Why do we need upper and lower limits in definite integration?

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Upper and lower limits in definite integration are essential for defining the specific interval over which the integration occurs, allowing for the calculation of the area under a curve. This contrasts with differentiation, which focuses on instantaneous rates of change at a single point. Definite integration is linked to the concept of summing infinitesimal contributions over a range, as described by the Fundamental Theorem of Calculus. The discussion highlights that while indefinite integration serves as the inverse of differentiation, definite integration requires boundaries to quantify changes over an interval. Understanding these concepts is crucial for applying calculus in physics and other fields.
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My question is that why is their a need for both upper and lower while calculating Definite Integrals.

The question arose when i thought of Definite integration as something related to Differentiation. Or is it that only Indefinite Integration is directly related to differentiation.

In differentiation, we get the slope or rate of change.

So if i differentiate s(displacement) w.r.t. t(time), i get ds/dt = v(Velocity). By putting just one value of t, i get a value of velocity at that instant.

But to get back that one value of s(displacement), why do we need two values of t(time)?
 
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Integration is obviously related to differentiation, both definite and indefinite ones. However, the inverse of differentiation is indefinite integration. Definite integration is an infinite sum of infinitely small things that just so happens to be computed using antiderivatives, see the Fundamental Theorem of Calculus. We need limits to determine over what interval we are going to perform the sum on.
 
What, exactly, is your understanding of the derivative and integral? Most texts introduce the derivative as "slope" of the curve (actually slope of the tangent line) at a given point, integral as area under a curve. "At a given point" is necessarily a single value of x while "area" has to have bounds- the curve as upper bound, y= 0 as lower bound, and two x values as left and right bounds.

In Physics, we can think of the derivative (of the distance function) as the speed at a given instant- one value of t. The derivative of the velocity function is the increase in distance. Increase over what time interval. We have to have a "beginning" time as well as a final time in order to talk about an increase.
 
Of course definite integration is something related to differentiation.
$$\mathop{f}(b)-\mathop{f}(a)=\int_a^b \! {\mathop{f} }^ \prime (x) \, \mathop{dx}$$
I would not focus on the number of values it is not a central idea.
Traditionally integration is introduced as associating a value to a function and interval and differentiation associates value to a point and function. These are mostly the same as we can think of a point as a small interval and we can think of an interval as a point.
 
HallsofIvy said:
What, exactly, is your understanding of the derivative and integral? Most texts introduce the derivative as "slope" of the curve (actually slope of the tangent line) at a given point, integral as area under a curve. "At a given point" is necessarily a single value of x while "area" has to have bounds- the curve as upper bound, y= 0 as lower bound, and two x values as left and right bounds.

My understanding of the topic is what is mentioned by you and other in this thread.
Differentiation is the process of finding the derivative. Derivative being nothing but the instantaneous (limit x->0) rate of change of y(function of x) w.r.t. x By this we see that derivative is slope of tangent to the curve at the point in consideration.

Indefinite integration is the inverse of differentiation. Definite Integration is the process of finding area under the curve as the area corresponds to the product of the Slope and the independent variable which helps us find the change in The dependent variable( Am i right?)

HallsofIvy said:
In Physics, we can think of the derivative (of the distance function) as the speed at a given instant- one value of t. The derivative of the velocity function is the increase in distance. Increase over what time interval. We have to have a "beginning" time as well as a final time in order to talk about an increase.

Nice. Exactly what i wanted.
Edit : sorry but i think you meant the integration of velocity function as increase in distance?

One more thing. Sometimes d(any variable here) is a change like in dv(velocity) and sometimes it is the just infinitesimally small quantity like dW. Sometimes i feel both are the same and sometimes i dont. Which one is true?
 
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Also can you tell me why Fundamental Theorem of Calculus is Useful in Physics.
I mean if we take the example of ds/dt= V , then we can say that ds = V.dt which indicates that the area of the product of V and dt gives us a number which is equal to the small change is s all from the equation. I don't think i used any F.T.C here , so why is it considered so important atleast in this case?
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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