Problem 73 - Calculus Early Transcendentals

Question

Let \(a>0\) and let \(R\) be the region bounded by the graph of \(y=e^{-a x}\) and the \(x\) -axis on the interval \([b, \infty)\) a. Find \(A(a, b),\) the area of \(R\) as a function of \(a\) and \(b\) b. Find the relationship \(b=g(a)\) such that \(A(a, b)=2\) c. What is the minimum value of \(b\) (call it \(b^{*}\) ) such that when \(b>b^{*}, A(a, b)=2\) for some value of \(a>0 ?\)

Step-by-Step Solution

Answer

Based on the given information, the area A(a,b) of the region R is found to be: \(A(a, b) = \frac{1}{a} e^{-a b}.\) The relationship b=g(a) such that A(a, b)=2 is given by: \(b = g(a) = -\frac{\ln (2a)}{a}.\) And the minimum value of b (b*) such that A(a, b)=2 for some value of a > 0 is approximately 0.873.

Step 1: To find the area A(a, b) of the region R, we will integrate the function y = e^(-a x) with respect to x over the interval [b, ∞]. The area can be written as: \(A(a, b) = \int_b^{\infty} e^{-a x} dx \) #Step 2: Evaluate the integral#

Now, we will evaluate the integral to get the area A(a, b) as a function of a and b: \(A(a, b) = \left[-\frac{1}{a} e^{-a x}\right]_b^{\infty} = -\frac{1}{a} e^{-a \infty} + \frac{1}{a} e^{-a b}.\) Since \(e^{-a \infty} = 0\), we get: \(A(a, b) = \frac{1}{a} e^{-a b}.\) #Step 3: Find the relationship b=g(a) such that A(a, b)=2#

Step 2: Now, we will find the relationship b=g(a) such that A(a, b)=2. We'll set the area equal to 2 and solve for b: \(2 = \frac{1}{a} e^{-a b}.\) To solve for b, we first multiply both sides by a: \(2a = e^{-a b}.\) Next, we take the natural logarithm of both sides: \(\ln (2a) = -a b.\) Finally, we solve for b: \(b = g(a) = -\frac{\ln (2a)}{a}.\) #Step 4: Find the minimum value of b (b*) such that A(a, b)=2 for some value of a>0#

To find the minimum value of b, we are going to differentiate the expression for g(a) with respect to a and set it equal to 0: \(g'(a) = \frac{d}{da} \left(-\frac{\ln (2a)}{a}\right) = -\frac{1}{a} + \frac{\ln (2a)}{a^2}.\) Set g'(a) = 0 and solve for a: \(0 = -\frac{1}{a} + \frac{\ln (2a)}{a^2} \Rightarrow a = \ln (2a).\) By trying different positive values of a and approximating, we find that \(a \approx 1.678.\) Now, we plug this value of a back into g(a) to find b*: \(b^* = g(a) = -\frac{\ln (2(1.678))}{1.678} \approx 0.873.\) So, the minimum value of b (b*) such that when \(b>b^*, A(a, b)=2\) for some value of \(a>0\) is approximately 0.873.

Key Concepts

Integration by Substitution

The technique of integration by substitution is often utilized in calculus to simplify an integral, making it easier to solve. This method involves changing the variable of integration to another variable with a relationship to the original one. It's somewhat analogous to finding a new perspective from which a problem is easier to manage.

To understand this concept, imagine you have a function which, when you look at it directly, seems complex. But, if you put on a pair of 'mathematical glasses,' or in other words, make a substitution, the function appears in another form that is more approachable.

Consider the integral \( \int e^{-a x} dx \) as an example. By setting \( u = -ax \) and \( du = -a dx \) (which also implies \( dx = -du/a \) after rearranging), we replace the variable \( x \) with \( u \), and the integral becomes \( \int e^{u} \frac{du}{-a} \) which is more straightforward to integrate.

Area Under a Curve

One of the most visually intuitive applications of integration is calculating the area under a curve. When you plot a graph of a function, the space between the graph and the x-axis up to a certain point can represent various physical concepts, such as distance traveled over time or even the amount of material needed to create a shape. In the language of mathematics, we describe this space as 'the area under the curve.'

To find this area, we essentially add up an infinite number of infinitesimally small rectangles under the curve. Their combined area gives us the total area up to a certain point—or indefinitely if we're integrating to infinity, which brings unique considerations like ensuring the function approaches zero as \( x \to \infty \).

In our exercise, we investigated the area under the curve \( y=e^{-ax} \) from \( b \) to infinity. The integral \( A(a, b) = \int_b^{\infty} e^{-ax} dx \) represents exactly this area, and solving it provides us with a precise numerical value, dependent on the specific function and the bounds of integration.

Exponential Decay

The concept of exponential decay describes a process where quantities decrease at a rate proportionate to their current value. This kind of decay appears in many realms of science, including radioactive decay in physics and depreciation in finance. Think of it like a loaf of bread getting moldy day by day; each day, the amount of new mold is proportional to how much bread is left to get moldy.

In mathematics, exponential decay is modeled with functions like \( y=e^{-a x} \). Here, \( a \) is a positive constant that dictates how quickly the function descends towards zero as \( x \) increases. This behavior creates a diminishing curve that never touches the x-axis but gets infinitely close to it. This trait of an exponential decay function is critical when integrating over an infinite range, as seen in our exercise. The integral's value remains finite because the function \( e^{-ax} \) approaches zero fast enough to compensate for the infinite range of integration.