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The 'King's property' of definite integration is:
A$\int_a^b f(x)\,dx = \int_a^b f(b-x)\,dx$
B$\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$
C$\int_a^b f(x)\,dx = \int_a^b f(x+a)\,dx$
D$\int_a^b f(x)\,dx = \int_a^b f(b/x)\,dx$
Answer & Solution
Correct answer: B. $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$
Property V: substitute $t = a + b - x$. The interval $[a, b]$ maps to itself reversed, giving $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$. Useful for symmetric simplification.
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