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Chapter 8: Problem 73

Estimating the value of a convergent improper integral whose domain is infinite a. Show that $$\int_{3}^{\infty} e^{-3 x} d x=\frac{1}{3} e^{-9}<0.000042$$ and hence that \(\int_{3}^{\infty} e^{-x^{2}} d x<0.000042 .\) Explain why this means that \(\int_{0}^{\infty} e^{-x^{2}} d x\) can be replaced by \(\int_{0}^{3} e^{-x^{2}} d x\) without introducing an error of magnitude greater than 0.000042 . b. Evaluate \(\int_{0}^{3} e^{-x^{2}} d x\) numerically.

Short Answer

Expert verified
The integral \( \int_{0}^{3} e^{-x^2} dx \approx 0.886 \). The error introduced is less than 0.000042.

Step by step solution

01

Evaluate the First Integral

First, consider the integral \( \int_{3}^{\infty} e^{-3x} dx \). To evaluate this improper integral, set it up as a limit: \[ \lim_{b \to \infty} \int_{3}^{b} e^{-3x} dx. \] The antiderivative of \( e^{-3x} \) is \( -\frac{1}{3}e^{-3x} \). Thus, \[ \begin{align*} \int_{3}^{b} e^{-3x} dx & = \left[ -\frac{1}{3}e^{-3x} \right]_{3}^{b} \ & = \left( -\frac{1}{3}e^{-3b} + \frac{1}{3}e^{-9} \right).\end{align*} \] Taking the limit as \( b \to \infty \), we are left with: \[ \lim_{b \to \infty} \left( -\frac{1}{3}e^{-3b} + \frac{1}{3}e^{-9} \right) = \frac{1}{3}e^{-9}, \] where \( \frac{1}{3}e^{-3b} \to 0 \) as \( b \to \infty \).
02

Compare with 0.000042

Next, calculate \( \frac{1}{3}e^{-9} \) to ensure that it is less than 0.000042. First, compute \( e^{-9} \approx 0.000123 \). Then, \( \frac{1}{3}e^{-9} \approx \frac{1}{3} \times 0.000123 = 0.000041 \). Since 0.000041 < 0.000042, we confirm that \( \int_{3}^{\infty} e^{-3x} dx < 0.000042 \).
03

Estimate Second Integral

Now, consider the integral \( \int_{3}^{\infty} e^{-x^2} dx \). Since the decay of \( e^{-x^2} \) is faster than that of \( e^{-3x} \), it follows that: \[ \int_{3}^{\infty} e^{-x^2} dx < \int_{3}^{\infty} e^{-3x} dx < 0.000042 . \] This inequality holds because \( e^{-x^2} \) decreases at least as fast as or faster than \( e^{-3x} \). Therefore: \[ \int_{3}^{\infty} e^{-x^2} dx < 0.000042 . \]
04

Justify Integral Approximation

From Step 3, we know that \( \int_{3}^{\infty} e^{-x^2} dx < 0.000042 \). Thus, the error introduced by replacing \( \int_{0}^{\infty} e^{-x^2} dx \) with \( \int_{0}^{3} e^{-x^2} dx \) is less than 0.000042, which is acceptable for precise evaluations and approximations.
05

Numerically Evaluate \( \int_{0}^{3} e^{-x^2} dx \)

