Section 11.10: fix sign in Exercise 11.10.3 statement#533
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The reflection x ↦ -x preserves interval length, so
∫_{[-b,-a]} f(-x) dx = ∫_{[a,b]} f, with no sign change.
Counterexample to the current statement (f = 1, a = 1, b = 2):
LHS = integ 1 [-2,-1] = 1, but RHS = -integ 1 [1,2] = -1.
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The reflection x ↦ -x preserves interval length, so ∫_{[-b,-a]} f(-x) dx = ∫_{[a,b]} f, with no sign change.
Counterexample to the current statement (f = 1, a = 1, b = 2): LHS = integ 1 [-2,-1] = 1, but RHS = -integ 1 [1,2] = -1.