Show that the value of a definite integral is unity

Solution 1:

We show that $I=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}\,dx$ equals $1$, where $f(x)=\sqrt{\log(9-x)}$.

Proof. By making a substitution $y=6-x$, we get $$I=\int_4^2 -\frac{f(6-y)}{f(y)+f(6-y)}\,dy=\int_2^4 \frac{f(6-y)}{f(y)+f(6-y)}\,dy.$$

Therefore $$\begin{align*}2I&=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}dx+\int_2^4 \frac{f(6-y)}{f(y)+f(6-y)}dy\\ &=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}dx+\int_2^4 \frac{f(6-x)}{f(x)+f(6-x)}dx=\int_2^4 1\,dx =2.\end{align*}$$

Edit. In the last part, $y$ is a dummy variable and can be changed to $x$ or any other variable you like.

Edit2. If you don't like the same $x$ being used, you could use another variable, say $s$, so that with two substitutions $$\begin{align*}2I&=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}dx+\int_2^4 \frac{f(6-y)}{f(y)+f(6-y)}dy\\ &=\int_2^4 \frac{f(s)}{f(s)+f(6-s)}ds+\int_2^4 \frac{f(6-s)}{f(s)+f(6-s)}ds=\int_2^4 1\,ds =2.\end{align*}$$

Solution 2:

$ \int^a_b f(x)\,dx $ = $ \int^a_b f(a+b-x)\,dx $ We can prove this by changing the dummy variable x to a+b-x, we get the integrand as $-f(a+b-x)\,dx$ and by changing the limits accordingly with $a+b-b$ and $a+b-a$, i.e. Now the integral becomes $-\int^b_af(a+b-x)dx$=$ \int^a_b f(a+b-x)dx $

The function given in the question satisfies the property that $f(x)+f(6-x)=1$. Let the integral be $I$ then $2I= \int^a_b f(x)\,dx+ \int^a_b f(a+b-x)dx=\int^a_b f(x)+f(a+b-x)\,dx=\int^a_b\,dx=a-b$, In your case $a-b=2$, hence $2I=2$. So $I=1$.