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We say that a function $f:[a,b]\to \mathbb{R}$ is Riemann integrable if for every $\epsilon>0$, there are two step functions $g_1,g_2$ such that $g_1 \leq f \leq g_2$ and $\int_a^b g_2(x)-g_1(x)<\epsilon.$

Assuming that $f$ is Reimann integrable, then using this above definition, I want to show that $$\int_{a}^{b}f(x)=\int_{-b}^{-a}f(-x)$$

In fact, one can show this using "substitution", but I am not sure how to prove it using minimal theory (i.e. based on definition)

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  • $\begingroup$ What happens if $f(x) = {{\rm ln(x)}}$ and $a = 1, b=2$ ? $\endgroup$ – MikeTeX Apr 26 '15 at 8:29
  • $\begingroup$ @MikeTeX wolframalpha.com/input/?i=integral+ln%28-x%29+from+-2+to+-1 $\endgroup$ – user231725 Apr 26 '15 at 8:35
  • $\begingroup$ Sorry, I misread your question. Indeed, the only basic way to prove this is by a simple change of variables $y=-x$. This IS minimal theory. $\endgroup$ – MikeTeX Apr 26 '15 at 8:40
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Simply consider the new functions $$\check f(t):=f(-t),\quad \check g_1(t):=g_1(-t),\quad \check g_2(t):=g_2(-t)\qquad(-b\leq t\leq -a)\ .$$ Then $\check g_1\leq\check f\leq\check g_2$, and $\int_{-b}^{-a}\bigl(\check g_2(t)-\check g_1(t)\bigr)\>dt$ is essentially the same sum of rectangles as before.

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