Question

A family of exponentials The curves \(y=x e^{-a x}\) are shown in the figure for \(a\)=1,2, and 3. Figure cannot copy 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 \(A(a, b)\). d. Use part (c) to show that \(A(1, \ln b)=4 A\left(2, \frac{\ln b}{2}\right)\). e. Does this pattern continue? Is it true that \(A(1, \ln b)=a^{2} A(a,(\ln b) / a) ?\)

Short answer

The relationship between the areas under different exponential curves is given by the equation \(A(1, \ln b) = a^2 A(a, \dfrac{\ln b}{a})\), where \(A(a, b)\) is the area under the curve \(y = xe^{-ax}\) from \(x=0\) to \(x=b\). This equation holds true for any positive values of \(a\) and \(b\).

Step-by-step solution

Part a: Area under \(y = x e^{-x}\) from \(x=0\) to \(x=4\)

To find the area under the curve \(y = x e^{-x}\) between \(x=0\) and \(x=4\), we will integrate the function with respect to \(x\) over that interval. \(\int_{0}^{4} x e^{-x}\,dx\) Now, integrate by parts, where: \(u=x\), \(dv=e^{-x}\,dx\), \(du=dx\), and \(v=-e^{-x}\). \(\left[-xe^{-x}\right]_{0}^{4}-\int_{0}^{4} -e^{-x}\,dx\) \(\left[-4e^{-4}\right]-\left[\left[-e^{-x}\right]_{0}^{4}\right]\) \(-4e^{-4}+\left[-e^{-4}+e^0\right]\) Now, simplify and find the solution: \(-4e^{-4}+1 - e^{-4} \approx 0.981684\). So the area under the curve from \(x=0\) to \(x=4\) is approximately \(0.981684\).

Part b: Area under \(y = x e^{-ax}\) from \(x=0\) to \(x=4\) where \(a>0\)

Now, we want to generalize the area under the curve \(y = x e^{-ax}\) between \(x=0\) and \(x=4\) for any \(a>0\). \(\int_{0}^{4} x e^{-ax}\,dx\) Again, integrate by parts, where: \(u=x\), \(dv=e^{-ax}\,dx\), \(du=dx\), and \(v=-\dfrac{1}{a} e^{-ax}\). \(\left[-\dfrac{1}{a}xe^{-ax}\right]_{0}^{4}-\int_{0}^{4} -\dfrac{1}{a} e^{-ax}\,dx\) \(\left[-\dfrac{1}{a}4e^{-4a}\right]-\left[\left[-\dfrac{1}{a^2}e^{-ax}\right]_{0}^{4}\right]\) \(-\dfrac{4}{a}e^{-4a}+\left[\left[-\dfrac{1}{a^2}e^{-4a}\right] + \dfrac{1}{a^2}\right]\) The area under the curve with general \(a>0\) is: \(-\dfrac{4}{a}e^{-4a} + \dfrac{1}{a^2} - \dfrac{1}{a^2} e^{-4a}\)

Part c: Area under \(y = x e^{-ax}\) from \(x=0\) to \(x=b\) where \(a,b>0\)

Now, we want to generalize the area under the curve \(y = xe^{-ax}\) between \(x=0\) and \(x=b\) for any \(a,b > 0\). This area will be represented as \(A(a, b)\). \(A(a, b) = \int_{0}^{b} x e^{-ax}\, dx\) Integrate again by parts, where: \(u=x\), \(dv=e^{-ax}\,dx\), \(du=dx\), and \(v=-\dfrac{1}{a} e^{-ax}\). \(\left[-\dfrac{1}{a}xe^{-ax}\right]_{0}^{b}-\int_{0}^{b} -\dfrac{1}{a} e^{-ax}\,dx\) \(\left[-\dfrac{1}{a}be^{-ab}\right]-\left[\left[-\dfrac{1}{a^2}e^{-ax}\right]_{0}^{b}\right]\) \(-\dfrac{b}{a}e^{-ab}+\dfrac{1}{a^2}-\dfrac{1}{a^2} e^{-ab}\) So, \(A(a, b) = -\dfrac{b}{a}e^{-ab} + \dfrac{1}{a^2} - \dfrac{1}{a^2} e^{-ab}\).

