Animated Solution for Mathematics - Definite Integration: Let $f$ and $g$ be continuous functions on $[0,a]$ such that $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$, then ${\int }_0^{a}f(x)g(x)dx$ is equal to :-

Mastering the King's Property

Welcome, future engineer. Today, we are not just solving an integral; we are conducting a symphony of functions.

When you look at an integral like ${\int }_0^{a}f(x)g(x)dx$, your first instinct might be to panic because you don't know what $f(x)$ or $g(x)$ are. But in the world of JEE Advanced, the absence of explicit functions is not a roadblock—it is an invitation to use properties.

Let us embark on this journey to uncover the hidden elegance of this problem.

The King's Property—Your Magic Wand

Whenever you encounter a definite integral with limits from $0$ to $a$, I want you to immediately think of the 'King's Property.' It is the most powerful tool in your calculus toolkit.

It states that for any continuous function $h(x)$, the integral ${\int }_0^{a}h(x)dx$ is identical to ${\int }_0^{a}h(a-x)dx$.

Think of this as a reflection. If you have a curve, and you flip it horizontally across the vertical line $x=\frac{a}{2}$, the area under the curve remains unchanged. This is the geometric soul of the property.

The Transformation

Let us define our target integral as $I={\int }_0^{a}f(x)g(x)dx$. Now, let us apply our magic wand.

By the King's Property, we can replace every $x$ with $(a-x)$. This gives us a new representation of the same integral:

At first glance, this looks like we have just made the expression more complicated. But look closer—we have been given clues!

The Algebraic Dance

The problem statement provides two vital pieces of information: $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$. These are not random numbers; they are the keys to the lock.

Because $f(x)=f(a-x)$, we can simplify our transformed integral immediately. The $f(a-x)$ term simply becomes $f(x)$.

Now, look at the $g(a-x)$ term. From the second condition, we know that $g(a-x)=4-g(x)$. Let us substitute these back into our integral:

Do you see the beauty of this? We have transformed a product of two unknown functions into a linear combination of the original integral. Let us expand this expression:

Using the linearity of the integral, we can split this into two parts:

The Elegant Cancellation

Pause for a moment. Look at the second term: ${\int }_0^{a}f(x)g(x)dx$. That is exactly our original integral $I$!

We have arrived at a beautiful, self-referential equation:

This is the moment of triumph. We have successfully isolated the unknown integral. By adding $I$ to both sides, we get $2I=4{\int }_0^{a}f(x)dx$.

Finally, dividing by $2$, we arrive at our destination:

We did not need to know the specific formulas for $f(x)$ or $g(x)$. We only needed to understand their symmetry and the properties of the integral.

This is the essence of JEE Advanced mathematics: it is not about brute-force calculation; it is about recognizing patterns and using the right tools to let the complexity cancel itself out.

Keep this mindset, and you will find that even the most intimidating problems have a simple, elegant heart waiting to be discovered.