Notion of an integral as a summation

sachin
Messages
61
Reaction score
7
Homework Statement
the notion of an integral is a special form of summation of differentials and an indefinite integral is also an integral with limits 0 to x what is conventionally written without the limits,
Relevant Equations
∫f(x)dx with limits 0 to x = ∫f(x)dx
While analyzing the foundation of calculus,
I am finding that the notion of an integral is a special form of summation of differentials and an indefinite integral is also an integral with limits 0 to x what is conventionally written without the limits,

the notation is given in the image,

Pl confirm if my assumption is correct,
thanks.
notion of an integral.png
 
Physics news on Phys.org
That is not correct. The antiderivative of the function, ##f(x)##, with no specified limits indicates a function of ##x## whose derivative is ##f##. As such, any constant can be added and it would still be an antiderivative. That constant is sometimes not indicated, but it should be. In other words, if ##d/dx\ g(x) = f(x)##, then ##\int f(x) dx = g(x)+C##, where ##C## is an (as yet) unknown, arbitrary constant. There are many problems where the correct value of ##C## must be determined.

UPDATE: See @pasmith 's comment #5 about the possibility of the constant ##C## being different in two sections of a domain that is not connected.
 
Last edited:
It is also poor notation to have the dummy variable identical with a bound.
 
  • Like
Likes PhDeezNutz and FactChecker
That's not correct. The indefinite integral is an equivalence class of functions whose derivative is the integrand.

Technically, you cannot use the same variable ##x## in the integrand and as a bound. We have,
$$\int_0^x f(t)dt = F(x) - F(0)$$That is a specific function of ##x##. And$$\int f(x)dx = F(x) + C$$That is a set of functions (equivalence class). Where ##F(x)## is any function where ##F'(x) = f(x)##. These are both examples of the Fundamental Theorem of Calculus.
 
PeroK said:
And$$\int f(x)dx = F(x) + C$$That is a set of functions (equivalence class). Where ##F(x)## is any function where ##F'(x) = f(x)##.

A point which is often missed is that if the domain of f is connected then C must indeed be a constant, but if the domain is not connected (eg. f(x) = x^{-1}, x \in \mathbb{R} \setminus \{0\}) then C can take different values on different connected components of the domain.
 
  • Like
Likes MidgetDwarf and FactChecker
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top