Animated Solution for Mathematics - Definite Integration: ${\lim }_{n\to \infty }\left(\frac{(n+1{)}^{1/3}}{n^{4/3}}+\frac{(n+2{)}^{1/3}}{n^{4/3}}+\cdots +\frac{(2n{)}^{1/3}}{n^{4/3}}\right)$ is equal to :

Imagine standing before a massive, intimidating wall of numbers. You see a series, a summation that stretches out to infinity, and your first instinct might be to panic.

You see terms like $\frac{(n+1{)}^{1/3}}{n^{4/3}}$ and $\frac{(2n{)}^{1/3}}{n^{4/3}}$, and it feels like a chaotic mess. But here is the secret of the JEE Advanced: chaos is just order waiting to be discovered.

This problem is not a test of your ability to add infinite numbers; it is a test of your ability to see the geometry hidden within the algebra.

Let us look at the general term, $T_r$. We have a numerator that changes as $r$ goes from $1$ to $n$, and a denominator that stays stubbornly fixed at $n^{4/3}$.

The term looks like $T_r=\frac{(n+r{)}^{1/3}}{n^{4/3}}$. Now, we need to perform some algebraic surgery. We want to extract a $\frac{1}{n}$ factor because that is the key to the Riemann Sum.

If we pull an $n$ out of the numerator's bracket, it comes out as $n^{1/3}$. So, we have:

Using the laws of exponents, we subtract the powers: $\frac{1}{3}-\frac{4}{3}=-1$. Suddenly, the expression simplifies beautifully to:

Do you see it now? The complexity has vanished, leaving behind a clean, elegant structure.

We are now looking at a sum $S={\lim }_{n\to \infty }{\sum }_{r=1}^n\frac{1}{n}(1+\frac{r}{n}{)}^{1/3}$. This is the standard form of a Riemann Sum.

Think of this geometrically. We are summing up the areas of $n$ tiny rectangles. Each rectangle has a width of $\frac{1}{n}$ and a height determined by the function $f(x)=(1+x{)}^{1/3}$, where $x=\frac{r}{n}$.

As $n$ approaches infinity, the width of these rectangles becomes infinitesimally small, and the sum of their areas perfectly converges to the area under the curve $y=(1+x{)}^{1/3}$ from $x=0$ to $x=1$. We have transformed a discrete, terrifying sum into a smooth, continuous integral:

Now, we simply apply the fundamental theorem of calculus. We need the integral of $(1+x{)}^{1/3}$.

Using the power rule $\int x^{n}dx=\frac{x^{n+1}}{n+1}$, we get $\frac{(1+x{)}^{4/3}}{4/3}$, which simplifies to $\frac{3}{4}(1+x{)}^{4/3}$. We evaluate this from $0$ to $1$.

Plugging in the upper limit, we get $\frac{3}{4}(1+1{)}^{4/3}=\frac{3}{4}(2{)}^{4/3}$. Plugging in the lower limit, we get $\frac{3}{4}(1+0{)}^{4/3}=\frac{3}{4}$.

Subtracting the two, our final answer is:

It is elegant, it is precise, and it is exactly what we were looking for. You have just mastered the art of converting a limit of a sum into a definite integral.