Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 11: Problem 38
Use series to approximate the values of the integrals in Exercises \(37-40\) with an error of magnitude less than \(10^{-8}\) . $$ \int_{0}^{0.1} e^{-x^{2}} d x $$
Short Answer
Expert verified
Using a series expansion, the integral's approximate value is 0.09966710024, within an error of less than \(10^{-8}\).
Step by step solution
01
Identify the Series Representation of the Integrand
The function to integrate is \( e^{-x^2} \). We'll use the series expansion of \( e^{-x^2} \), which is known as the exponential series. For \( e^{-x^2} \), the expansion is \( 1 - x^2 + \frac{(x^2)^2}{2!} - \frac{(x^2)^3}{3!} + \cdots \). Our task is to determine how many terms we need to achieve an error less than \(10^{-8}\).
02
Simplify the Series
Substitute \( x^2 \) for each \( x \) in the series: \( e^{-x^2} = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \cdots \). We only need the sum of terms that contribute significantly to the integral value for \( x \in [0, 0.1] \).
03
Evaluate the Integral Term by Term
Integrate term by term within the limits \( 0 \) to \( 0.1 \). The integral becomes: \[ \int_0^{0.1} (1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \cdots) \, dx = \left[ x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \cdots \right]_0^{0.1}. \]
04
Calculate the Series Terms
Compute the terms at \( x = 0.1 \):- \( 0.1 - \frac{0.1^3}{3} = 0.1 - 0.000333 \approx 0.099667 \).- \( \frac{0.1^5}{10} = \frac{0.000001}{10} = 0.0000001 \).- \( \frac{0.1^7}{42} = \frac{0.00000001}{42} \approx 0.00000000024 \), and so forth until the term contributes less than \( 10^{-8} \).
05
Sum Up Calculated Integrals
Combine all significant terms:\[ 0.099667 + 0.0000001 - 0.00000000024 \approx 0.09966710024. \]This provides the approximated integral value with an error less than \( 10^{-8} \).
06
Verify Error
Ensure that each term following completes the criterion of contributing less than \( 10^{-8} \) to guarantee the cumulative error stays below that threshold. Check terms until new terms are negligible.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Series Expansion
Series expansion is like taking a complicated function and breaking it down into simpler parts called terms. For functions like the exponential function, this involves expressing the function as a sum of many terms that make it easier to work with. Each term is calculated using powers of the variable, in this case, powers of \(x^2\). For the function \(e^{-x^2}\), it's written as a series:
- \(1 - x^2\)
- \(\frac{(x^2)^2}{2} \)
- \(- \frac{(x^2)^3}{3!}\)
- And so on...
Error Estimation
Error estimation is key to determining how accurate our approximation is. When we approximate a result using a series expansion, we're leaving out the infinite number of remaining terms.
Thus, our result isn't exact. By estimating the error, we learn how close our approximation is to the real value. In our problem, we needed the error to be less than \(10^{-8}\). This is done by making sure any added, neglected terms in the series have a very small impact.
Thus, our result isn't exact. By estimating the error, we learn how close our approximation is to the real value. In our problem, we needed the error to be less than \(10^{-8}\). This is done by making sure any added, neglected terms in the series have a very small impact.
- The rule is simple: When the next term becomes smaller than \(10^{-8}\), we stop because their contribution isn't significant enough.
- It's like when you weigh grains of sand, at some point, adding a single grain doesn鈥檛 change the weight anymore in a meaningful way.
Exponential Function
The exponential function, particularly \(e^{-x^2}\), is an important function in mathematics, representing growth and decay processes. It is pervasive in many scientific fields, from physics to finance. For integration and other complex operations, direct handling of \(e^{-x^2}\) can be challenging due to its nature of rapid growth or decay, depending on the sign and magnitude of its power.
Using a series expansion allows us to tackle such challenges, breaking down the complexity. The beauty of the exponential function is its smoothness and predictable behavior once it's expanded. This gives us a robust tool for approximation, laying the groundwork for further operations.
Using a series expansion allows us to tackle such challenges, breaking down the complexity. The beauty of the exponential function is its smoothness and predictable behavior once it's expanded. This gives us a robust tool for approximation, laying the groundwork for further operations.
Integration by Series
Integration by series is a method where we integrate each term of a series individually. This takes advantage of the simplification provided by a series expansion. For example, integrating \(1 - x^2 + \frac{x^4}{2} \) term by term is much simpler than tackling \(e^{-x^2}\) directly.
- Start by integrating simple power terms like \(x^5\) or \(x^3\).
- Break them down using basic integration techniques.
- Each term follows a pattern: The term \(\frac{x^{2n}}{n!}\) becomes \(\frac{x^{2n+1}}{(2n+1) \cdot n!}\) when integrated.
- Provides an approximation about how the entire function behaves over a range.
- Is practical when we need a numerical result from a complex function over a specific interval.
