Animated Solution for Mathematics - Definite Integration: The value of ${\lim }_{n\to \infty }\frac{1}{n}{\sum }_{r=0}^{2n-1}\frac{n^2}{n^2+4r^2}$ is:

Welcome, future engineer. Today, we are going to demystify one of the most elegant concepts in calculus: the Riemann Sum.

When you look at a problem like

it is easy to feel overwhelmed. It looks like a messy, infinite pile of fractions, but in the world of JEE Advanced, this is a beautiful geometric construction waiting to be unveiled.

Whenever you see a limit as $n\to \infty$ paired with a $\frac{1}{n}$ outside a summation, your mathematical intuition should scream: "Riemann Sum!"

Think of it this way: we are summing up the areas of infinitely thin rectangles. As $n$ approaches infinity, the width of these rectangles, $\frac{1}{n}$, becomes the infinitesimal $dx$, and the summation symbol transforms into the integral sign $\int$.

Our goal is to force the expression inside the sum to look like a function of $\frac{r}{n}$.

Let us look at our general term: $\frac{n^2}{n^2+4r^2}$. To make this look like a function of $\frac{r}{n}$, we need to perform some algebraic surgery.

Let us factor out $n^2$ from the denominator:

By factoring out $n^2$, the numerator and denominator cancel out perfectly, leaving us with a clean function $f(x)=\frac{1}{1+4x^2}$, where $x=\frac{r}{n}$. We have successfully translated the discrete variable $r$ into the continuous variable $x$.

Now, here is where the battle is won or lost. Many students blindly assume the integral goes from $0$ to $1$. But look closely at the summation index: $r$ goes from $0$ to $2n-1$.

We must calculate the limits of integration carefully:

Lower limit: ${\lim }_{n\to \infty }\frac{0}{n}=0$.

Upper limit: ${\lim }_{n\to \infty }\frac{2n-1}{n}=2$.

If you had assumed the limit was $1$, you would have missed the entire point of the upper bound. The integral is not from $0$ to $1$; it is from $0$ to $2$. Always check your bounds!

We have arrived at our integral:

We know that $\int \frac{1}{1+u^2}du={\tan }^{-1}(u)$. Here, our $u$ is $2x$. When we integrate with respect to $x$, we must account for the coefficient of $x$ by dividing by it.

Thus, the integral becomes:

Substituting the limits, we get $\frac{1}{2}{\tan }^{-1}(2\times 2)-\frac{1}{2}{\tan }^{-1}(2\times 0)$. This simplifies beautifully to $\frac{1}{2}{\tan }^{-1}(4)-0$.

The final value of the expression is:

You started with a terrifying limit of a sum and ended with a clean inverse trigonometric value. This is the power of calculus: you translated a complex discrete system into a simple, continuous geometric area.