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JEE Main 2019
Definite Integration
Definite Integration
Hard

Question

limn3n{4+(2+1n)2+(2+2n)2++(31n)2}\lim\limits_{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\} is equal to :

Options

Solution

Key Concepts and Formulas

  • Definite Integral as a Limit of a Sum (Riemann Sum): A definite integral can be represented as the limit of a sum. The fundamental formula is: limn1nr=abf(rn)=01f(x)dx\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=a}^{b} f\left(\frac{r}{n}\right) = \int_0^1 f(x) \, dx More generally, for a sum of the form limn1nr=mkf(c+rn)\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=m}^{k} f\left(c + \frac{r}{n}\right), it can be converted to c+m/nc+k/nf(x)dx\int_{c + m/n}^{c + k/n} f(x) \, dx as nn \to \infty.
  • Integral Evaluation: Using the power rule for integration: ukdu=uk+1k+1+C\int u^k \, du = \frac{u^{k+1}}{k+1} + C.

Step-by-Step Solution

Step 1: Analyze the Given Expression and Convert to Standard Summation Form

The given limit is: L=limn3n{4+(2+1n)2+(2+2n)2++(31n)2}L = \lim_{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\} Our goal is to express the sum inside the curly braces in sigma notation. Let's examine the terms: The first term is 44. We can write 4=22=(2+0n)24 = 2^2 = \left(2+\frac{0}{n}\right)^2. This corresponds to the case when r=0r=0. The subsequent terms are of the form (2+rn)2\left(2+\frac{r}{n}\right)^2 for r=1,2,r=1, 2, \ldots. The last term is (31n)2\left(3-\frac{1}{n}\right)^2. We need to express this in the form (2+rn)2\left(2+\frac{r}{n}\right)^2. 31n=2+11n=2+n1n3 - \frac{1}{n} = 2 + 1 - \frac{1}{n} = 2 + \frac{n-1}{n} So, the last term is (2+n1n)2\left(2+\frac{n-1}{n}\right)^2. This corresponds to the case when r=n1r=n-1.

Thus, the sum inside the curly braces can be written as: r=0n1(2+rn)2\sum_{r=0}^{n-1} \left(2+\frac{r}{n}\right)^2 Now, substitute this back into the limit expression: L=limn3nr=0n1(2+rn)2L = \lim_{n \rightarrow \infty} \frac{3}{n} \sum_{r=0}^{n-1} \left(2+\frac{r}{n}\right)^2 We can factor out the constant 33: L=3limn1nr=0n1(2+rn)2L = 3 \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} \left(2+\frac{r}{n}\right)^2 This expression is now in the standard form for conversion to a definite integral.

Step 2: Convert the Limit of Sum to a Definite Integral

We use the definition of the definite integral as the limit of a Riemann sum. We identify the following:

  • The term rn\frac{r}{n} is replaced by xx.
  • The term 1n\frac{1}{n} is replaced by dxdx.
  • The function f(x)f(x) is determined from the general term (2+rn)2\left(2+\frac{r}{n}\right)^2. Thus, f(x)=(2+x)2f(x) = (2+x)^2.

Now, we determine the limits of integration:

  • The sum starts with r=0r=0. The lower limit of integration is limn0n=0\lim_{n \rightarrow \infty} \frac{0}{n} = 0.
  • The sum ends with r=n1r=n-1. The upper limit of integration is limnn1n=limn(11n)=1\lim_{n \rightarrow \infty} \frac{n-1}{n} = \lim_{n \rightarrow \infty} \left(1 - \frac{1}{n}\right) = 1.

Therefore, the limit of the sum can be converted to the definite integral: L=301(2+x)2dxL = 3 \int_0^1 (2+x)^2 \, dx

Step 3: Evaluate the Definite Integral

We now evaluate the definite integral: L=301(2+x)2dxL = 3 \int_0^1 (2+x)^2 \, dx Let u=2+xu = 2+x. Then du=dxdu = dx. When x=0x=0, u=2u=2. When x=1x=1, u=3u=3. L=323u2duL = 3 \int_2^3 u^2 \, du Using the power rule for integration: L=3[u2+12+1]23L = 3 \left[ \frac{u^{2+1}}{2+1} \right]_2^3 L=3[u33]23L = 3 \left[ \frac{u^3}{3} \right]_2^3 The constant 33 cancels out: L=[u3]23L = \left[ u^3 \right]_2^3 Now, apply the limits of integration: L=(3)3(2)3L = (3)^3 - (2)^3 L=278L = 27 - 8 L=19L = 19

Common Mistakes & Tips

  • Incorrectly identifying the summand: Ensure the entire expression inside the summation is correctly represented as f(rn)f(\frac{r}{n}) or f(c+rn)f(c + \frac{r}{n}). Pay close attention to algebraic manipulations, especially for the first and last terms.
  • Errors in determining integration limits: The limits of integration are derived from the minimum and maximum values of rr divided by nn as nn \to \infty.
  • Forgetting constant factors: Any constant multiplier outside the summation and limit, such as the 3n\frac{3}{n} in this problem, must be carried through the conversion to the definite integral.

Summary

The problem involves evaluating a limit of a sum, which can be converted into a definite integral using the definition of a Riemann sum. By carefully rewriting the given sum in the standard sigma notation and identifying the function and integration limits, we transformed the limit into a definite integral. Evaluating this integral yielded the final result.

The final answer is 19\boxed{19}, which corresponds to option (C).

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