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Magazine Chapter 7: Problem 41
In Problems \(38-41,\) evaluate \(f(3)\). $$f(x)=\int_{0}^{\infty} 2 t x e^{-t x^{2}} d t$$
Short Answer
Expert verified
\(f(3) = 1\).
Step by step solution
01
Understand the exercise
We need to evaluate the function \(f(x)\) at \(x = 3\). The function is given as a definite integral from 0 to infinity.
02
Substitute into the integral
Substitute \(x = 3\) into the expression for \(f(x)\). This gives \(f(3) = \int_{0}^{\infty} 2t \, 3 \, e^{-3t^{2}} \, dt\). Simplify this to get \(f(3) = 6 \int_{0}^{\infty} t e^{-3t^{2}} \, dt\).
03
Simplify the integrand
Recognize that the expression inside the integral, \(t e^{-3t^{2}}\), is suitable for a substitution method. This integral can be simplified by substitution.
04
Use a substitution
Use the substitution \(u = 3t^{2}\). Then, \(du = 6t \, dt\) or \(t \, dt = \frac{1}{6} \, du\). Transform the limits: when \(t = 0\), \(u = 0\), and when \(t \rightarrow \infty\), \(u \rightarrow \infty\). Now, the integral becomes \(\frac{1}{6} \int_{0}^{\infty} e^{-u} \, du\).
05
Evaluate the integral
The integral \(\int_{0}^{\infty} e^{-u} \, du\) is a standard integral that evaluates to 1. Hence, \(\frac{1}{6} \int_{0}^{\infty} e^{-u} \, du = \frac{1}{6} \times 1 = \frac{1}{6}\).
06
Multiply by the constant
Recall that \(f(3) = 6 \times \frac{1}{6}\). Simplifying gives: \(f(3) = 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
A definite integral is a fundamental concept in integral calculus. It allows us to find the accumulation of quantities, such as area under a curve, from one point to another. In the exercise, we have the definite integral:
\[\int_{0}^{\infty} 2txe^{-tx^2} dt\]It calculates the total change or accumulated quantity from 0 to infinity. With definite integrals, the limits of integration (in this example, 0 to infinity) define the region of concern, and they help find precise numerical values, like our result of 1 in this problem. This critical tool is applied in physics, economics, statistics, and more to derive important quantities.
\[\int_{0}^{\infty} 2txe^{-tx^2} dt\]It calculates the total change or accumulated quantity from 0 to infinity. With definite integrals, the limits of integration (in this example, 0 to infinity) define the region of concern, and they help find precise numerical values, like our result of 1 in this problem. This critical tool is applied in physics, economics, statistics, and more to derive important quantities.
- Computes total measurement over an interval
- Requires limits of integration
- Results in a specific numerical value
Substitution Method
The substitution method is a popular technique for solving integrals. It makes complex integrals more manageable by transforming them into simpler forms. In this exercise, we use the substitution:
\[\text{Let } u = 3t^2 \quad \text{so } du = 6t \, dt\]This step reorganizes the integral into a form where variables and constants are easier to work with:
\[\frac{1}{6} \int_{0}^{\infty} e^{-u} \, du\]The substitution method effectively changes the variable and integrand, simplifying the integration process. This method is powerful and often essential when dealing with exponential, trigonometric, and logarithmic functions, as it cleverly manipulates the integral to a known and more straightforward form.
\[\text{Let } u = 3t^2 \quad \text{so } du = 6t \, dt\]This step reorganizes the integral into a form where variables and constants are easier to work with:
\[\frac{1}{6} \int_{0}^{\infty} e^{-u} \, du\]The substitution method effectively changes the variable and integrand, simplifying the integration process. This method is powerful and often essential when dealing with exponential, trigonometric, and logarithmic functions, as it cleverly manipulates the integral to a known and more straightforward form.
- Simplifies the original integrand
- Involves variable and limit changes
- Transforms complicated integrals into standard forms
Exponential Function
Exponential functions play a vital role in calculus due to their consistent rate of increase. They appear frequently in integrals, differential equations, and real-world applications like population growth, radioactive decay, and compound interest. In this exercise, the integrand:
\[2txe^{-tx^2}\]contains an exponential function, \(e^{-tx^2}\), which is characteristic of exponential decay. Exponential functions have simple differentiation and integration properties, making them easily manageable when handled with appropriate calculus techniques, like substitution. They naturally model phenomena that have a multiplicative rate of change.
\[2txe^{-tx^2}\]contains an exponential function, \(e^{-tx^2}\), which is characteristic of exponential decay. Exponential functions have simple differentiation and integration properties, making them easily manageable when handled with appropriate calculus techniques, like substitution. They naturally model phenomena that have a multiplicative rate of change.
- Model rapid growth or decay
- Standard functions with well-known properties
- Common in a variety of scientific models
