Animated Solution for Mathematics - Definite Integration: The value of integral, ${\int }_3^6\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx$ is

Welcome, future engineers! Today, we are going to embark on a journey through one of the most elegant problems in integral calculus.

When you first look at the integral

your instinct might be to panic. You see nested square roots, a fraction, and limits that don't immediately suggest a simple antiderivative.

You might be tempted to try a complex substitution, perhaps setting $u=\sqrt{x}$ or $u=\sqrt{9-x}$. I want you to pause and take a deep breath. In the JEE Advanced, the most difficult-looking problems often have the most beautiful, simple solutions hidden just beneath the surface. This problem is a classic example of symmetry as a weapon.

If you were to attempt direct integration, you would quickly find yourself in a labyrinth. You would be dealing with terms like $\sqrt{x}$ and $\sqrt{9-x}$ that simply do not play nicely with standard integration rules.

The algebra would balloon, the terms would become unmanageable, and you would likely lose your way. This is the trap. The examiners want to see if you can recognize the structure of the problem rather than just blindly applying mechanical rules. We are not here to calculate; we are here to observe.

Your Secret Weapon

In the world of definite integrals, there is a theorem so powerful, so transformative, that we call it the King's Property. It states that for any continuous function $f(x)$, the integral

Think about what this means geometrically. We are essentially flipping the function across the midpoint of the interval $[a,b]$. If the function has a specific symmetry, this flip leaves the area under the curve unchanged.

In our case, the lower limit $a=3$ and the upper limit $b=6$. Their sum is $a+b=9$. This is the "aha!" moment. The number $9$ is the key that unlocks this problem.

Let us apply this property. We define our integral as

Now, we replace every instance of $x$ with $(9-x)$. The integral becomes

Look closely at the denominator. The term $\sqrt{9-(9-x)}$ simplifies beautifully to $\sqrt{x}$. So, our transformed integral is

Notice something incredible? The denominator is identical to our original integral!

Now, we perform the masterstroke. We add our original integral $I$ to our transformed integral $I$. This gives us

Because the denominators are the same, we can combine the numerators:

The numerator and the denominator are now exactly the same! They cancel out to leave us with the integral of $1$.

We are left with

This is the beauty of mathematics. We started with a terrifying expression involving square roots and ended with the simplest possible integral.

The integral of $1$ from $3$ to $6$ is simply $[x{]}_3^6$, which is $6-3=3$. Thus, $2I=3$, which means $I=\frac{3}{2}$.

You have successfully navigated the trap, utilized the King's Property, and arrived at the solution with elegance and precision. Remember this: whenever you see symmetric limits and complementary radical terms, do not rush to calculate. Pause, look for the symmetry, and let the King's Property do the heavy lifting for you.