what's the meaning of integral from b to a, b>a?

Question

we know that
integral from b to a of f = - integral from a to b of f (1), where a But definite integral is defined as in RHS(right hand side).
Now LHS can appear, when we do substitutions of varibles, and we can switch using formula (1) to RHS form of integral.
So LHS doesn't have sense ouside of formula (1), in other words is defined by (1).
Or is there something else?

Answer 1

Actually, it can be defined the same way as the RHS definite integral, as a limit of Riemann sums, where the individual sub-areas are effectively calculated with negative width.

You can formalise this definition in an order-neutral way. Rather than looking at the integral from b to a, keep it as the integral from a to b, but we'll define it in such a way that it works whether a is greater than b or less than b.

Let a and b be distinct real numbers. Let x_0 = a, and pick a sequence x_1, ..., x_(n-1) of real numbers such that x_(i+1) is strictly between x_i and b (NB: this doesn't imply that x_i is less than b). Let x_n = b. Pick z_1, ..., z_n such that z_i is between x_(i-1) and x_i. The Riemann sum is then Σ(i=1..n) f(z_i).(x_(i+1)-x_i). Take the limit of this as max(|x_(i+1)-x_i|) -> 0. If this limit exists, we define it to be the integral.

You can then prove from this definition that if ∫(a to b) f(x) dx exists, then ∫(b to a) f(x) dx exists and is equal to -∫(a to b) f(x) dx.

Alternatively, you can look at it this way: let f have an anti-derivative F and let u = a+b-x. Then
∫(b to a) f(x) dx = ∫(a to b) f(a+b-u) (-du)
= -∫(a to b) f(a+b-u) du
= -[F(a+b-u)/(-1)][a to b]
= F(a+b-b) - F(a+b-a)
= F(a) - F(b)
= -[F(b) - F(a)]
= -∫(a to b) f(x) dx

This has the advantage of being completely independent of whether a or b is greater.

Answer 2

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