Problem 377 - Calculus Volume 1

Question

Compute the integral of \(f(x)=x e^{-x^{2}}\) and find the smallest value of \(N\) such that the area under the graph \(f(x)=x e^{-x^{2}}\) between \(x=N\) and \(x=N+10\) is, at most, 0.01.

Step-by-Step Solution

Answer

The smallest value for \(N\) is approximately 3.

Step 1: Set Up the Integral

We need to compute the integral \(\int x e^{-x^2} \, dx\).

Step 2: Select Integration Technique

Use substitution method. Let \( u = -x^2 \). Then \( du = -2x \ dx \). Thus, substituting we get \( x \, dx = -\frac{1}{2} \, du \).

Step 3: Substitute and Simplify

The integral \( \int x e^{-x^2} \, dx \) becomes \(-\frac{1}{2} \int e^u \, du \).

Step 4: Solve the Integral

The integral \(\int e^u \, du = e^u + C\). Therefore, \(-\frac{1}{2} \int e^u \, du = -\frac{1}{2} e^u + C = -\frac{1}{2} e^{-x^2} + C\).

Step 5: Set Up Condition for Area

We need to solve \( \left| \int_N^{N+10} x e^{-x^2} \, dx \right| \leq 0.01\).

Step 6: Evaluate Definite Integral

Evaluating \(\int_N^{N+10} x e^{-x^2} \, dx \) for complementary error function applications, \(\frac{1}{2} \left[ \text{erf}(-N) - \text{erf}(-(N+10)) \right]\).

Step 7: Approximate using Error Function

Using properties of the error function, as \(N\) increases, \(\text{erf}(N+10) \rightarrow 1\) and \(\text{erf}(N) \rightarrow 0\). Find \(N\) such that \( \left|\frac{1}{2} \left[\text{erf}(-N) - \text{erf}(-(N+10))\right]\right| \leq 0.01\).

Step 8: Estimate Necessary N

Through numerical approximations, we find that \(N\) around 3 yields a small enough area for the integral from \(N\) to \(N+10\).

Step 9: Verify Less than 0.01

Check that the area calculated for \(N = 3\) does indeed satisfy \(\leq 0.01\).

Key Concepts

Substitution Method

In integral calculus, the substitution method is a powerful technique used to simplify integrals, especially when dealing with composite functions. This method involves substituting a part of the integrand with a new variable, making the integral easier to solve. Consider the integral \( \int x e^{-x^2} \, dx \). Here, direct integration is complex because the function \( e^{-x^2} \) resists elementary antidifferentiation.
To apply the substitution method, we choose a substitution that simplifies the integrand. In this case, let \( u = -x^2 \). The derivative \( du = -2x \, dx \) helps express the differential \( x \, dx \) in terms of \( u \). By substituting \( x \, dx = -\frac{1}{2} \, du \), the integral transforms into something more manageable: \( -\frac{1}{2} \int e^u \, du \).
This method works effectively because it reduces the complexity by focusing on an inner function. The substitution aligns with the derivative of \( u \), making \( e^{u} \) straightforward to integrate. Once integrated, we revert the substitution by replacing \( u \) with \( -x^2 \), obtaining \( -\frac{1}{2} e^{-x^2} + C \).
The substitution method is particularly useful when integrands contain compositions like \( e^{g(x)} \) or trigonometric functions, supported by their derivatives. This method not only simplifies integration but also provides a systematic way to attack otherwise complex problems.

Error Function

The error function, denoted as \( \text{erf}(x) \), emerges often in the study of probability, statistics, and integral calculus. It is a special function that quantifies the probability that a Gaussian random variable takes a value between zero and \( x \). Its connection to the study of integrals comes from problems involving the normal distribution.
In our original problem, evaluating \( \int x e^{-x^2} \, dx \) leads to using the complementary error function \( \text{erfc}(x) = 1 - \text{erf}(x) \). The integral involves \( e^{-x^2} \), resembling part of the Gaussian function, which needs \( \text{erf}(x) \) for its antiderivative, rather than standard elementary functions.
By expressing the definite integral \( \int_N^{N+10} x e^{-x^2} \, dx \) in terms of the error function, we have \( \frac{1}{2} [ \text{erf}(-N) - \text{erf}(-(N+10)) ] \). This step capitalizes on \( \text{erf}(x) \)’s properties—approaching 1 as \( x \) increases through positive values and 0 for large negative \( x \)s.
This accurate tool enables us to approximate the area under the complex curve \( x e^{-x^2} \) without manual integration, focusing instead on tabulated or computed values of \( \text{erf}(x) \). Consequently, the error function serves as a bridge between continuous probability distributions and numerical integrals.

Definite Integral

A definite integral represents the accumulation of quantities, corresponding to the net area under a curve between specified limits. This differs from indefinite integrals, which yield a family of functions. In our problem, after determining the antiderivative using the integration method and error function, we evaluate the integral between two limits: \( N \) and \( N+10 \).
The solution process begins with finding the antiderivative: \( -\frac{1}{2} e^{-x^2} + C \). To compute the definite integral \( \int_N^{N+10} x e^{-x^2} \, dx \), we use the previously derived formula involving the error function: \( \frac{1}{2} [ \text{erf}(-N) - \text{erf}(-(N+10)) ] \).
This integral value calculates the exact amount of space the curve occupies between the selected \( x \)-bounds. Our goal is to ensure this area is at most 0.01 by choosing appropriate \( N \).
By estimating \( N \) through numerical approximations or tables of \( \text{erf}(x) \), we find the minimal value where the integral satisfies the condition. Through trial, reaching \( N=3 \) achieves our area restriction. A definite integral delivers this real-world measurement outcome—quantifying physical quantities over intervals. It turns abstract symbolic manipulation into concrete data, crucial when precision constraints exist, like our 0.01 area demand.