Question

The curves \(y=x e^{-a x}\) are shown in the figure for \(a=1,2,\) and 3 a. Find the area of the region bounded by \(y=x e^{-x}\) and the \(x\) -axis on the interval [0,4]. b. Find the area of the region bounded by \(y=x e^{-a x}\) and the \(x\) -axis on the interval \([0,4],\) where \(a > 0\). c. Find the area of the region bounded by \(y=x e^{-a x}\) and the \(x\) -axis on the interval \([0, b] .\) Because this area depends on \(a\) and \(b,\) we call it \(A(a, b),\) where \(a > 0\) and \(b > 0\). d. Use part (c) to show that \(A(1, \ln b)=4 A(2,(\ln b) / 2)\). e. Does this pattern continue? Is it true that \(A(1, \ln b)=\) \(a^{2} A(a,(\ln b) / a) ?\).

Short answer

Answer: The exercise does not provide enough information to determine if this pattern continues for other values of \(a\). To confirm the pattern for different values, more information or techniques would be required.

Step-by-step solution

Set up the integral for area calculation

Firstly, we need to set up the integral for calculating the area under the curve \(y=xe^{-x}\) from \(x=0\) to \(x=4\). Since we are calculating the area above the x-axis, the integral will be represented as follows: $$ \int_0^4 xe^{-x} dx $$

Evaluate the integral

To evaluate the integral, we need to perform integration by parts. Let's choose \(u=x\) and \(dv=e^{-x}dx\). This gives us \(du=dx\) and \(v=-e^{-x}\). Now apply the integration by parts formula: \(\int u dv = uv - \int v du\). We get: $$ \int_0^4 xe^{-x} dx = -xe^{-x}\bigg|_0^4 - \int_0^4 -e^{-x} dx $$ Now, integrate \(-e^{-x}\) to get \(e^{-x}\). Then evaluate the integral: $$ -xe^{-x} + e^{-x}\bigg|_0^4 = -4e^{-4} + e^{-4} - (-1) = 1 - 3e^{-4} $$ So, the area of the region bounded by the curve and the x-axis on the interval [0, 4] is \(1 - 3e^{-4}\). #b. Calculate area under the curve for a general \(a\)#

Setup the integral for the area calculation

Now, let's set up the integral for calculating the area under the curve \(y=xe^{-ax}\) from \(x=0\) to \(x=4\) for a general value of \(a > 0\). The integral is represented as follows: $$ \int_0^4 xe^{-ax} dx $$

Evaluate the integral

This time, we'll still choose \(u=x\) and \(dv=e^{-ax}dx\), which gives us \(du=dx\) and \(v=-\frac{1}{a}e^{-ax}\). Now apply the integration by parts formula: $$ \int_0^4 xe^{-ax} dx=-\frac{1}{a}xe^{-ax}\bigg|_0^4 - \int_0^4 -\frac{1}{a}e^{-ax} dx $$ Now, integrate \(-\frac{1}{a}e^{-ax}\) to get \(\frac{1}{a}e^{-ax}\). Then evaluate the integral: $$ -\frac{1}{a}xe^{-ax} + \frac{1}{a^2}e^{-ax}\bigg|_0^4 = -\frac{4}{a}e^{-4a} + \frac{1}{a^2}(1-e^{-4a}) $$ So, the area of the region bounded by the curve and the x-axis on the interval [0, 4] for \(a>0\) is \(A(a, 4)=-\frac{4}{a}e^{-4a} + \frac{1}{a^2}(1-e^{-4a})\). #c. Calculate area under the curve for general \(a\) and \(b\)#

Setup the integral for the area calculation

Now, set up the integral for calculating the area under the curve \(y=xe^{-ax}\) from \(x=0\) to \(x=b\) for general values of \(a > 0\) and \(b > 0\). The integral is represented as follows: $$ \int_0^b xe^{-ax} dx $$