Part d: Show \(A(1, \ln b) = 4A\left(2, \frac{\ln b}{2}\right)\) using part (c)

Use the result from part (c) to find \(A(1, \ln b)\) and \(A(2, \dfrac{\ln b}{2})\) and show that \(A(1, \ln b) = 4A(2, \dfrac{\ln b}{2})\). First, calculate \(A(1, \ln b)\): \(A(1, \ln b)=-\ln b \cdot e^{-\ln b}+\dfrac{1}{1^2}-\dfrac{1}{1^2}e^{-\ln b}\) \(-\ln b \cdot \dfrac{1}{b} + 1 - \dfrac{1}{b}\) Next, calculate \(A(2, \dfrac{\ln b}{2})\): \(A(2, \dfrac{\ln b}{2})= -\dfrac{\ln b}{4} e^{-\ln b} + \dfrac{1}{4} - \dfrac{1}{4} e^{-\dfrac{\ln b}{2}}\) Now, we want to show that \(A(1, \ln b) = 4A(2, \dfrac{\ln b}{2})\): \(4A(2, \dfrac{\ln b}{2}) = 4 \left(-\dfrac{\ln b}{4} e^{-\ln b} + \dfrac{1}{4} - \dfrac{1}{4} e^{-\dfrac{\ln b}{2}}\right)\) \(= -\ln b \cdot e^{-\ln b} + 1 - e^{-\ln b}\) Since \(A(1, \ln b) = 4A(2, \dfrac{\ln b}{2})\), the relationship is proved.

Part e: Investigating whether the pattern continues for \(A(1, \ln b) = a^2 A(a, \dfrac{\ln b}{a})\)

We will use the result from part (c), replacing \(A(a, b)\) with the proposed pattern: \(A(1, \ln b) = a^2 A(a, \dfrac{\ln b}{a})\). \(A(a, \dfrac{\ln b}{a}) = -\dfrac{\ln b}{a^2} e^{-\ln b} + \dfrac{1}{a^2} - \dfrac{1}{a^2} e^{-\ln b}\) Now, compute \(a^2 A(a, \dfrac{\ln b}{a})\): \(a^2 A(a, \dfrac{\ln b}{a}) = a^2\left[-\dfrac{\ln b}{a^2} e^{-\ln b} + \dfrac{1}{a^2} - \dfrac{1}{a^2} e^{-\ln b}\right]\) \(= -\ln b e^{-\ln b} + 1 - e^{-\ln b}\) We see that this expression is equal to \(A(1, \ln b)\) from part (d), which means that the pattern \(A(1, \ln b) = a^2 A(a, \dfrac{\ln b}{a})\) holds true.

Key concepts

Integration by Parts

Integration by Parts is a powerful technique in calculus used to find integrals that are products of functions. It is especially useful when dealing with polynomial and exponential functions. The method derives from the product rule for differentiation and can turn a complex integral into simpler ones. The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \), where you identify parts of the integrand as \(u\) and \(dv\).

• Choose \(u\) to be a function whose derivative \(du\) is simpler than \(u\).
• Choose \(dv\) to be a function whose integral \(v\) is easy to calculate.

For example, in the integral \( \int x e^{-ax} \, dx \), select \( u = x \) (since its derivative \( du = dx \) simplifies the expression), and \( dv = e^{-ax} \, dx \) (since its integral \( v = -\frac{1}{a} e^{-ax} \) is straightforward). After applying the integration by parts formula, these selections lead to break down the problem into a simpler integration task. Remember, the key is in choosing the correct \(u\) and \(dv\) appropriately to facilitate the easiest solution.

Exponential Functions

Exponential functions are incredibly common in calculus problems due to their unique properties. They are functions of the form \( y = e^{kx} \), where \(e\) is Euler's number (approximately 2.718), and \(k\) is a constant. Their derivative and integral properties are significant because they maintain their form, making them predictable and easy to manipulate.Some key points about exponential functions:

• The derivative of \( e^{kx} \) with respect to \( x \) is \( k e^{kx} \).
• The integral of \( e^{kx} \) with respect to \( x \) is \( \frac{1}{k} e^{kx} + C \), where \( C \) is the constant of integration.

In definite integrals, such as those encountered in various calculus problems, these properties allow for straightforward computation of areas under curves described by exponential functions.
For instance, any function of the form \( y = x e^{-ax} \) combines both linear and exponential elements, requiring techniques like integration by parts for their evaluation. Knowing these derivative and integral relationships is crucial for effectively solving calculus problems involving exponential functions.

Definite Integrals

Definite integrals are used to calculate the area under a curve from one point to another on a graph. Represented as \( \int_{a}^{b} f(x) \, dx \), they have limits of integration from \( a \) to \( b \). This concept is essential in real-world applications where you need to find the total accumulation of a quantity, such as the area under an exponential function on a certain interval.Here are a few essential aspects to remember about definite integrals:

• They give a fixed numerical answer, representing the bounded area under the curve of a function between two points.
• The Fundamental Theorem of Calculus links definite integrals with antiderivatives, stating \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \), where \( F \) is any antiderivative of \( f \).
• They can also be interpreted as the net area, accounting for areas above and below the x-axis, with areas below being subtracted.

In practice, when you compute an integral like \( \int_{0}^{4} x e^{-ax} \, dx \), you are actually finding the net area of the region bounded by this curve and the x-axis, from 0 to 4 units along the x-axis. Understanding definite integrals is crucial for accurately modeling and solving a wide range of scientific and engineering problems.