Problem 4 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r}{n^{2}} \cdot \sec ^{2} \frac{r^{2}}{n^{2}}\) is equal to (A) \(\tan 1\) (B) \(\frac{1}{3} \tan 1\) (C) \(\frac{1}{2} \tan 1\) (D) None of these

Step-by-Step Solution

Answer

(C) \(\frac{1}{2} \tan 1\).

Step 1: Recognize the limit of a Riemann Sum

The given expression is an infinite sum as a limit, which suggests that it's a Riemann sum approximating a definite integral. The form is \( \lim_{n \to \infty} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) \cdot \frac{1}{n} \).

Step 2: Identify function and interval

In the given Riemann sum, compare the expressions to identify the function and interval. Here, \( f\left(\frac{r}{n}\right) = \frac{r/n}{n} \sec^2\left(\frac{r^2}{n^2}\right) \) and the differential element is \( \frac{1}{n} \). The variable changes from \( \frac{r^2}{n^2} \) suggests the rule of substitution and integration over \([0, 1]\).

Step 3: Perform substitution for integration

To simplify the integration, denote \( x = \frac{r}{n} \), and notice \( dx\approx \frac{1}{n} \). Hence, the expression can be rewritten as \( x \sec^2(x^2) \). The integration will be over \( x \in [0,1] \), recognizing that \( x^2 \) changes limits from \( 0 \) to \( 1 \) as \( r \) runs from \( 1 \) to \( n \).

Step 4: Integrate the function

Now integrate \( x \sec^2(x^2) \) over \((0,1)\). Integrating by parts, let \( u = x \) and \( dv = \sec^2(x^2) dx \), then differentiate \( du = dx \) and integrate \( v = \tan(x^2) / 2 x \); this evaluates to \( \int_{0}^{1} x \sec^{2}(x^2) dx = \frac{1}{2}\tan(1\cdot 1) \). By integration by parts, this will evaluate to \( \frac{1}{2}\tan(1) \).

Step 5: Match the result with the options

Now compare the computed value \( \frac{1}{2}\tan(1) \) to the given options: A) \( \tan 1 \), B) \( \frac{1}{3} \tan 1 \), C) \( \frac{1}{2} \tan 1 \), or D) none of these. The correct choice is (C).

Key Concepts

Definite Integral

When we talk about a definite integral, we're discussing the finding of the accumulated area under a curve or function within a set boundary on an axis. This is a key concept in calculus, as it allows us to quantify spaces that are irregular in shape. The basic notation for a definite integral is \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the limits of integration, \( f(x) \) is the function you're integrating, and \( dx \) indicates the variable of integration.

In the context of Riemann sums, the sum approximates this integration by adding up small areas under the curve within the interval \( [a, b] \). The more intervals or partitions we have, the closer our approximation gets to the real value of the definite integral. In this exercise, the definite integral of \( x \sec^2(x^2) \) from \( 0 \) to \( 1 \) helps us find the precise area under the curve, which is exactly what the limit of the Riemann sum represents.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It is a strategic way to evaluate integrals where the standard integration methods might not be straightforward. The formula for integration by parts is \( \int u \cdot dv = uv - \int v \cdot du \), where \( u \) and \( dv \) are parts of the original function.

In this exercise, we apply integration by parts to the function \( x \sec^2(x^2) \). Choosing \( u = x \) and \( dv = \sec^2(x^2)dx \), we differentiate \( du = dx \) and integrate \( v \), getting that \( v = \frac{1}{2} \tan(x^2) \). By plugging these into our integration by parts formula, we can solve the integral over the desired interval. This clever manipulation turns a complex problem into smaller, more manageable pieces.

Limits of Summation

The concept of limits of summation is rooted in assessing how a sum behaves as its range gets infinitely large. It's closely tied with the idea of Riemann sums and definite integrals. Here \( \lim _{n \rightarrow \infty} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) \frac{1}{n} \) represents a Riemann sum, which becomes the definite integral as \( n \) approaches infinity, reflecting an increasingly accurate area estimation.

As \( n \) increases, the width of each partition (or strip) in our summation decreases, leading to more partitions approximating the integral's curve more closely. The exercise highlights this through \( \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r}{n^2} \sec ^2\left(\frac{r^2}{n^2}\right) \), where finishing with the actual function integration gives us a clearer understanding of the limits of summation in practice.