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We define $\int_{a}^{b} f(x)dx=-\int_{b}^{a} f(x)dx$ because defining it like this gives nice formulas later. Because this is definition, can we change it to another one? Are we going to lose every good property we know about integrals or we can work around it?

LearningMath
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  • What would you like to change it to? – zhw. Feb 21 '18 at 19:05
  • I don't know, that's also a part of my question. – LearningMath Feb 21 '18 at 19:09
  • Somewhat related: https://math.stackexchange.com/questions/856654/why-absolute-values-of-jacobians-in-change-of-variables-for-multiple-integrals-b (discusses orientation and change-of-variables in higher dimensions). – Air Conditioner Feb 21 '18 at 19:25
  • If you want to prove something, you need definitions (and maybe for the sake of efficacy, some lemmas). Have you made the connection between the definite integrals you asked about and the Fundamental Theorem of Calculus. – hardmath Feb 22 '18 at 00:44

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This relation is not defined, but a direct result of the fundamental theorem of calculus. Actully, from the Newton–Leibniz axiom $$\int _{a}^{b}f(x)\,dx=F(b)-F(a),$$ and $$\int _{b}^{a}f(x)\,dx=F(a)-F(b),$$ we get immediatly the relation $$\int_{a}^{b} f(x)dx=-\int_{b}^{a} f(x)dx.$$

Jordan He
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    It's not a theorem. See the accepted answer here: https://math.stackexchange.com/questions/1769597/interchange-of-limits-of-definite-integral – LearningMath Feb 21 '18 at 19:17
  • Maybe my writing. I did not say it is a theorem. I'm saying it can be derived from the fundamental theorem of calculus, or the Newton-Leibniz axiom. – Jordan He Feb 21 '18 at 19:27
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The reason for this definition becomes clearer when you start to talk about integration on manifolds, especially curves. We may imagine that a curve $\Gamma$ in e.g. three-dimensional space to be the trajectory of some particle in space, and we may define a choice of orientation on the curve to be a choice of a direction in which said particle is going to trace out the curve (from endpoint $A$ to endpoint $B$ or viceversa). Then we define the integral of a function over the curve to be $$\int_\Gamma f ds = \int_a^b f(g(t)) g’(t) dt$$ where $g : [a,b] \to \Gamma$ is a parametrization of the curve. This definition makes it so that the value of the integral does not depend on the parametrization we choose (you use the change of variables theorem to prove this) but its sign does depend on the orientation we choose on $\Gamma$ (the sign of $g’$, which of course depends on $g$).

This definition is central to physics. Suppose $\mathbf f$ is a conservative field (meaning there exists a real function $g$ such that $\nabla g = \mathbf f$) and take a curve $\Gamma \subseteq \mathbb R^3$ parametrized by a function $\varphi$ like before. Then the definition of integral we gave above (which extends easily to vector functions like $\mathbf f$) makes it so that if $\Gamma$ is closed, that is its endpoints are the same point, then the integral of $\mathbf f$ over $\Gamma$ vanishes. Indeed you may imagine splitting the curve into two curves, one from a point $A$ to $B$ and one from $B$ to $A$, then employing the fact that $\mathbf f$ is conservative to state that the value of the integral calculated over both curves depends only on their endpoints, and finally using the fact that the sign of the integral depends on the orientation of the curves to conclude that one integral is the opposite of the other and so they add up to zero, QED. Without this result, much of physics is useless.

giobrach
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