Animated Solution for Mathematics - Definite Integration: The value of ${\int }_2^3\frac{\sqrt{x}}{\sqrt{5-x}+\sqrt{x}}dx$ is $\dots \dots \dots$

Imagine you are standing before a mountain, a seemingly insurmountable peak of calculus. The problem before us is to evaluate the definite integral:

At first glance, this expression looks intimidating. The square roots, the fraction, and the limits from $2$ to $3$ feel like a trap designed to lead you into a labyrinth of complex substitutions. But in the world of JEE Advanced, complexity is often a mask for elegance.

A Secret Weapon

Whenever you encounter an integral with a complex algebraic fraction involving square roots, your intuition might scream for substitution. But pause. Before you dive into the deep end, look at the limits of integration.

The lower limit is $a=2$, and the upper limit is $b=3$. Their sum is $a+b=5$. This is the magic key. We are going to use the King's Property of definite integrals:

This property is a profound statement about the nature of area under a curve. It tells us that if we flip the function across the midpoint of the interval, the total area remains unchanged.

The Magic of $5-x$

Let us apply this property to our integral. We replace every instance of $x$ in our integrand with $5-x$.

The numerator $\sqrt{x}$ becomes $\sqrt{5-x}$. The denominator $\sqrt{5-x}+\sqrt{x}$ transforms into $\sqrt{5-(5-x)}+\sqrt{5-x}$.

Simplifying the denominator, $\sqrt{5-(5-x)}$ becomes $\sqrt{x}$. Thus, our transformed integral becomes:

Notice the elegance here. The denominator is now $\sqrt{x}+\sqrt{5-x}$, which is identical to our original denominator, just with the terms swapped.

Adding the Integrals

Now, we have two expressions for the same area $I$. Let us call the original integral Equation (1) and our transformed integral Equation (2). If we add them together, we get:

Because the limits of integration are the same, we can combine these into a single integral:

Look at that fraction! The numerator is exactly the same as the denominator. The entire expression collapses into the integral of a constant:

We have arrived at the finish line. The integral of $1$ with respect to $x$ is simply $x$. Evaluating this from $2$ to $3$:

So, $2I=1$. To find our original integral $I$, we simply divide by $2$:

This is the power of mathematical insight. By recognizing the symmetry and applying the right tool, we transformed a complex problem into a simple arithmetic calculation. Remember, in JEE Advanced, the most beautiful solutions are often the ones that reveal the hidden simplicity beneath the surface.