Problem 69 - Calculus Early Transcendentals

Question

Graph the integrands and then evaluate and compare the values of \(\int_{0}^{\infty} x e^{-x^{2}} d x\) and \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x\).

Step-by-Step Solution

Answer

Answer: Yes, both integrals have the same value, which is $\frac{1}{2}$.

Step 1: Graph the integrands

First, we need to graph the functions \(f(x) = xe^{-x^2}\) and \(g(x) = x^2e^{-x^2}\). You can use a graphing calculator or an online graphing tool to plot these functions. Observe their behavior, especially as \(x\) goes to infinity.

Step 2: Evaluate the first integral

We want to evaluate: $$\int_{0}^{\infty} xe^{-x^{2}} dx$$ To do this, use substitution method. Let \(u = x^2\), then \(\frac{du}{dx} = 2x\) or \(dx = \frac{du}{2x}\). The new limits for \(u\) will be \(0\) and \(\infty\): $$\int_{0}^{\infty} xe^{-x^{2}} dx = \int_{0}^{\infty} e^{-u} \frac{du}{2}$$ Now, integrate \(e^{-u}\): $$\int_{0}^{\infty} e^{-u} \frac{du}{2} = \frac{1}{2}\int_{0}^{\infty} e^{-u} du$$ $$\frac{1}{2}\left[ -e^{-u} \right]_{0}^{\infty} = \frac{1}{2}\left(-e^{-\infty} + e^0\right) = \frac{1}{2}(1) = \frac{1}{2}$$ So, the value of the first integral is \(\frac{1}{2}\).

Step 3: Evaluate the second integral

Now, we want to evaluate: $$\int_{0}^{\infty} x^2e^{-x^{2}}dx$$ Use the substitution method again. Let \(v = x^2\), then \(\frac{dv}{dx} = 2x\) or \(dx = \frac{dv}{2x}\). The new limits for \(v\) will be \(0\) and \(\infty\): $$\int_{0}^{\infty} x^2e^{-x^{2}} dx = \int_{0}^{\infty} ve^{-v} \frac{dv}{2x^2}$$ Cancel out \(x^2 = v\) from the numerator and denominator: $$\int_{0}^{\infty} ve^{-v} \frac{dv}{2x^2} = \int_{0}^{\infty} e^{-v} \frac{dv}{2}$$ Now integrate \(e^{-v}\): $$\int_{0}^{\infty} e^{-v} \frac{dv}{2} = \frac{1}{2}\int_{0}^{\infty} e^{-v} dv$$ $$\frac{1}{2}\left[ -e^{-v} \right]_{0}^{\infty} = \frac{1}{2}\left(-e^{-\infty} + e^0\right) = \frac{1}{2}(1) = \frac{1}{2}$$ So, the value of the second integral is also \(\frac{1}{2}\).

Step 4: Compare the values

We have found the values of both integrals as follows: $$\int_{0}^{\infty} xe^{-x^{2}} dx = \frac{1}{2}$$ $$\int_{0}^{\infty} x^2e^{-x^{2}} dx = \frac{1}{2}$$ Both integrals have the same value, which is \(\frac{1}{2}\).

Key Concepts

Graphing Functions

Graphing functions is a powerful way to visually understand the behavior of mathematical expressions. In this exercise, we're dealing with two integrands: \( f(x) = xe^{-x^2} \) and \( g(x) = x^2e^{-x^2} \). By graphing these functions, we can observe how they behave as \( x \) approaches infinity.

• As \( x \) increases, both \( xe^{-x^2} \) and \( x^2e^{-x^2} \) approach zero. This happens because the exponential term \( e^{-x^2} \) decays very quickly, overpowering the polynomial terms in the functions.
• The shape of the graph is crucial. For instance, \( xe^{-x^2} \) will peak at some point before sharply dropping off as \( x \) increases. This tells us that the effect of \( e^{-x^2} \) is dominant, especially in the regions away from the origin.
• Graphing these functions can also provide insight into the areas under the curve, which is essentially what the definite integrals are calculating.

Additionally, graphing tools or calculators can be highly beneficial to students trying to get an intuitive grasp of these functions and their limits.

Substitution Method

The substitution method is a fundamental technique used to evaluate integrals, especially when dealing with complex expressions. This approach helps simplify the problem by transforming the variables involved.

In this exercise, the substitution method is applied to handle the integrals involving \( xe^{-x^2} \) and \( x^2e^{-x^2} \). Here’s how it works:

• For the integral \( \int_{0}^{\infty} xe^{-x^2} \, dx \), we substitute \( u = x^2 \). The differential \( du = 2x \, dx \) leads to \( dx = \frac{du}{2x} \). This transforms our integral into \( \int_{0}^{\infty} e^{-u} \frac{du}{2} \), making it more straightforward to solve.
• Similarly, for \( \int_{0}^{\infty} x^2e^{-x^2} \, dx \), use \( v = x^2 \) as the substitution, which simplifies to \( \int_{0}^{\infty} ve^{-v} \frac{dv}{2x^2} \). With further simplification, you focus on \( \int_{0}^{\infty} e^{-v} \frac{dv}{2} \).

These substitutions turn complex integrals into a more familiar form, allowing us to apply basic integral calculations to reach our solution.

Evaluate Integrals

Evaluating integrals is about finding the area under a curve defined by a function. In our case, we're looking at two separate integrals, both spanning from 0 to infinity.

When we evaluate \( \int_{0}^{\infty} xe^{-x^2} \, dx \), after using substitution, the integral simplifies to \( \frac{1}{2} \int_{0}^{\infty} e^{-u} \, du \). This evaluates to \( \frac{1}{2} \), since the signum integral \( \int_{0}^{\infty} e^{-u} \, du \) is known to converge to 1.

• Apply similar steps to \( \int_{0}^{\infty} x^2e^{-x^2} \, dx \) after substitution to show that it also simplifies to an integral that evaluates to \( \frac{1}{2} \).
• Both of these calculations highlight the methodical approach: recognizing substitution, transforming limits, simplifying the expression, and then computing the integral.
• These results also tell us something fascinating about symmetry and decay rate in functions involving exponentials, where despite different starting modus forms, the integrals produced from both functions inform us of similar area values under their curves.

Understanding this evaluation process helps in tackling improper integrals and builds confidence in handling similar problems in calculus.