Animated Solution for Mathematics - Definite Integration: If $I(m,n)={\int }_0^1{x}^{m-1}(1-x{)}^{n-1}dx,$ m, $n>0$, then $I(9,14)+I(10,13)$ is

Unlocking the Beta Function

Welcome, future engineers. Today, we are not just solving an integral; we are embarking on a journey into the heart of mathematical symmetry.

When you first look at the expression $I(m,n)={\int }_0^1{x}^{m-1}(1-x{)}^{n-1}dx$, it is natural to feel a sense of hesitation. It looks like a standard calculus problem, but it is actually a gateway to the beautiful world of the Beta function.

In the JEE Advanced examination, the examiners often test your ability to see beyond the surface. They don't want you to calculate; they want you to observe.

The Setup

Let us begin by demystifying the given function. We are asked to evaluate $S=I(9,14)+I(10,13)$.

Instead of diving into the deep end of integration, let us write out exactly what these terms represent. For $I(9,14)$, we substitute $m=9$ and $n=14$ into our definition. The exponent $m-1$ becomes $8$, and $n-1$ becomes $13$.

Thus, we have:

Now, let us do the same for $I(10,13)$. Here, $m=10$ and $n=13$. The exponents become $9$ and $12$, respectively:

Look at these two integrals side-by-side. They are not just random functions; they are siblings. They share the same limits of integration, from $0$ to $1$, which is our first clue that we can combine them.

The Power of Linearity

One of the most powerful tools in your arsenal is the linearity of the definite integral. It tells us that the sum of two integrals is the integral of their sum.

So, let us bring them together:

Now, I want you to pause. Look at the integrand: $x^8(1-x{)}^{13}+x^9(1-x{)}^{12}$. This is where the magic happens. We have common factors hidden in plain sight.

The Algebraic Revelation

To simplify this, we must extract the highest common power. For the $x$ terms, the common factor is $x^8$. For the $(1-x)$ terms, the common factor is $(1-x{)}^{12}$.

Let us pull these out:

Do you see it? The expression inside the square brackets is $(1-x)+x$. The $-x$ and $+x$ cancel each other out with perfect precision, leaving us with just $1$.

This is the moment where the complexity collapses. The entire expression simplifies to:

The Final Identification

We have arrived at the finish line. We need to map this back to our original definition $I(m,n)={\int }_0^1{x}^{m-1}(1-x{)}^{n-1}dx$.

By comparing the exponents, we see that $m-1=8$, which implies $m=9$, and $n-1=12$, which implies $n=13$. Therefore, our sum is simply $I(9,13)$.

This result is not just an answer; it is a fundamental property of the Beta function:

By recognizing this, you save precious minutes that you can spend on more challenging problems. Remember, in JEE Advanced, the most elegant solution is often the one that relies on deep conceptual understanding rather than brute force. Keep practicing, keep observing, and keep falling in love with the beauty of mathematics.