Problem 191 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}}\) (A) \(e\) (B) \(e-1\) (C) \(1-e\) (D) \(e+1\)

Step-by-Step Solution

Answer

(B) \(e-1\)

Step 1: Recognize the Expression as a Riemann Sum

The given limit and sum expression \( \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}} \) is a Riemann sum. It approximates the integral of a function by dividing the interval into \(n\) sub-intervals. Here, \( \frac{1}{n} \) represents the width of each sub-interval, and \( e^{\frac{r}{n}} \) represents the function value at specific points.

Step 2: Identify the Function and Interval

Observe that this sum can be seen as a Riemann sum for the integral of the function \( f(x) = e^x \) over the interval \([0, 1]\). The terms \( \frac{r}{n} \) are points within the interval, and the function evaluated at these points is \( e^{\frac{r}{n}} \). As \( n \rightarrow \infty \), these points become dense across the interval \[0, 1\].

Step 3: Set up the Integral

The Riemann sum \( \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}} \) is equivalent to the integral \( \int_{0}^{1} e^x \, dx \). This means we will calculate the definite integral of \(e^x\) over the interval from 0 to 1.

Step 4: Compute the Integral

To find \( \int_{0}^{1} e^x \, dx \), we need to find the antiderivative of \(e^x\), which is simply \(e^x\) itself. Using the fundamental theorem of calculus, evaluate \(e^x\) from 0 to 1:\[ \int_{0}^{1} e^x \, dx = \left[ e^x \right]_{0}^{1} = e^1 - e^0 = e - 1. \]

Step 5: Conclusion

The Riemann sum converges to the value of the definite integral \(e-1\) as \(n\) approaches infinity.

Key Concepts

Definite integral

A definite integral is a key mathematical concept used to calculate the total area under a curve within a specific interval on the x-axis. This concept is instrumental in understanding the accumulation of quantities. For example, it can represent the total distance traveled over time given a velocity function.

• In the context of the original exercise, the definite integral is represented by \int_{0}^{1} e^x \, dx.
• This integral calculates the area under the exponential function \( e^x \) over the interval from \(0\) to \(1\).
• Finding the definite integral involves finding the antiderivative of the function, a process which often uses the fundamental theorem of calculus.

The result of the integration provides a precise number, which in this case is \( e-1\). This precise number helps us understand the behavior of functions across a specified domain.

Exponential function

The exponential function, expressed as \(e^x \), is a fundamental mathematical concept that describes growth and decay processes in various scientific fields. It is characterized by a constant base \(e\), a transcendental number approximately equal to \2.718\.

• In our problem, the function \(e^x\) is being integrated, showing the versatile application of exponentials in calculus.
• Exponential functions grow significantly faster than polynomial and linear functions, making them crucial for modeling rapid growth or decay.

In the integral from 0 to 1, the exponential function \(e^x \) depicts growth from \(e^0 = 1\) to \(e^1 = e\). This aids in understanding the impact of continuous compounding in various real-life scenarios, like finance and population dynamics.

Limit of a sequence

In mathematics, the limit of a sequence refers to the value that the terms of a sequence approach as the index goes to infinity. Understanding limits is essential for working with sequences and series, particularly in the evaluation of definite integrals and Riemann sums.

• In this exercise, the limit \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}} is a Riemann sum converter into a definite integral.
• The result of the limit connects the concept of summing infinitely small quantities to determine the total area under a curve.

As the number of partitions (n) becomes infinite, the sum \(\sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}} \) approaches the definite integral \(\int_{0}^{1} e^x \, dx \), leading to the result \( e-1\). This illustrates how the limit of a sequence forms a bridge to integral calculus.