Problem 5 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \sin ^{2 k} \frac{r \pi}{2 n}\) is equal to (A) \(\frac{2 k !}{2^{2 k}(k !)^{2}}\) (B) \(\frac{2 k !}{2^{k}(k !)}\) (C) \(\frac{2 k !}{2^{k}(k !)^{2}}\) (D) None of these

Step-by-Step Solution

Answer

The limit converges to (A) \(\frac{2k!}{2^{2k}(k!)^{2}}\)."

Step 1: Interpret the Limit Expression

The given limit expression is \( \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \sin ^{2 k} \frac{r\pi}{2 n} \). This is similar to an integral approximation of a function over the interval \([0, 1]\). It can be thought of as a Riemann sum, and we want to find the continuous limit as \( n \) approaches infinity.

Step 2: Identify the Riemann Sum

Recognize that \( \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) \) approximates \( \int_{0}^{1} f(x) \, dx \). Here, \( f\left(x\right) = \sin^{2k} \left( \frac{x\pi}{2} \right) \) for \( x = \frac{r}{n} \), suggesting the integral \( \int_{0}^{1} \sin^{2k} \left( \frac{x \pi}{2} \right) \ dx \).

Key Concepts

Limit approximation

In mathematics, especially in calculus, limits are fundamental to understanding the concept of continuity and change. The exercise we are dealing with revolves around the limit of a sum as a way to approximate an integral, a method known as the Riemann Sum. As you approach the limit, you're essentially letting the number of subdivisions, denoted by \( n \), go to infinity. This creates increasingly finer partitions of the function you've graphed over a defined interval.

For example, if we have \( \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f(x) \), this expression states that as \( n \) tends to infinity, the sum is supposed to cover every point in the interval \([0, 1]\).

• The role of \( \frac{1}{n} \) is to suggest the width of each subinterval, which becomes infinitesimally small as \( n \) approaches infinity.
• The sum \( \sum_{r=1}^{n} \) approximates adding up the areas of these infinitesimally small regions under the function \( f(x) \).

Using a large \( n \), the Riemann sum becomes a better approximation of the integral.

Integral calculus

Integral calculus is a major part of calculus which focuses on the accumulation of quantities and the areas under and between curves. In the context of the given exercise, we are interested in the definite integral, which has bounds and gives the net area between the x-axis and the curve over an interval. The specific problem uses the trigonometric function \( \sin\), raised to an even power and integrated over a set interval, which in this case is \([0, 1]\).

The connection between limits and integrals is established through the method of Riemann sum. Here, with \( f(x) = \sin^{2k} \left( \frac{x\pi}{2} \right) \), the integral is \[ \int_{0}^{1} \sin^{2k} \left( \frac{x \pi}{2} \right) \ dx \].

• This sets up the intuition that integrals are limits of sums and can be visualized as the sum of infinitely many infinitesimally small elements.
• This approach allows us to capture the complete area under the curve, starting from 0 to 1 in this particular problem.

Trigonometric integrals

Trigonometric integrals are a special class of integrals that involve trigonometric functions such as sine or cosine. These come with distinct techniques and identities to solve them. In this exercise, we deal with an integral involving \( \sin^{2k} \), which belongs to this category. Evaluating such an integral often requires the use of trigonometric identities or transformations to simplify the powers and allow for easier integration.

To tackle \( \int_{0}^{1} \sin^{2k} \left( \frac{x \pi}{2} \right) \ dx \), one might leverage the power-reduction formulas or symmetry properties of the sine function to express these powers in lower terms.

• Trigonometric identities help express higher powers in terms of first powers, making integration more straightforward.
• The focus is often on recognizing patterns or forms that allow for the application of standard integration techniques.

Establishing the relationship between limits and these trigonometric integrals deepens our understanding of how calculus describes and resolves complex areas under curves.