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Without using the Fundamental Theorem of Calculus, I want to show that $$ \int_a^bf(x)dx=-\int_{b}^{a} f(x)dx $$ when $b < a$. Apparenly it follows from the rule $$ \int_a^c f(x)dx=\int_a^b f(x)dx+\int_b^c f(x)dx $$ which holds for all real numbers $a, b \ \mathrm{and} \ c$ by setting $c=a$. But where does this rule come from? Or is it some kind of axiom of integrals? And don't we also have to define separately that $$ \int_a^a f(x)dx=0? $$

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mathslover
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  • Does this help https://math.stackexchange.com/questions/232455/is-integration-from-a-to-b-same-or-b-to-a-or-is-negative ? – Jakeup Jul 30 '21 at 17:26
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    $\int_a^bf(x)dx=-\int_{b}^{a} f(x)dx$ is more a matter of definition. So it is not something to be proved. The Riemann integral $\int_a^b$ is defined for $a<b$, and we extend that to the case $b<a$ using this definition. – GEdgar Jul 30 '21 at 17:30
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    As mentioned in other comments, it is more or less a definition. But of course, FToC (or the additivity of integral) justifies why the definition $\int_{a}^{b}=-\int_{b}^{a}$ should be useful. – Sangchul Lee Jul 31 '21 at 02:06

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