Problem 70
Question
The error function $$ \displaystyle \text{erf}(x) = \frac{2}{\sqrt{\pi}} \int^x_0 e^{-t^2} \, dt $$ is used in probability, statistics, and engineering. (a) Show that \( \displaystyle \int^b_a e^{-t^2} \, dt = \frac{1}{2} \sqrt{\pi} [ \text{erf}(b) - \text{erf}(a) ] \). (b) Show that the function \( y = e^{x^2} \text{erf}(x) \) satisfies the differential equation \( y' = 2xy + 2/\sqrt{\pi} \).
Step-by-Step Solution
Verified Answer
(a) Use the definition of erf to express the definite integral.
(b) Differentiate using the product rule and verify terms cancel as needed.
1Step 1: Use properties of definite integrals for part (a)
To show that \( \int^b_a e^{-t^2} \, dt = \frac{1}{2} \sqrt{\pi} [ \text{erf}(b) - \text{erf}(a) ] \), recall the definition of the error function: \( \text{erf}(x) = \frac{2}{\sqrt{\pi}} \int^x_0 e^{-t^2} \, dt \). Start by writing the definite integral from \(a\) to \(b\):\[ \int^b_a e^{-t^2} \, dt = \int^b_0 e^{-t^2} \, dt - \int^a_0 e^{-t^2} \, dt. \]Using the definition of the error function, this can be rewritten as:\[ \int^b_a e^{-t^2} \, dt = \frac{\sqrt{\pi}}{2} \text{erf}(b) - \frac{\sqrt{\pi}}{2} \text{erf}(a). \]Thus, \( \int^b_a e^{-t^2} \, dt = \frac{1}{2} \sqrt{\pi} [ \text{erf}(b) - \text{erf}(a) ] \).
2Step 2: Differentiate y = e^{x^2} erf(x) for part (b)
Define the function \( y = e^{x^2} \text{erf}(x) \). We need to find \( y' \) and show it satisfies the given differential equation. Differentiate using the product rule: \[ y = u \cdot v, \text{ where } u = e^{x^2}, \text{ and } v = \text{erf}(x). \]The derivative is \[ y' = u'v + uv'. \]Here, \( u' = 2xe^{x^2} \) because the derivative of \( e^{x^2} \) is \( 2xe^{x^2} \), and \( v' = \frac{2}{\sqrt{\pi}} e^{-x^2} \) from the definition of \( \text{erf}(x) \) where differentiating inside the integral gives this expression. So,\[ y' = (2xe^{x^2}) \text{erf}(x) + e^{x^2} \left( \frac{2}{\sqrt{\pi}} e^{-x^2} \right). \]
3Step 3: Simplify terms and verify the equation
The expression for \( y' \) is \[ y' = 2x(e^{x^2} \text{erf}(x)) + e^{x^2} \cdot \left( \frac{2}{\sqrt{\pi}} e^{-x^2} \right). \]Notice that \( e^{x^2} \cdot \left( \frac{2}{\sqrt{\pi}} e^{-x^2} \right) = \frac{2}{\sqrt{\pi}} \) because the exponent terms cancel out \( e^{x^2} e^{-x^2} = 1 \).Plug this back into \( y' \):\[ y' = 2xy + \frac{2}{\sqrt{\pi}}. \]This confirms that \( y' = 2xy + \frac{2}{\sqrt{\pi}} \), which satisfies the given differential equation.
Key Concepts
ProbabilityDifferential EquationCalculusIntegration Techniques
Probability
The error function, often denoted as \( ext{erf}(x) \), plays a significant role in the field of probability and statistics. It helps in calculating the probability that a random variable following a normal distribution falls within a certain range. This function is particularly useful because many natural phenomena follow a normal distribution, making \( ext{erf}(x) \) a handy tool for engineers, scientists, and statisticians.
- The error function is linked to the cumulative distribution function (CDF) of the normal distribution, which is commonly used in statistics to assess probabilities related to the normal curve.
- In probability theory, \( ext{erf}(x) \) is used to determine the likelihood of a variable falling within the tails of the distribution, which is crucial for risk assessment and decision making.
Differential Equation
The study of differential equations involves understanding equations that describe how things change. These are crucial in modeling the behavior of complex systems in engineering and physics. The exercise demonstrates how differential equations can be derived and solved using the error function.
In the given problem, we have the equation \( y = e^{x^2} ext{erf}(x) \) satisfying a differential equation \( y' = 2xy + 2/\sqrt{\pi} \). This showcases how calculus and differential equations intertwine:
In the given problem, we have the equation \( y = e^{x^2} ext{erf}(x) \) satisfying a differential equation \( y' = 2xy + 2/\sqrt{\pi} \). This showcases how calculus and differential equations intertwine:
- We apply the product rule to differentiate the function \( y = e^{x^2} ext{erf}(x) \).
- Understanding the concept of derivatives is essential, as it helps describe how \( y \) changes with respect to \( x \).
- With the derived expression for \( y' \), we verify that it fulfills the requirements of the given equation, illustrating the properties and behaviors of the system modeled by our equation.
Calculus
Calculus is a branch of mathematics focused on change and motion, dealing extensively with differentiation and integration. In this exercise, calculus plays a pivotal role, particularly with respect to the differentiation and integration of functions related to the error function.
Here’s how calculus principles are applied:
Here’s how calculus principles are applied:
- We use integration to express the error function, \( ext{erf}(x) \), by integrating the exponential function \( e^{-t^2} \) over a variable range.
- Calculus also helps in the derivation of the differential equation by applying the product rule and understanding differential calculus to express \( y' \).
Integration Techniques
Integration techniques involve strategies to determine the integral of functions, which helps calculate areas under curves or find cumulative quantities. In the error function, integration is central, as shown by its definition involving the integral of \( e^{-t^2} \).
To solve the given exercise, specific techniques are used:
To solve the given exercise, specific techniques are used:
- The definite integral from \( a \) to \( b \) is evaluated using the properties of the error function, defining how one can transform the integral of an exponential function into a recognizable form.
- These techniques include understanding and applying limits of integration and manipulating them based on known functions like \( ext{erf}(x) \).
Other exercises in this chapter
Problem 69
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Evaluate the definite integral. \( \displaystyle \int^1_0 \frac{e^z + 1}{e^z + z} \, dz \)
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