Why is integration the inverse of differentiation, I mean why do I get the same function when I integrate and then differentiate the result?
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$\begingroup$I'm going to assume that you want an intuitive explanation rather than the mathematical details. An intuitive explanation is that
integration is like saying "give me a running total of all the values this function has had up to a certain point"
differentiation is like saying "how much is the value of this function going to increase from a given point to the next point"
So you can see if I ask for a running total, I'm adding up values the function takes one by one, and integration can be roughly read as accumulation. On the other hand if I ask how much a function is going to increase from one point to the next I'm asking for the difference between one value and the next.
The intuitive relationship that differentiation undoes integration can then be seen as follows. If you take the difference between one value of a running total and the next value of the running total, you get what you added to the current value of the running total to get the next value of the running total, which is the function you had in the first place before you integrated it.
This is illustrated pictorially below where the simple function $y=1$ at the top of the diagram is accumulated over an interval size (or step size) of 1 to give the step function at the bottom left. You can see that each time you go from $x$ to $x+1$ we accumulate a yellow box on the bottom left graph. Conversely to go from the bottom left graph to the top graph we just take the difference between the total at $x$ and the total at $x+1$, which is a yellow box.
What happens in calculus is that the points are moved closer and closer together until the intervals over which functions are accumulated (integration) or differenced (differentiation) are infinitesimally small. As the interval size is made smaller and smaller the steps gradually get closer and closer together until in our example you get the smooth function $y=x$ at the bottom right of the diagram. This is equivalent to taking the area under the curve to get to $y=x$, and taking the gradient of the curve to go back from $y=x$ to $y=1$.
The relationship between integration and differentiation is a very important relationship in calculus, so important that it is called the Fundamental Theorem of Calculus.
$\endgroup$ $\begingroup$I tried to make this explanation as intuitive and simple as possible.
The figure below summarizes everything
The derivative of a function $y = f(x)$ is $dy/dx$, i,e., difference in $y$ over a very small difference in $x$ for each point.
One can draw a graph from derivative function $y = f’(x)$
One can compute the definite integral, i.e., the area from curve to x-axis between any 2 values of $x$ . How to interpret this number?
Suppose one divides the area of the curve to the x-axis into thousands of small rectangles between $x =x_{A}$ and $x =x_{B}$.
Each rectangle has width $dx$ (as small as one likes) and height $dy / dx$ , which is the derivative of the function at each point. For instance, $x=x_{P}$ in function $f(x)$ corresponds in $f’(x)$ to the point $(x_{P}, dy/dx(x_{P}))$, i.e., the derivative in point $x=x_{P}$.
The area of any rectangle between $x =x_{A}$ and $x =x_{B}$ is $dy / dx \times dx = dy$, which is the microvariation of $y$ of the original function $f$ from the start of this rectangle to the end of it.
Thus, if one adds all areas of the rectangles between $x =x_{A}$ and $x =x_{B}$, one has the total variation of $y $ in the original function $f$ between $x =x_{A}$ and $x =x_{B}$, that is, $f (x_{B}) - f (x_{A})$, that can be written as $y_{B}-y_{A}$
In the end, the integral of the derivative of a function returns to the original function difference between any 2 points the one selects.
For that reason, the indefinite integral of $f’(x)$ is the function $f(x)$. A constant appears because is it coherent with a general function (for calibration objectives) and it disappears in any definite integral calculation, that is a difference between two function evaliations.
$\endgroup$ $\begingroup$I will suggest you to go and see the definition of integration (indefinite). There you will see that integration is a method to find the function when at any point in domain, its differentiation is provided to you. And so it becomes the inverse of differentiation.
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