$\displaystyle\lim_{n\to\infty} \left(\frac{\left(n+1\right)^{\frac{1}{3}} }{n^{\frac{4}{3}}} + \frac{\left(n+2\right)^{\frac{1}{3}}}{n^{\frac{4}{3}}} + ..... + \frac{\left(2n\right)^{\frac{1}{3}}}{n^{\frac{4}{3}}}\right) $ equal to :

Question

Evaluate the integral: $ \int \sin^5 x \, dx $

Answer 1

We are given the limit expression:

$\lim_{n\to\infty} \left(\frac{\left(n+1\right)^{\frac{1}{3}} }{n^{\frac{4}{3}}} + \frac{\left(n+2\right)^{\frac{1}{3}}}{n^{\frac{4}{3}}} + \ldots + \frac{\left(2n\right)^{\frac{1}{3}}}{n^{\frac{4}{3}}}\right)$

This problem can be approached by looking at it as a sum:

$\sum_{k=1}^{n} \frac{(n+k)^{1/3}}{n^{4/3}}$

The term $\frac{(n+k)^{1/3}}{n^{4/3}}$ can be rewritten as:

$= \frac{1}{n} \cdot \frac{(n+k)^{1/3}}{n^{1/3}} = \frac{1}{n} \left(1 + \frac{k}{n}\right)^{1/3}$

As $n \to \infty$, this becomes a Riemann sum, similar to the integral:

$\int_{0}^{1} (1+x)^{1/3} dx$ for $x = \frac{k}{n}$.

Let's calculate this integral:

$\int_{0}^{1} (1 + x)^{1/3} \, dx$

Using the integration formula, let $u = 1 + x, \, du = dx$. The limits of integration change from $1$ to $2$.

Thus, we have:

$\int_{1}^{2} u^{1/3} \, du$

The integral becomes:

$= \left[ \frac{u^{4/3}}{4/3} \right]_{1}^{2}$

$= \frac{3}{4} \left[ u^{4/3} \right]_{1}^{2}$

$= \frac{3}{4} \left(2^{4/3} - 1\right)$

Hence, the value of the original limit is:

$\frac{3}{4} \left(2\right)^{\frac{4}{3}} - \frac{3}{4}$

This matches the provided correct answer:

$\frac{3}{4} \left(2\right)^{\frac{4}{3}} - \frac{3}{4}$.

Answer 2