Problem 88 - A Complete Resource Book in Mathematics for JEE Main

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 is \( \frac{2 k !}{2^{2 k}(k !)^{2}} \), which is option (A).

Step 1: Understanding the Problem

The problem asks us to evaluate the limit \( \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \sin^{2k} \frac{r\pi}{2n} \). This involves the average value of \( \sin^{2k} x \) over a specific interval.

Step 2: Express the Sum as a Riemann Sum

Recognize that \( \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r\pi}{2n}\right) \) resembles a Riemann sum. Our function is \( \sin^{2k} x \) with \( x = \frac{r\pi}{2n} \), and the interval is \([0, \frac{\pi}{2}]\). The limit as \( n \to \infty \) corresponds to evaluating the integral of this function.

Step 3: Set Up the Integral

According to the definition of Riemann sums, this limit is equivalent to the integral \( \int_{0}^{\frac{\pi}{2}} \sin^{2k} x \, dx \).

Step 4: Solve the Integral

The integral \( \int_{0}^{\frac{\pi}{2}} \sin^{2k} x \, dx \) can be solved using the reduction formula for \( \sin^{n} x \), given as \( \int \sin^{n} x \, dx = -\frac{1}{n} \sin^{n-1} x \cos x + \frac{n-1}{n} \int \sin^{n-2} x \, dx \). Using this formula iteratively helps compute the exact integral.

Step 5: Apply Specific Formulas

For \( \int_{0}^{\frac{\pi}{2}} \sin^{2k} x \, dx \), it is known that \( \int_{0}^{\pi/2} \sin^{2k} x \, dx = \frac{(2k-1)!!}{(2k)!!} \frac{\pi}{2} \). The double factorial \((2k-1)!! = (2k-1)(2k-3)...(1)\) and \((2k)!! = (2k)(2k-2)...(2)\).

Step 6: Simplify Using Known Values

For large \( n \), the integral over \([0, \frac{\pi}{2}]\) corresponds to \( \frac{2k!}{2^{2k}(k!)^{2}} \) based on symmetry and known integral results for the powers of sine functions.

Step 7: Conclusion

Based on these integral calculations, we find that the answer corresponds to option (A).

Key Concepts

Riemann Sum

Riemann sums help us approximate the area under a curve by dividing it into a series of rectangles. This method serves as a bridge to finding definite integrals, especially when dealing with continuous functions.
To form a Riemann sum, select a function and partition its domain into equal sub-intervals. The value of each rectangle's height is typically the function's value at a certain point within that interval.

• This approach estimates the total area under the curve over the desired interval.
• In our problem, the interval is \([0, \frac{\pi}{2}]\) and involves the trigonometric function \(\sin^{2k}(x)\).
• Using Riemann sums, the given expression approximates the integral of \(\sin^{2k}(x)\), as seen in limit form as \(n \to \infty\).

This leads us naturally to integral calculus as these sums provide a practical stepping stone to evaluate definite integrals.

Integral Calculus

Integral calculus focuses on finding the total size or value, such as area under a curve. In the realm of the calculus, integrals stand as pivotal structures, known for reversing differentiation.
Evaluating the integral \(\int_{0}^{\frac{\pi}{2}} \sin^{2k} x \, dx\) finds the exact area under \(\sin^{2k}(x)\) over our defined interval.

• The transformation from a Riemann sum into this integral allows for detailed computation as \( n \) approaches infinity.
• Reduction formulas simplify computing such integrals, incrementally decreasing the exponent through known steps.
• Applying the reduction formula results in recognizing repetitive patterns that aid in solving complex trigonometric integrals.

Hence, integral calculus in this context gives us the precise outcome, answering the integral value of \(\sin^{2k}(x)\) on the stipulated range.

Trigonometric Integrals

Trigonometric integrals, like those involving sine or cosine functions, require unique techniques. These functions have periodic properties that must be considered.
Specifically, \(\int \sin^n x \, dx\) uses repeated integration by parts or reduction formulas to break down the integral efficiently.

• The integral \(\int_{0}^{\frac{\pi}{2}} \sin^{2k} x \, dx\) reveals this when the integral is reduced progressively, using known values of factorized products.
• Double factorial expressions settle the integral values based on prior calculated identities.
• Such expressions involve sequences that skip every other number, like \( (2k - 1)!!\) and \( (2k)!! \).

Mastering such concepts helps tackle many problems involving trigonometric forms, allowing the simplification of complex exponential trigonometric components.