Animated Solution for Mathematics - Definite Integration: If $f$ and $g$ are continuous function on $[0,a]$ satisfying $f(x)=f(a-x)$ and $g(x)+g(a-x)=2$, then show that ${\int }_0^{a}f(x)g(x)dx={\int }_0^{a}f(x)dx$

Welcome, future engineer. Today, we are not just solving an integral; we are uncovering a hidden symmetry. You are presented with an integral,

and you are asked to prove that it equals ${\int }_0^{a}f(x)dx$.

The best problems in JEE Advanced are not about brute-force calculation; they are about observation. We are given the conditions $f(x)=f(a-x)$ and $g(x)+g(a-x)=2$.

If you were to graph $f(x)$, you would see a curve that is perfectly symmetric about the vertical line $x=a/2$. The function $g(x)$ satisfies a point symmetry, indicating that these functions are 'aware' of the interval $[0,a]$.

The Magic Wand

We introduce the 'King's Property' of definite integrals, which is rooted in the substitution $u=a-x$. The property states:

This is powerful because it allows us to transform the integrand without changing the area under the curve. We apply this to our integral $I$ by replacing every $x$ with $a-x$:

Now, we use our given conditions to simplify the expression. We know $f(a-x)=f(x)$, so we substitute this into our integral:

Next, we use the condition $g(x)+g(a-x)=2$ to isolate $g(a-x)=2-g(x)$. Substituting this into the integral yields:

Let us distribute $f(x)$ into the bracket:

By the linearity of integrals, we can split this into two parts:

Observe that the second term is exactly the original integral $I$. We have created a self-referential equation:

The algebra is now trivial. We add $I$ to both sides to obtain:

Dividing by $2$, we arrive at the beautiful, clean result:

We have proven the statement without needing the specific forms of $f(x)$ or $g(x)$. This is the essence of mathematics: finding the underlying structure that governs the behavior of the system.