Question

For each \(x>0 .\) let \(G(x)\) \(=\int_{0}^{\infty} \mathrm{e}^{-\mathrm{xt}} \mathrm{dt}\). Prove that \(\mathrm{xG}(\mathrm{x})=1\) for each \(\mathrm{x}>0\).

Short answer

Question: Prove that for the integral \(G(x) = \int_{0}^{\infty} e^{-xt} dt\) and for each \(x > 0\), the condition \(xG(x) = 1\) holds true. Answer: We derived that \(G(x) = \frac{1}{x}\) for each \(x > 0\). After checking the condition by multiplying x to G(x), we found that \(xG(x) = 1\) for each \(x > 0\). Therefore, the statement is true.

Step-by-step solution

Evaluate the integral of G(x)

To evaluate the integral, we use the fact that the integral of \(e^{-xt}\) with respect to \(t\) is \(-\frac{1}{x}e^{-xt}\), provided \(x \neq 0\). Hence, using the fundamental theorem of calculus, we have: $$ G(x) = \int_{0}^{\infty} e^{-xt} dt = \left[-\frac{1}{x}e^{-xt}\right]_{0}^{\infty} $$

Evaluate the integral's limits

Now, we need to evaluate the limit of the expression as t approaches infinity and when t is equal to 0 to find G(x): $$ G(x) = \lim_{t \to \infty} \left(-\frac{1}{x}e^{-xt}\right) + \left(\frac{1}{x}e^{0}\right) $$ As \(t\) approaches infinity, the term \(e^{-xt}\) will tend to 0 since \(x > 0\). So the integral becomes: $$ G(x) = 0 + \frac{1}{x} = \frac{1}{x} $$

Check the condition xG(x) = 1

Now, we want to check if \(xG(x) = 1\). So multiply x to G(x): $$ xG(x) = x\left(\frac{1}{x}\right) $$ Notice that the \(x\) in the numerator and denominator will cancel out, leaving the following expression: $$ xG(x) = 1 $$ The condition \(xG(x)=1\) holds true for each \(x > 0\).

Key concepts

Definite Integration

Definite integration is a fundamental concept within integral calculus that involves evaluating the integral of a function over a specific interval. The process gives us the accumulation of the function's values across the interval between the lower and upper limits of integration.

To illustrate with a simple example, if we're integrating a function from point A to point B, the definite integral tells us the 'total' that adds up under the curve of the function between these two points. It is a crucial technique in various scientific fields, from physics to economics, as it allows the calculation of areas, volumes, and other quantities that can be accumulated.

In the context of the exercise provided, the definite integral of the exponential function given by \( G(x) = \int_{0}^{\infty} e^{-xt} \ dt \) represents the total accumulation of the function's values from 0 to infinity, accounting for the function's exponential decay as controlled by the parameter \(x\).

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration and plays a pivotal role in our understanding of calculus. It states that if a function is continuous over an interval, the definite integral of the derivative of the function over that interval is equal to the difference in the values of the original function at the boundaries of the interval.

In simpler terms, this theorem allows us to evaluate definite integrals using antiderivatives. For our example, the integral \( G(x) = \int_{0}^{\infty} e^{-xt} \ dt \) can be evaluated by finding the antiderivative of the exponential function \( e^{-xt} \), then applying the limits of integration. Essentially, it transforms the seemingly complex problem of finding areas under curves into a more straightforward procedure of evaluating functions at specific points.

Exponential Function Integration

Exponential function integration involves finding the integral of functions that have the exponential function, typically denoted as \(e^{rt}\), where \(r\) can be any real number. In these functions, the variable is in the exponent, which results in a rapidly changing function that can describe growth or decay processes, such as radioactive decay or population growth.

In our exercise's context, \( e^{-xt} \) is the exponential function with a negative rate, implying decay. To integrate it, we seek its antiderivative, which results in \( -\frac{1}{x}e^{-xt} \) with respect to \(t\). This expression is paramount in calculating the definite integral over the range from 0 to infinity, as it encapsulates the entire process of exponential decay. The negative sign signifies that the function is decreasing over time—another critical feature of exponential decay functions and their integrals.

Limits of Integration

The limits of integration determine the interval over which we are calculating the integral. They are essential because they define the section of the function we are interested in, and changing these limits can dramatically affect the result of the integral.

In our specific example, the limits of integration are from 0 to infinity (\(0\) to \(\infty\)), denoting that we're considering the entire positive t-axis. When evaluating at infinity, we're not calculating a numeric value but rather understanding the behavior of the function as it extends towards an infinitely large value. For positive values of \(x\), as \(t\) increases, the exponential function \(e^{-xt}\) decreases rapidly towards zero. Upon integrating, this behavior impacts the evaluation at the upper limit, which, in many cases including this one, contributes nothing to the integral's total value. The lower limit, being zero, typically provides the initial value of the function's antiderivative for the evaluation process.