Animated Solution for Mathematics - Definite Integration: The value of $5050\frac{{\int }_0^1{\left(1-x^{50}\right)}^{100}dx}{{\int }_0^1{\left(1-x^{50}\right)}^{101}dx}$ is.

Taming the Monster Integral

Welcome, warriors of JEE! Today, we are going to face a problem that, at first glance, looks like a mathematical monster. We are asked to evaluate:

Just looking at those powers—50, 100, 101—is enough to make anyone want to skip the question. But here is the secret: in JEE Advanced, whenever you see high powers in a definite integral that seem impossible to expand, your mind should immediately pivot to the concept of Reduction Formulas. We are not here to calculate; we are here to find a pattern.

Defining the General Integral

To tame this beast, let us define a general integral, $I_n$, to represent the structure we are dealing with. Let:

By doing this, our original expression becomes much simpler: $5050\frac{I_{100}}{I_{101}}$. Now, our goal is clear: we need to find a relationship between $I_n$ and $I_{n-1}$. This is the heart of the reduction formula technique.

The Art of Integration by Parts

We need to relate $I_n$ to $I_{n-1}$. The most powerful tool in our calculus toolkit for this is Integration by Parts.

But wait, we only have one function here. The trick is to introduce a 'dummy' function. We treat the integrand as $1\cdot (1-x^{50}{)}^n$.

Let $u=(1-x^{50}{)}^n$ and $dv=dx$. Now, we differentiate $u$ using the chain rule: $du=n(1-x^{50}{)}^{n-1}\cdot (-50x^{49})dx$. And for $v$, we simply integrate $dv$ to get $v=x$.

The Vanishing Boundary

Now, let's apply the Integration by Parts formula: $\int udv=uv-\int vdu$. Our integral becomes:

Let's look at that boundary term $[x(1-x^{50}{)}^n{]}_0^1$. At $x=1$, we have $1(1-1^{50}{)}^n=0$. At $x=0$, we have $0(1-0^{50}{)}^n=0$.

The boundary term vanishes completely! This is the elegance of the problem—the math is clearing the path for us.

The Algebraic Bridge

We are left with $I_n=50n{\int }_0^1{x}^{50}(1-x^{50}{)}^{n-1}dx$. We have $x^{50}$ inside, but we want $(1-x^{50})$ to match our original definition.

Here is the classic algebraic trick: rewrite $x^{50}$ as $1-(1-x^{50})$. Substituting this back, we get:

Distributing the terms, we get:

This is exactly $I_n=50n(I_{n-1}-I_n)$.

The Final Victory

Now, we just solve for the ratio. $I_n=50n{I}_{n-1}-50n{I}_n$, which rearranges to $I_n(1+50n)=50n{I}_{n-1}$.

Therefore:

For $n=101$, we get:

Finally, multiplying by the original coefficient 5050, we get $5050\times \frac{5051}{5050}=5051$. The 5050s cancel out, and we are left with the beautiful, clean answer: 5051.