Animated Solution for Mathematics - Definite Integration: If $I(m,n)={\int }_0^1{t}^m(1+t{)}^{n}dt$, then the expression for $I(m,n)$ in terms of $I(m+1,n-1)$ is

Mastering the Integral

Welcome, future engineer. Today, we are not just solving an integral; we are embarking on a journey of mathematical transformation. We are looking at the integral $I(m,n)={\int }_0^1{t}^m(1+t{)}^{n}dt$.

At first glance, it looks like a standard calculus problem, but it is actually a gateway to understanding how we can manipulate complex expressions into simpler, recursive forms. This is the heart of reduction formulas—a technique that turns a daunting problem into a manageable sequence of steps.

The Strategy of the Trade-off

Imagine you are standing on a bridge, and you have two heavy weights: $t^m$ and $(1+t{)}^n$. You need to move some weight from one side to the other.

You want to transform the integral into a form involving $I(m+1,n-1)$. Notice the shift: the power of $t$ increases from $m$ to $m+1$, and the power of $(1+t)$ decreases from $n$ to $n-1$.

This is not a coincidence; it is a design. When you see a product of functions where you need to shift exponents, your intuition should immediately scream: Integration by Parts (IBP).

Recall the IBP formula: $\int udv=uv-\int vdu$. This formula is a 'trade'. It allows us to take the derivative of one part (to reduce its power) and the integral of the other (to increase its power).

The Setup

Now, we must choose our $u$ and $dv$ wisely. We have two candidates: $(1+t{)}^n$ and $t^m$.

We want to reduce the power of $(1+t{)}^n$, so we must differentiate it. Thus, we set $u=(1+t{)}^n$. Consequently, we must integrate $t^m$, so we set $dv=t^{m}dt$.

Let's execute this. Differentiating $u$ using the chain rule gives us $du=n(1+t{)}^{n-1}dt$. See how the power dropped from $n$ to $n-1$?

Now, integrating $dv$ gives us $v=\frac{t^{m+1}}{m+1}$. We have successfully increased the power of $t$ from $m$ to $m+1$. We have set the stage perfectly.

The Execution

Now, we plug these into the IBP formula. The integral becomes:

Take a deep breath. This looks complex, but look at the structure. We have a boundary term and a new integral.

Let's tackle the boundary term first. We evaluate $\left[(1+t{)}^n\frac{t^{m+1}}{m+1}\right]$ at the limits $0$ and $1$.

At $t=1$, we get $(1+1{)}^n\frac{1^{m+1}}{m+1}=\frac{2^n}{m+1}$. At $t=0$, the term becomes $(1+0{)}^n\frac{0^{m+1}}{m+1}=0$.

So, the boundary term simplifies beautifully to $\frac{2^n}{m+1}$. This is the constant we were looking for!

The Final Synthesis

Now, look at the remaining integral: ${\int }_0^1\frac{t^{m+1}}{m+1}n(1+t{)}^{n-1}dt$. We can pull the constants $n$ and $\frac{1}{m+1}$ outside the integral.

What remains inside is ${\int }_0^1{t}^{m+1}(1+t{)}^{n-1}dt$. Does that look familiar? It is exactly the definition of $I(m+1,n-1)$!

Putting it all together, we get:

This is the elegance of calculus. We started with a complex integral and, through the strategic application of Integration by Parts, we reduced it to a simpler form.

You have successfully navigated the reduction. Remember this process—it is the same logic you will use for more complex problems in your JEE journey. Keep practicing, keep visualizing, and most importantly, keep falling in love with the process.