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Magazine Chapter 30: Problem 31
Solve the given problems. Evaluate \(\int_{0}^{1} e^{x} d x\) directly and compare the result obtained by using four terms of the series for \(e^{x}\) and then integrating.
Short Answer
Expert verified
The direct integration gives \(e-1\) and using the series gives \(\frac{41}{24}\), which is approximately equal.
Step by step solution
01
Recognize the Integration Task
You are asked to evaluate the definite integral \(\int_{0}^{1} e^{x} \, dx\). This represents the area under the curve of \(e^{x}\) from \(x=0\) to \(x=1\).
02
Evaluate the Integral Directly
To solve \(\int e^{x} \, dx\), remember that the antiderivative of \(e^{x}\) is itself. Thus, the definite integral is evaluated as follows:\[\int_{0}^{1} e^{x} \, dx = \left[e^{x}\right]_{0}^{1} = e^{1} - e^{0} = e - 1.\]
03
Express \(e^{x}\) as a Series
The function \(e^{x}\) can be expanded using its Maclaurin series, which is:\[e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots\] We will use only the first four terms: \[e^{x} \approx 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6}.\]
04
Integrate the Series Term by Term
Integrate each term of the series from \(0\) to \(1\):\[\int_{0}^{1} \left(1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6}\right) \ dx = \int_{0}^{1} 1 \ dx + \int_{0}^{1} x \ dx + \int_{0}^{1} \frac{x^{2}}{2} \ dx + \int_{0}^{1} \frac{x^{3}}{6} \ dx.\]
05
Evaluate Each Integral
Calculate each integral separately:1. \[\int_{0}^{1} 1 \, dx = \left[x\right]_{0}^{1} = 1\]2. \[\int_{0}^{1} x \, dx = \left[\frac{x^{2}}{2}\right]_{0}^{1} = \frac{1}{2}\]3. \[\int_{0}^{1} \frac{x^{2}}{2} \, dx = \left[\frac{x^{3}}{6}\right]_{0}^{1} = \frac{1}{6}\]4. \[\int_{0}^{1} \frac{x^{3}}{6} \, dx = \left[\frac{x^{4}}{24}\right]_{0}^{1} = \frac{1}{24}\]
06
Sum the Series Results
Add the results of the individual integrals:\[1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} = \frac{24}{24} + \frac{12}{24} + \frac{4}{24} + \frac{1}{24} = \frac{41}{24}.\]
07
Compare Both Results
The direct integration gives \(e-1\), and the series approximation gives \(\frac{41}{24}\). Calculating numerically, \(e \approx 2.718\), so \(e - 1 \approx 1.718\). Evaluating \(\frac{41}{24}\approx 1.708\), provides a close approximation but slightly less than \(e-1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
The concept of a definite integral is foundational in calculus. It refers to the calculation of the area under a curve, given by a function, between two points on the x-axis. In this exercise, the definite integral is represented as \(\int_{0}^{1} e^{x} \, dx\).To understand this:
- The function \(e^x\) refers to the curve whose area we are examining.
- The limits 0 to 1 specify the interval on the x-axis for which the area under this curve is calculated.
Maclaurin Series
The Maclaurin series is a powerful tool for representing functions as infinite sums of their derivatives at a single point, often zero. This technique is useful in approximating functions and integrating more complex calculations. For the function \(e^x\), its Maclaurin series can be written as:\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\]In this exercise:
- We approximate \(e^x\) by using only the first four terms of its series: \(1 + x + \frac{x^2}{2} + \frac{x^3}{6}\).
Integral Approximation
Integral approximation involves estimating the value of an integral when finding an exact solution is challenging. By using the Maclaurin series for \(e^x\):
- We break down the function into simpler terms that can be integrated individually.
- We integrate these terms term by term over the desired interval.
Exponential Function
The exponential function, denoted as \(e^x\), is a critical component in calculus due to its unique properties. It is widely used across different fields, including finance, physics, and biology, because of its continuous growth behavior. Key characteristics include:
- Its derivative is \(e^x\) itself, which simplifies many calculus operations.
- It grows exponentially, meaning it rapidly increases as \(x\) becomes larger.
