Problem 48 - Calculus Early Transcendentals: Pearson New International Edition

Question

Consider \(f(x)=n^{2} x e^{-n x}\). (a) Graph \(f(x)\) for \(n=1,2,3,4,5,6\) on \([0,1]\) in the same graph window. (b) For \(x>0\), find \(\lim _{n \rightarrow \infty} f(x)\). (c) Evaluate \(\int_{0}^{1} f(x) d x\) for \(n=1,2,3,4,5,6\). (d) Guess at \(\lim _{n \rightarrow \infty} \int_{0}^{1} f(x) d x\). Then justify your answer rigorously.

Step-by-Step Solution

Answer

(a) Plot shows peaks shifting leftwards. (b) Limit is 0. (c) Integrals approach 1. (d) Limit is 1 due to diminishing \((n+1)e^{-n}\).

Step 1: Graphing Multiple Functions

Graph the function \(f(x) = n^2 x e^{-nx}\) for each value \(n = 1, 2, 3, 4, 5, 6\) over the interval \([0, 1]\). Plotting these curves will show how the function changes with different values of \(n\). As \(n\) increases, the peak of the function moves to the left and becomes steeper because the factor \(e^{-nx}\) declines more rapidly.

Step 2: Finding the Limit as n Approaches Infinity

To find \(\lim_{n \to \infty} f(x)\) for \(x > 0\), note that \(f(x) = n^2 x e^{-nx}\). As \(n\) increases, \(e^{-nx}\) approaches zero very quickly for any \(x > 0\). Thus, the whole function tends toward zero: \(\lim_{n \to \infty} n^2 x e^{-nx} = 0\).

Step 3: Evaluating the Integral

Compute the integral \(\int_0^1 n^2 x e^{-nx} dx\) for each \(n = 1, 2, 3, 4, 5, 6\). Use integration by parts to solve this. Let \(u = x\) and \(dv = n^2 e^{-nx} dx\). Then \(du = dx\) and \(v = -n e^{-nx}\) (after integrating \(dv\)). The integration by parts formula \(\int u \, dv = uv - \int v \, du\) leads to: \[ \int_0^1 n^2 x e^{-nx} \, dx = \left[ -n x e^{-nx} \right]_0^1 + \int_0^1 n e^{-nx} \, dx = -n e^{-n} - n\left[ \frac{-e^{-nx}}{n} \right]_0^1 \]This simplifies to \(1 - (n+1)e^{-n}\). Evaluate this for each \(n = 1, 2, 3, 4, 5, 6\).

Step 4: Analyzing the Limit of the Integral

Look at the sequence of evaluated integrals from Step 3 as \(n\) increases. The term \((n+1) e^{-n}\) becomes very small as \(n\) grows, effectively approaching zero. Thus, the integral \(\int_0^1 n^2 x e^{-nx} \, dx\) approaches 1 as \(n \to \infty\). Hence, \(\lim_{n \to \infty} \int_0^1 f(x) \, dx = 1\). The limit relies on the fact that exponential decay in \(e^{-n}\) dominates the linear growth in \(n+1\).

Key Concepts

Exponential Decay

Exponential decay in mathematical functions often refers to how a quantity decreases rapidly over time or with an increase in a variable, typically denoted by an exponent. In the function given, \(f(x) = n^2 x e^{-nx}\), the term \(e^{-nx}\) decreases rapidly as \(x\) or \(n\) increases.
This characteristic is crucial to understand how the function behaves for different values of \(n\). As \(n\) gets larger, the decay becomes sharper, impacting how the function looks graphically and influences the results of limits and integrals.
Exponential decay is critical when modeling real-world scenarios like radioactive decay or cooling processes, where reducing quantities over a distance or time interval is optimal.

Integration by Parts

Integration by parts is a useful technique to solve integrals that are products of two different functions. It is derived from the product rule of differentiation and is particularly valuable when dealing with complex products like \(n^2 x e^{-nx}\).
The formula for integration by parts is given by:\[\int u \, dv = uv - \int v \, du\]Here, we choose \(u = x\) and \(dv = n^2 e^{-nx} dx\).

• Differentiate \(u\) to get \(du = dx\).
• Integrate \(dv\) to find \(v = -n e^{-nx}\).

Substitute these into the integration by parts formula to solve the integral:\[\int_0^1 n^2 x e^{-nx} \, dx = \left[ -n x e^{-nx} \right]_0^1 + \int_0^1 n e^{-nx} \, dx\]This step simplifies the calculation, making it easier to evaluate the improper integral over a specified interval.

Function Graphing

Graphing functions is an effective way to visualize how they change with different parameters. For the function \(f(x) = n^2 x e^{-nx}\) for \(n = 1, 2, 3, 4, 5, 6\), plotting helps illustrate the impact of exponential decay visually.
It's observed that

• as \(n\) increases,
  the peak of the function shifts to the left,
• and the slope becomes steeper.

This behaviour results from the \(e^{-nx}\) term decaying more rapidly with larger \(n\) values.
Graphing offers insights into the practical applications of mathematics by showcasing these changes and supporting solutions involving limits and integrals.

Limits

In calculus, a limit helps describe the behavior of a function as a variable approaches a particular value or infinity. For the function \(f(x) = n^2 x e^{-nx}\), finding the limit as \(n\) approaches infinity is fundamental.
For \(x > 0\), observe that \(e^{-nx}\) becomes very small rapidly, thus making \(n^2 x e^{-nx}\) tend towards zero.
This is formally expressed as:\[\lim_{n \to \infty} n^2 x e^{-nx} = 0\]This \(zero\) result demonstrates how exponential decay dominates, ensuring the function value decreases to an infinitesimally small amount as \(n\) becomes very large.

Improper Integrals

Improper integrals occur when the boundaries of an integral stretch to infinity or the function being integrated becomes infinite within the interval. For this exercise, the integral \(\int_0^1 n^2 x e^{-nx} dx\) concerns an infinite parameter \(n\) eventually approximating an improper integral as \(n \rightarrow \infty\).
Often using integration techniques and limits, as done here, the indefinite growth of \(n\) is tempered by the rapid exponential decay of \(e^{-nx}\).
The particular integral simplifies to:\[1 - (n+1)e^{-n}\] as \(n\) increases.
The exponential part \((n+1)e^{-n}\), approaches zero fast, leading to the integral approaching 1, validating how improper integrals behave and why exponential decay plays such a significant role in their convergence.