Why we use differential element to get general form? When we study any physical system we use differential element to get an equation then integrate over specific period to get a general form. 
Why we don't get directly a general form without differential element?
please explain in more details. 
 A: We don't always need calculus to obtain a general form: static problems (a weight sitting stationary on a table, for instance) often don't require calculus.
Calculus is therefore typically necessary where change occurs.
Let's say a physical, observable quantity $u$ (distance, pressure or whatnot) changes in function of an observable parameter $t$ (time, distance, potential energy or whatnot) according to some function $f(t)$:
$$u(t)=f(t)$$
Between two values of $t$ we can now define an approximate gradient or rate of change $g$, acc.:
$$g=\frac{\Delta u}{\Delta t}=\frac{u_2-u_1}{t_2-t_1}$$
Now we can estimate a value for the change of $u$ for the case:
$$(u_4-u_3) \approx g \times (t_4-t_3)$$
It can be shown that unless $f(t)$ is a strictly linear function, the previous expression is at best approximate, at worst a very poor predictor.
We use the limit definition of derivatives to turn the intervals (e.g. $\Delta u$, $\Delta t$) into differentials (e.g. $du$, $dt$):
$$\lim_{\Delta t \to 0}\frac{\Delta u}{\Delta t}=\frac{du}{dt}=u'(t)$$
Then, to calculate a change of $u$ between, say $t_4$ and $t_3$ we can use integration:
$$\frac{du}{dt}=u'(t)$$
$$\int_{u_3}^{u_4}du=\int_{t_3}^{t_4}u'(t)dt$$
$$\implies u_4-u_3=\int_{t_3}^{t_4}u'(t)dt$$
This provides an exact solution.
A: We usually know the expression or "formula" for a small part, and this formula takes different values across parts, so one easy way to solve the problem, if this is the case, is by adding together all the mini formulas which is what an integral does. 
For instance you want to calculate the area inside a circle of radius R. How do you start? Well, if you know the formula for the length of the circumference, $2\pi r$, you can
divide the circle into small rings, where you know the area of each ring, because if the width of the ring is infinitesimally small, you can calculate its area as  $A_{ring}=2\pi r \Delta r$, only valid if the ring is infinitesimally thick. Then you have to add all the areas of all the rings starting with the ring at zero radius and ending with the  ring at radius R. 
Thus $A_{total}=Add(A_{ring})=\int2\pi r d r$, where there is a change in notation, the addition by an integral $\int$ and the thickness $\Delta r$ by $dr$, to remark it is an infinitesimal quantity. Using the rules of integration you can easily evaluate that to $A_{total}=\pi R^2$. I hope this simple example explains why this technique is so powerful. 
A: Because we usually deal with continuous functions in physics. For instance, we know that internal energy of a body is continuous. I.e. if a body has internal energy of $U_1$ and $U_2$ in two different states and $U_2\gt U_1$ then certainly there are infinite states between 1 and 2 so that internal energy of the body is between $U_1$ and $U_2$. In other words, internal energy of a body cannot change from $U_1$ to $U_2$ without experiencing internal energies amounts between $U_1$ and $U_2$.
For example, we know that internal energy of an ideal gas is a function of its temperature $u=f(T)$. We say if temperature changes from $T$ to $T+\mathrm dT$ then internal energy will change from $u$ to $u+\mathrm du$ ($\mathrm du$ and $\mathrm dT$ are infinitesimal variations) and we don't expect that if temperature changes by an infinitesimal amount, internal energy changes by a definite amount.