To evaluate \( \int_{0}^{3} e^{-x^2} dx \) numerically, we can use integration techniques such as Simpson's Rule or numerical integration tools like a calculator or software:Let's assume we use a numerical calculator that gives us the value for this integral, which is approximately \( 0.886 \). Thus, \( \int_{0}^{3} e^{-x^2} dx \approx 0.886 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Improper Integral Approximation
Improper integrals are those where the function's domain extends to infinity or the function has an infinite discontinuity. Calculating these integrals often involves limits.
When working with such integrals, we use approximations to handle them practically. This was exemplified in the estimation of \( \int_{3}^{\infty} e^{-x^{2}} dx \).
To find this approximation effectively, it is critical to compare the integral with another function that decays just as fast or slower. In this case, we used \( \int_{3}^{\infty} e^{-3x} dx \).
Since \( e^{-3x} \) decays slower than \( e^{-x^{2}} \), comparing these integral values provides an upper bound. The result \( \int_{3}^{\infty} e^{-x^{2}} dx < 0.000042 \) confirms that its contribution to \( \int_{0}^{\infty} e^{-x^{2}} dx \) is minimal, allowing for an approximation with \( \int_{0}^{3} e^{-x^{2}} dx \) without significant error.
Exponential Decay
Exponential decay is an essential concept, especially in the context of improper integrals. It describes the rapid decrease of exponential functions over increasing values.
In the given problem, both \( e^{-3x} \) and \( e^{-x^{2}} \) represent exponential decay functions.
  • \( e^{-3x} \) decays exponentially but at a constant rate provided by the exponent.
  • \( e^{-x^{2}} \) decays faster due to its quadratic exponent.
This faster decay rate makes \( e^{-x^{2}} \) smaller than \( e^{-3x} \) for all \( x > 3 \) and underlines the logic behind using \( e^{-3x} \) as a form of comparison. Recognizing these decay properties assists in setting up more informed integral bounds and making more accurate estimates when comparing or approximating integrals.
Numerical Integration
Numerical integration is a technique to approximate the value of integrals when they cannot be solved analytically. For integrals like \( \int_{0}^{3} e^{-x^{2}} dx \), exact solutions might be infeasible or impossible, thus numerical methods come handy.
Common techniques include:
  • Simpson's Rule: Uses parabolic segments to estimate the area under the curve.
  • Trapezoidal Rule: Approximates the area using trapezoids.
  • Numerical software gives quick results for complex integrals, saving time and effort.
In our example, a numerical method provided an approximate value of 0.886, ensuring practical and efficient computation. Understanding when and how to apply these numerical methods is critical to functioning well in areas involving advanced calculus and applied mathematics.

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Most popular questions from this chapter

Elliptic integrals The length of the ellipse $$ x=a \cos t, \quad y=b \sin t, \quad 0 \leq t \leq 2 \pi $$ turns out to be $$ =4 a \int_{0}^{\pi / 2} \sqrt{1-e^{2} \cos ^{2} t} d t $$ where \(e\) is the ellipse's eccentricity. The integral in this formula, called an elliptic integral, is non elementary except when \(e=0\) or 1 a. Use the Trapezoidal Rule with \(n=10\) to estimate the length of the ellipse when \(a=1\) and \(e=1 / 2\) . b. Use the fact that the absolute value of the second derivative of \(f(t)=\sqrt{1-e^{2} \cos ^{2} t}\) is less than 1 to find an upper bound for the error in the estimate you obtained in part (a).

Evaluate the integrals in Exercises \(1-28\). $$ \int \frac{\left(1-x^{2}\right)^{3 / 2}}{x^{6}} d x $$

In Exercises \(89-92\) , use a CAS to explore the integrals for various values of \(p\) (include noninteger values). For what values of \(p\) does the integral converge? What is the value of the integral when it does converge? Plot the integrand for various values of \(p .\) $$ \int_{-\infty}^{\infty} x^{p} \ln |x| d x $$

In Exercises \(29-36\) , use an appropriate substitution and then a trigonometric substitution to evaluate the integrals. $$ \int \frac{d x}{x \sqrt{x^{2}-1}} $$

As we mentioned at the beginning of the section, the definite integrals of many continuous functions cannot be evaluated with the Fundamental Theorem of Calculus because their antiderivatives lack elementary formulas. Numerical integration offers a practical way to estimate the values of these so-called nonelementary integrals. If your calculator or computer has a numerical integration routine, try it on the integrals in Exercises \(39-42\) . $$ \int_{0}^{\pi / 2} 40 \sqrt{1-0.64 \cos ^{2} t} d t $$
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