The 'King's property' of definite integration is:
- $\int_a^b f(x)\,dx = \int_a^b f(b-x)\,dx$
- $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$
- $\int_a^b f(x)\,dx = \int_a^b f(x+a)\,dx$
- $\int_a^b f(x)\,dx = \int_a^b f(b/x)\,dx$
Show answer and 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.