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 \(a>0 ?\)

Short answer

#Summary# In this problem, we were given a region R bounded by the exponential function \(y=e^{-ax}\) and the x-axis. We found the area A(a, b) of the region as a function of a and b by integrating and evaluating: \(A(a, b) = \frac{1}{a}e^{-ab}\). Then, we found the relationship between a and b when the area is 2: \(b = \frac{1}{a}\ln{\frac{1}{2a}}\). Lastly, we determined the minimum value of b such that the area is 2 for some value of a. The minimum value of b is 0, so when \(b>0\), the area A(a, b) = 2 for some value of a > 0.

Step-by-step solution

Calculating the Area A(a, b) using Integration

Using definite integration, we can calculate the area between the curve \(y=e^{-ax}\) and the x-axis. The integral will be from b to infinity, as our region goes to infinity along the x-axis. The resulting formula is: \(A(a, b) = \int_b^\infty e^{-ax} dx\) Now we need to find the antiderivative of \(e^{-ax}\) with respect to x and evaluate the integral.

Finding the Antiderivative of \(e^{-ax}\)

To find the antiderivative of \(e^{-ax}\), we can apply the u-substitution method by letting \(u=-ax\). The derivative of \(u\) with respect to x is \(-a\). Therefore, we have \(dx = -\frac{1}{a}du\). Replacing the u and dx values in the integral, we get: \(A(a, b) = -\frac{1}{a}\int e^u du\) The antiderivative of \(e^u\) is \(e^u\), which gives us: \(A(a, b) = -\frac{1}{a}e^u\) Now we need to substitute back the u in terms of x: \(A(a, b) = -\frac{1}{a}e^{-ax}\) Now we will evaluate this antiderivative from the limits b to \(\infty\).

Evaluating the Definite Integral

We have: \(A(a, b) = -\frac{1}{a}[e^{-a\infty} - e^{-ab}]\) As the exponential term involving infinity goes to 0, we are left with the following expression: \(A(a, b) = \frac{1}{a}e^{-ab}\) Now we have A(a, b) as a function of a and b.

Finding the Relationship Between a and b When A(a, b) = 2

To find the relationship between a and b when the area equals 2, we use the equation \(A(a, b) = 2\): \(2 = \frac{1}{a}e^{-ab}\) Then, we solve for b: \(b = \frac{1}{a}\ln{\frac{1}{2a}}\) This equation represents the relationship between a and b, \(b = g(a)\).

Finding Minimum Value of b for A(a, b) = 2

To find the minimum value of b, we need to minimize the function \(g(a)\): \(b = g(a) = \frac{1}{a}\ln{\frac{1}{2a}}\) First, let's find the critical points of g(a) by taking its derivative: \(g'(a) = \frac{d}{da}[\frac{1}{a}\ln{\frac{1}{2a}}]\) Here, we apply the quotient rule and product rule, getting: \(g'(a) = \frac{2a\ln{\frac{1}{2a}} + 1}{a^2}\) Now let's find the critical points by setting \(g'(a) = 0\): \(\frac{2a\ln{\frac{1}{2a}} + 1}{a^2} = 0\) Solving for a, we find a critical value of \(a_c = \frac{1}{2}\). Now, we can plug this value of a_c into g(a) to find the minimum value of b: \(b^* = g(a_c) = g(\frac{1}{2}) = \frac{1}{\frac{1}{2}}\ln{\frac{1}{2\frac{1}{2}}} = 2\ln{1} = 0\) Thus, the minimum value of b is 0. Therefore, when \(b > 0\), the area A(a, b) = 2 for some value of a > 0.

Key concepts

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. A common type of exponential function is the natural exponential function, denoted as \(e^x\), where \(e\) is Euler's number, approximately 2.718.

This category of functions exhibits unique characteristics, such as a consistent rate of growth or decay, depending on whether the exponent is positive or negative, respectively. The function \(y = e^{-ax}\) from the exercise showcases exponential decay as the variable \(x\) increases.

Understanding exponential behavior is crucial when dealing with phenomena like population growth, radioactive decay, and financial calculations involving compound interest. In this context, we explore how the graph of \(y = e^{-ax}\) exponentially decreases, forming a curve above the \(x\)-axis that stretches towards infinity.

Area Under a Curve

The area under a curve is a key concept in calculus, often determining the accumulated quantity over a continuous interval. In this case, the challenge is to compute the area under the exponential decay function \(y = e^{-ax}\) from \(b\) to infinity.

To find this area, we use integration, a powerful tool in calculus. The definite integral of a function provides the sum of areas of infinitesimally small slices under the curve. More formally, it calculates the limit of these sums as the slices become infinitely narrow.

• In the exercise, setting up the integral \(\int_b^\infty e^{-ax} \, dx\) represents the area under the curve from \(b\) to infinity.
• The process transforms solving this area problem into evaluating an integral, which gives us the required measurement—in this case, \(A(a, b)\).

This outcome reveals the extent to which the region under the curve is stretched along the \(x\)-axis, capturing how it spans towards infinity.

Antiderivative

The antiderivative of a function, also known as the indefinite integral, unveils another important calculus concept. It identifies a function whose derivative returns the original function. In simpler terms, it's about finding a function that, when differentiated, yields the given function.

In the steps given, to find the antiderivative of \(e^{-ax}\), a common integration technique called "u-substitution" is deployed. This method simplifies the integration process through substitution. By letting \(u = -ax\), differentiation results in \(du = -a \, dx\), transforming the original integral into a simpler form.

Evaluating the integral using this antiderivative provides a formula needed to calculate the area under the curve. The exploration of such a derivative illuminates how seemingly complex functions are integrated. Moreover, antiderivatives offer insights into how various functions behave and their relationships with derivatives.