\int d[f(x)] \amp = \int 6x^2\, dx\\
F(x) \amp = 6\frac{x^3}{3}+C = 2x^3+C
\end{align*}
plug in initial condition:
\begin{align*}
F(x) \amp =2x^3+C \\
F(1) \amp =2 \cdot 1^3 +C \\
5 \amp = 2 +C \Rightarrow \, C= 5-2=3
\end{align*}
\begin{equation*}
\therefore \, f(x) = 2x^3+3
\end{equation*}
On differentiating \(f(x) = 2x^3+3 \) we get -
\begin{equation*}
\frac{d}{dx}f(x) = \frac{d}{dx}(2x^3+3) = 2\cdot 3x^2+0 = 6x^2
\end{equation*}
