The property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ is derived by making the substitution $u = a+b-x$. When $x=a$, $u=b$. When $x=b$, $u=a$. Also, $du = -dx$. Thus, $\int_{a}^{b} f(x) dx = \int_{b}^{a} f(a+b-u) (-du) = \int_{a}^{b} f(a+b-u) du$. Since the variable of integration is a dummy variable, we can replace $u$ with $x$ to get $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$. This property is particularly useful when the sum of the limits $a+b$ simplifies the integrand significantly.

Setting $a=0$ and $b=a$ in the King's Property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ yields $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(0+a-x) dx = \int_{0}^{a} f(a-x) dx$. This property is extremely useful for integrals with limits from 0 to a, especially when $f(x) + f(a-x)$ simplifies nicely.

Using the property $\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{a}^{2a} f(x) dx$. In the second integral, let $x = 2a - u$, so $dx = -du$. When $x=a$, $u=a$. When $x=2a$, $u=0$. Thus, $\int_{a}^{2a} f(x) dx = \int_{a}^{0} f(2a-u) (-du) = \int_{0}^{a} f(2a-u) du$. Replacing $u$ with $x$, we get $\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{0}^{a} f(2a-x) dx$. This is particularly useful because if $f(x)$ is odd about $x=a$ (i.e., $f(2a-x) = -f(x)$), the integral is 0. If $f(x)$ is even about $x=a$ (i.e., $f(2a-x) = f(x)$), the integral is $2\int_{0}^{a} f(x) dx$.