Problem 25 - Calculus Early Transcendentals

Question

Let \(g(x)=x e^{x}-e^{x} .\) Find \(f(x)\) so that \(f^{\prime}(x)=g^{\prime}(x)\) and \(f(1)=2\)

Step-by-Step Solution

Answer

\(f(x) = x e^x - e^x + 2\) satisfies the conditions.

Step 1: Differentiate \(g(x)\)

To find the derivative of \(g(x) = x e^x - e^x\), we use the product rule. Recall that the derivative of \(xe^x\) is the derivative of \(x\) times \(e^x\) plus the derivative of \(e^x\) times \(x\). \[ g'(x) = \frac{d}{dx}(x e^x) - \frac{d}{dx}(e^x) = (e^x + xe^x) - e^x = xe^x \]

Step 2: Integrate \(g'(x)\) to Find \(f(x)\)

Now that we have \(g'(x) = x e^x\), we need to find \(f(x)\) such that \(f'(x) = g'(x)\). This requires integrating \(g'(x)\). \[ f(x) = \int x e^x \, dx \]To integrate, use integration by parts where \(u = x\) and \(dv = e^x dx\).Calculate \(du = dx\) and \(v = e^x\), then apply:\[ \int u \, dv = uv - \int v \, du \]\[ \int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C \]Therefore, \(f(x) = x e^x - e^x + C\).

Step 3: Use Initial Condition \(f(1) = 2\) to Find \(C\)

To determine the constant \(C\), use the initial condition \(f(1) = 2\). Substitute 1 into the expression for \(f(x)\):\[ f(1) = 1\cdot e^1 - e^1 + C = 2 \]\[ e - e + C = 2 \]\[ C = 2 \]Thus, \(C = 2\) and the function is \(f(x) = x e^x - e^x + 2\).

Key Concepts

Derivative

The derivative is a fundamental concept in calculus that represents the rate of change of a function with respect to one of its variables. In simple terms, think of the derivative as a way to measure how a function changes. For example, the derivative of the distance traveled over time is velocity. It tells us how quickly the distance is changing at any given moment.

In the given exercise, the function to differentiate is

• \( g(x) = x e^x - e^x \)

To find the derivative\( g'(x) \), the product rule is essential. The product rule is used when you're finding derivatives of products of functions. Here, we have the term \( xe^x \), which consists of two expressions multiplied together: \( x \) and \( e^x \).

Apply the product rule: If you have two functions \( u \) and \( v \), then the derivative \( \frac{d}{dx}(uv) = u'v + uv' \). Thus for \( xe^x \), \( u = x \) and \( v = e^x \), where \( u' = 1 \) and \( v' = e^x \), so:

• \( \frac{d}{dx}(xe^x) = e^x + xe^x \)

By subtracting the derivative of \( e^x \), which is \( e^x \), from this result, we find that \( g'(x) = xe^x \). This gives us the rate of change for \( g(x) \). Understanding derivatives is crucial for solving many types of calculus problems.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It's especially helpful when simple integration methods won't work. The formula for integration by parts comes from the product rule for derivatives and is expressed as:

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

This formula may look complex, but it simplifies the integration by turning it into an equation involving both an integral and a simpler expression.

In the exercise, to find \( f(x) \) such that \( f'(x) = g'(x) \), we need to integrate \( g'(x) = x e^x \). Integration by parts is employed by choosing:

• \( u = x \), therefore \( du = dx \)
• \( dv = e^x \, dx \), therefore \( v = e^x \)

Substitute these into the integration by parts formula:

• \( \int x e^x \, dx = x e^x - \int e^x \, dx \)
• The integral \( \int e^x \, dx = e^x \), so the expression becomes \( x e^x - e^x \).

Don’t forget the integration constant \( C \), which accounts for any constant shift in the integrated function, resulting in:

• \( f(x) = x e^x - e^x + C \)

Integration by parts is a structured method to tackle complex integrals by deconstructing them into manageable steps.

Initial Condition

In calculus, an initial condition allows us to find the specific value of an unknown constant that results from integration. An initial condition provides additional information about a function at a particular point, ensuring only one unique solution fits this criterion.

After integrating, we determined:

• \( f(x) = x e^x - e^x + C \)

To find the constant \( C \), use the initial condition provided: \( f(1) = 2 \). Substitute \( x = 1 \) into the function:

• \( f(1) = 1 \, e^1 - e^1 + C = 2 \)
• Simplifies to \( e - e + C = 2 \)
• Since \( e - e = 0 \), this results in \( C = 2 \)

Substituting \( C \) back into our function gives the specific solution: \( f(x) = x e^x - e^x + 2 \).

Initial conditions are crucial because they provide the necessary data to solve for constants and ensure the resulting function precisely models the desired behavior at specific points.