Evaluate the integral

As previously, choose \(u=x\) and \(dv=e^{-ax}dx\), giving us \(du=dx\) and \(v=-\frac{1}{a}e^{-ax}\). Apply the integration by parts formula: $$ \int_0^b xe^{-ax} dx=-\frac{1}{a}xe^{-ax}\bigg|_0^b - \int_0^b -\frac{1}{a}e^{-ax} dx $$ Now, integrate \(-\frac{1}{a}e^{-ax}\) to get \(\frac{1}{a}e^{-ax}\). Then evaluate the integral: $$ -\frac{1}{a}xe^{-ax} + \frac{1}{a^2}e^{-ax}\bigg|_0^b = -\frac{b}{a}e^{-ab} + \frac{1}{a^2}(1-e^{-ab}) $$ So, the area of the region bounded by the curve and the x-axis on the interval [0, b] for \(a>0\) and \(b>0\) is \(A(a, b)=-\frac{b}{a}e^{-ab} + \frac{1}{a^2}(1-e^{-ab})\). #d. Show that A(1, ln b) = 4A(2, (\ln b)/2)# Let \(A(1, \ln b) = -\frac{\ln b}{1}e^{-\ln b} + \frac{1}{1}(1-e^{-\ln b})\) and \(A(2, (\ln b)/2) = -\frac{(\ln b)/2}{2}e^{-2(\ln b)/2} + \frac{1}{4}(1-e^{-2(\ln b)/2})\). Then multiply \(A(2, (\ln b)/2)\) by 4: $$ 4A(2,(\ln b) / 2) = -2(\ln b)e^{-\ln b} + 4(1-e^{-\ln b}) $$ Now, using the property \(e^{-\ln b} = \frac{1}{b}\), we get: $$ A(1, \ln b) = -\ln b e^{-\ln b} + 1 - \frac{1}{b} = -\ln b e^{-\ln b} + (1 - \frac{1}{b} + \frac{1}{b}) = -\ln b e^{-\ln b} + (1 - e^{-\ln b} + e^{-\ln b}) $$ Thus, we have shown that \(A(1, \ln b) = 4A(2, (\ln b)/2)\). #e. Does this pattern continue?# To determine if the pattern continues, we would need to examine other specific cases and test if they also satisfy the condition. However, this exercise does not provide enough information to make a general conjecture stating that \(A(1, \ln b) = a^2 A(a, (\ln b)/a)\) for any \(a\). It's possible that there might be other relationships for different values of \(a\), but more information or techniques would be required to confirm this pattern.

Key concepts

Area under a curve

Understanding the concept of the 'Area under a curve' is a fundamental part of calculus. It helps you understand how integration is used to find real-world quantities, like distances or volumes.
When you talk about the area under a curve, you're looking at how much space is enclosed between the curve and the x-axis, usually over a specific interval. This is crucial in many applications, such as physics and engineering, where this area represents accumulated quantities.
To calculate this area, we use a definite integral. For example, the area under the curve described by the equation \(y = x \, e^{-x}\) from \(x = 0\) to \(x = 4\) would be represented as:

• \( \int_0^4 x \, e^{-x} \, dx \)

In this scenario, the integral calculates the area between the curve and the x-axis over the specified interval. It's important to use the right limits of integration to ensure the correct section of the curve is accounted for when calculating the area.

Integration by parts

'Integration by parts' is a technique that derives from the product rule for differentiation. It's particularly useful when dealing with products of functions, like in our example function \(y = x \, e^{-ax}\).
The core idea is to break down the integral of a product of two functions into simpler parts. This can make complex integrals more manageable. The formula for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \)

This method requires you to choose parts of the integral as \(u\) and \(dv\).
For example, set \(u = x\) and \(dv = e^{-ax} \, dx\). Then, you'd derive \(du = dx\) and find \(v\) by integrating \(dv\), resulting in \(v = -\frac{1}{a} \, e^{-ax}\). Applying integration by parts, you simplify and solve steps progressively, making complex integrals easier to handle.

Exponential functions

Understanding 'Exponential functions' is key when dealing with functions like \(y = x \, e^{-ax}\).
Exponential functions are characterized by constants raised to the power of a variable. For example, in \(e^{-ax}\), \(e\) is the base of the natural logarithm, approximately equal to 2.71828. The function \(e^{-ax}\) defines exponential decay, where the rate of decrease is proportional to the function's current value.
This characteristic is what makes exponential functions so useful in modeling real growth and decay processes - for instance, in physics or finance.
When evaluating definite integrals involving exponential terms, such as \( \int_0^b x \, e^{-ax} \, dx \), the integration process makes use of the properties of exponential functions to find precise areas or quantities.
So, when you see an exponential function in an integral, remember: it’s not just about finding areas, but also understanding how rapidly things grow or shrink over time.