Animated Solution for Mathematics - Definite Integration: Let $\beta (m,n)={\int }_0^1{x}^{m-1}(1-x{)}^{n-1}dx,m,n>0$. If ${\int }_0^1(1-x^{10}{)}^{20}dx=a\times \beta (b,c)$, then $100(a+b+c)$ equals

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an integral; we are uncovering the hidden architecture of the Beta function.

Many students see an integral like ${\int }_0^1(1-x^{10}{)}^{20}dx$ and feel a sense of dread. It looks like a binomial expansion nightmare, doesn't it?

But I want you to take a deep breath. In the world of advanced calculus, complexity is often just a mask for a deeper, more elegant structure waiting to be revealed.

The Vision

Let us look at the definition of the Beta function:

This is our North Star. It is a beautiful, symmetric, and powerful tool.

Now, look at our target integral:

Notice the dissonance? The Beta function demands a simple variable $x$ inside the parenthesis, but we have $x^{10}$.

This is the 'trap'—the moment where most students try to expand $(1-x^{10}{)}^{20}$ using the binomial theorem, which would lead to a mountain of terms. We must resist that urge. Instead, we must perform a transformation that aligns our integral with the Beta function's DNA.

The Substitution Strategy

If the problem is the $x^{10}$, then the solution is to make $x^{10}$ disappear. Let us define a new variable, $t$, such that $t=x^{10}$.

This is the spark of genius. By doing this, we instantly transform the term $(1-x^{10}{)}^{20}$ into $(1-t{)}^{20}$.

Now, we are halfway to the Beta function form! But we must be rigorous. If we change the variable, we must change the differential $dx$.

Differentiating $t=x^{10}$ with respect to $x$, we get $dt=10x^{9}dx$. This means:

The Differential Dance

We are not done yet. We have $dx$ in terms of $x$, but we need it in terms of $t$. Since $t=x^{10}$, it follows that $x=t^{1/10}$.

Therefore, $x^9=(t^{1/10}{)}^9=t^{9/10}$. Substituting this back into our expression for $dx$, we get:

This is the hidden beauty of the substitution! That $t^{-9/10}$ term is exactly what we need to complete the $x^{m-1}$ part of the Beta function definition.

The Transformation

Now, let us assemble our masterpiece. The limits of integration, as we checked, remain from $0$ to $1$.

Substituting everything into our integral $I$, we get:

Pulling the constant $\frac{1}{10}$ outside, we have:

Look at this! It is almost identical to the Beta function definition. We just need to express the powers as $(m-1)$ and $(n-1)$.

The Final Identification

We need $m-1=-9/10$, which implies $m=1/10$. We need $n-1=20$, which implies $n=21$.

Thus, our integral is:

Comparing this to the form $a\times \beta (b,c)$, we identify $a=1/10$, $b=1/10$, and $c=21$.

The final step is a simple arithmetic calculation:

And there you have it. We didn't fight the integral; we danced with it. We transformed a seemingly impossible problem into a standard form using the elegance of substitution. Keep this mindset, and no JEE problem will ever be too daunting for you.