Flipping the limits the of the integral.

Question

If my initial point is $a$ and final point is $b$ and there is a function $ f(x) $ continuous in this interval. I have to calculate the integral $$I= \int_ {a}^{b} f(x) dx~~~~~~~~~~~~~(1)$$ Now, if I want my $b$ to be the lower limit and $a$ to be the upper limit, then by the property of definite integral $$ I= - \int_{b}^{a} f(x) dx~~~~~~~~~~(2)$$Here comes a misconception, in this equation (2) can I say that $b$ is the initial point. I mean in my actual problem I defined $b$ to be the final point but due to some mathematics I got $b$ in a place where it should be considered the initial point.
I myself reasoned that although $b$ may be the starting point but what I'm doing going in opposite direction as earlier and this may cause problem in vector integrals.
Let's assume $\mathbf{F}$ is a force of some kind and $d\mathbf{x}$ is the displacement and I have to calculate the work done from $ a $ to $ b$, then $$ W = \int_{a}^{b} \mathbf{F} \cdot d\mathbf{x}~~~~~~~(i)$$ $$W= -\int_{b}^{a} \mathbf{F} \cdot d\mathbf{x}~~~~~~~(ii)$$ In the equation $(i)$ my force and displacements were in same directions (just assume that this was the scenario, that a force is in direction from $a$ to $b$) and in equation $(ii)$ force and displacement would be in different directions if I were to take $b$ as the starting point.

So, my question is can we treat $b$ as the initial point or will it only be the final point, because flipping the limits has caused the minus sign to take care of all things. I mean what will be the work, will it be $$ - (U_b - U_a)$$ or $$ - (U_a - U_b) $$ U being the potential energy and subscripts show the particular point of $U$

Thank you any help will be much appreciated.

Answer 1

Consider $\int_{a}^{b} {f(x)dx} = F(b) - F(a)$. ($\int {f(x)dx} = F(x) + C$)
So, $\int_{a}^{b} {f(x)dx} = F(a) - F(b) = - (F(b) - F(a)) = -\int_{a}^{b} {f(x)dx}$

Please imagine we're thinking about areas, not "directions".

Answer 2

Sorry for late.

Consider

$\int f(x) d x$ = $F(x) + C$.

I=$\int^b_a f(x) d x$=$F(b)-F(a)$

I'=$\int^a_b f(x) d x$=$F(a)-F(b)$ $=-(F(b)-F(a))$=−I

We're thinking of areas.

The work will be −(−). We can treat as the initial point. Why don't you treat it?

I referred to this cite https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-3-the-definite-integral-and-its-applications/part-a-definition-of-the-definite-integral-and-first-fundamental-theorem/session-50-combining-the-fundamental-theorem-and-the-mean-value-theorem/.

Could you read?

Answer 3

On the contrary, in your second form of the integral of differential work $$-\int_b^a\mathbf F\cdot\mathrm d\mathbf x,$$ the negative sign is not due to the fact that the force is now going in opposite direction. If you think of the parameter as time, then what you've done is not flipping the interval from $[a,b]$ to $[b,a]$ as you think, but rather from $[a,b]$ to $-[b,a],$ which is the same as before, only written differently.

Thus, whereas you can think of the force now working from the end of the path to its beginning ($b\to a$), the $-$ sign would undo this effect, and you'd still have the same value as before. So this is still equal to $$\int_a^b\mathbf F\cdot\mathrm d\mathbf x.$$

Answer 4

As far as I know that flipping the limits of the integrals works when the integrand in a function and not a vector or a vector dot product. $$ \int_{a}^{b} \mathbf{F} \cdot d\mathbf{x} = \int_{a}^{b} F~dx ~cos0 = \int_{a}^{b}F~dx$$ Now if we flip the limits then we won't need to bother about the directions anymore because we have already taken that into our account $$ -\int_{b}^{a} F~dx$$ and this gonna equal as before.