Question

Evaluate each improper integral or show that it diverges. \(\int_{1}^{\infty} x e^{-x} d x\)

Short answer

The improper integral converges to \( \frac{2}{e} \).

Step-by-step solution

Set Up the Integral

To evaluate the improper integral \( \int_{1}^{\infty} x e^{-x} \, dx \), we first need to express it using limits. The upper bound is infinity, so we rewrite the integral as \( \lim_{t \to \infty} \int_{1}^{t} x e^{-x} \, dx \).

Integration by Parts

To solve the integral \( \int x e^{-x} \, dx \), we'll use integration by parts. Let \( u = x \) and \( dv = e^{-x} \, dx \). Then, \( du = dx \) and \( v = -e^{-x} \). Integration by parts gives us \( \int u \, dv = uv - \int v \, du \).

Apply Integration by Parts

Substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula: \( \int x e^{-x} \, dx = -x e^{-x} - \int -e^{-x} \, dx \, = -x e^{-x} + \int e^{-x} \, dx \).

Integrate \( e^{-x} \, dx \)

Evaluate the integral \( \int e^{-x} \, dx \): This is a basic exponential integral, which is \( -e^{-x} \).

Combine Results

Substitute \( \int e^{-x} \, dx = -e^{-x} \) back into our expression: \( -x e^{-x} + (-e^{-x}) = -x e^{-x} - e^{-x} \). Simplifying gives \( -(x+1)e^{-x} \).

Evaluate the Definite Integral

Substitute back into \( \lim_{t \to \infty} \int_{1}^{t} x e^{-x} \, dx \), which becomes \( \lim_{t \to \infty} \left[ -(x+1)e^{-x} \right]_{1}^{t} \).

Apply the Limits

Evaluate \( \left[ -(x+1)e^{-x} \right]_{1}^{t} \) by calculating: \( \lim_{t \to \infty} \left( -(t+1)e^{-t} + 2e^{-1} \right) \). As \( t \to \infty \), \((t+1)e^{-t} \to 0 \) since exponential decay outweighs linear growth, leaving us with \( 2e^{-1} \).

Final Result

The limit evaluates to \(2e^{-1} = \frac{2}{e}\). Thus, the integral \( \int_{1}^{\infty} x e^{-x} \, dx = \frac{2}{e} \), which means it converges.

Key concepts

Integration by Parts

Integration by parts is a useful technique for solving integrals that involve products of functions. It comes from the product rule for differentiation and essentially reverses this rule to handle integration. Think of it as a way to simplify an integral that seems too complex at first glance.

When using integration by parts, we choose two components from the integral: one to differentiate, called **u**, and one to integrate, called **dv**. For our exercise, we let **u = x** and **dv = e^{-x} dx**. But why these choices?

Here's what you need to remember:

• Choose **u** so that its derivative (**du**) simplifies the integral.
• Choose **dv** such that the integration of **v** is straightforward.

After selecting **u** and **dv**, find **du** and **v** as follows:

• Differentiate **u** to get **du**: if **u = x**, then **du = dx**.
• Integrate **dv** to find **v**: if **dv = e^{-x} dx**, then **v = -e^{-x}**.

Once these are set, use the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] In the exercise, this formula breaks down to: \[ \int x e^{-x} dx = -x e^{-x} - \int -e^{-x} dx \] This simplifies to \[ -x e^{-x} + \int e^{-x} dx \] By repeating this process, you'll find that integration by parts helps greatly in tackling seemingly difficult integrals.

Exponential Functions

Exponential functions, especially when coupled with their unique properties, feature prominently in calculus. They're important because they are continuous, smooth, and always positive. An exponential function like **\( e^{-x} \)** decreases as **x** increases, which makes it a powerful tool in evaluating improper integrals due to its decay behavior.

For our exercise, solving \( \int e^{-x} dx \) gives the function \(-e^{-x} + C\), where C is the constant of integration. This function decreases rapidly towards zero as **x** increases, which significantly affects how improper integrals behave.

Understanding this decay is crucial:

• Since \( e^{-x} \) approaches 0 as **x** tends to infinity, it implies that \( e^{-x} \) decays fast enough to allow convergence in many cases.
• This property is leveraged in improper integrals to examine if the integral converges to a finite value or diverges.

Not every integral involving **e** is straightforward, but knowing how exponential functions behave helps in determining the behavior at infinity, an essential aspect for evaluating limits and convergence.

Limits and Convergence

Understanding limits and convergence is essential in calculus, especially when dealing with improper integrals. An integral is considered **improper** when it has one or more infinite limits or undefined points within the interval.

In our exercise, the integral is improper due to its upper bound of infinity. Therefore, we need to modify our approach with limits. We rewrite the integral from a finite point to a variable limit (like \( t \)), and then evaluate what happens as \( t \to \infty \).

Proper application of limits allows us to determine if an improper integral converges or diverges:

• If the limit exists and is finite, the integral **converges**.
• If the limit does not exist or is infinite, the integral **diverges**.

In our case, we evaluate \( \int_{1}^{t} x e^{-x} dx \) and take the limit as \( t \to \infty \). After substituting the expression \(-(x+1)e^{-x}\), and analyzing its behavior, we see that \( (t+1)e^{-t} \to 0 \). This shows the integral converges to a finite value of \( \frac{2}{e} \), as the exponential decay dominates the term.