Problem 51 - Calculus Single Variable

Question

Determine whether the given improper integral converges or diverges. If it converges, then evaluate it. $$ \int_{-\infty}^{0} x 2^{x} d x $$

Step-by-Step Solution

Answer

The integral converges with a value of \( \frac{1}{(\ln(2))^2} \).

Step 1: Set up the integral

First, we identify the improper integral limit running from -∞ to 0 of the function \( x \cdot 2^x \). We write this as: \[ \int_{-\infty}^{0} x \cdot 2^{x} \, dx \] Since the integral is improper due to running to -∞, we rewrite it using a limit: \[ \lim_{a \to -\infty} \int_{a}^{0} x \cdot 2^{x} \, dx \]

Step 2: Integration by Parts

We apply integration by parts to the integral, where \( u = x \) and \( dv = 2^x \, dx \). Then, \( du = dx \) and we compute \( v \) from integrating \( dv \): \[ v = \int 2^x \, dx = \frac{2^x}{\ln(2)} \] Using the integration by parts formula, \( \int u \, dv = uv - \int v \, du \), we have: \[ \int x \cdot 2^x \, dx = x \cdot \frac{2^x}{\ln(2)} - \int \frac{2^x}{\ln(2)} \, dx \]

Step 3: Evaluate the remaining integral

Continuing, the remaining integral is \[ \int \frac{2^x}{\ln(2)} \, dx = \frac{1}{\ln(2)} \int 2^x \, dx = \frac{2^x}{(\ln(2))^2} \] Substitute back into the integration by parts solution: \[ x \cdot \frac{2^x}{\ln(2)} - \frac{2^x}{(\ln(2))^2} \] So, the function for finding the limit is: \[ \lim_{a \to -\infty} \left[ \left( x \cdot \frac{2^x}{\ln(2)} - \frac{2^x}{(\ln(2))^2} \right) \Bigg|_a^0 \right] \]

Step 4: Evaluate at the limits

Evaluate the expression at 0 and \( a \), where \( a \to -\infty \): At \( x = 0 \): \[ 0 \cdot \frac{2^0}{\ln(2)} - \frac{2^0}{(\ln(2))^2} = -\frac{1}{(\ln(2))^2} \] At \( x = a \): The expression becomes: \[ a \cdot \frac{2^a}{\ln(2)} - \frac{2^a}{(\ln(2))^2} \] As \( a \to -\infty \), both terms containing \( 2^a \) approach zero because \( 2^a \to 0 \) as \( a \to -\infty \). Thus, the entire expression approaches zero.

Step 5: Conclude convergence and value

Now apply the limit: \[ \lim_{a \to -\infty} \left[ \left( x \cdot \frac{2^x}{\ln(2)} - \frac{2^x}{(\ln(2))^2} \right) \Bigg|_a^0 \right] = 0 - \left(-\frac{1}{(\ln(2))^2}\right) = \frac{1}{(\ln(2))^2} \]Therefore, the integral converges and its value is \( \frac{1}{(\ln(2))^2} \).

Key Concepts

Convergence and Divergence

When dealing with improper integrals, the concepts of convergence and divergence are key to understanding whether the integral has a finite value. An improper integral, like \( \int_{-\infty}^{0} x \cdot 2^{x} \, dx \), deals with the issue of infinity, either in its limits of integration or within the integrand. The main task is determining if it sums up to a finite number (converges) or not (diverges).

To decide, we utilize limits. If as we extend the integral to approach the defined infinite bounds, the result nears a specific number, we say the integral converges to that value. If not, it diverges. In our exercise, we examined the expression from \(-\infty\) to 0 by introducing \( \lim_{a \to -\infty} \) to see if the evaluated limit yields a numerical outcome, which it did. Thus, indicating convergence.

Your approach to testing convergence should always start by identifying if there are any indefinite bounds and subsequently enforcing limits to simulate how these bounds behave as they extend towards infinity or negative infinity.

Integration by Parts

Integration by parts is an essential technique in calculus, used primarily when dealing with products of functions. It's derived from the product rule for differentiation and can be a useful method to tackle integrals that aren't straightforward to solve otherwise.

The formula for integration by parts is:

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

In our exercise, we set \( u = x \) and \( dv = 2^x \, dx \). This meant deriving \( du = dx \) and integrating \( dv \) to find \( v = \frac{2^x}{\ln(2)} \).

Subsequently, we applied the integration by parts formula to break the integral into two parts, each simpler than the original problem. This method is highly effective when you need to peel away complexity in layers, enabling you to solve complicated integrals step-by-step.

Limits of Integration

The limits of integration define the boundaries over which the function is to be integrated. In improper integrals like \( \int_{-\infty}^{0} x \cdot 2^{x} \, dx \), one of the limits extends to infinity or negative infinity, requiring us to handle these cases differently as they can often lead to undefined behavior.

We address this ambiguity by replacing the infinite bound with a variable (such as \( a \)) and then introducing a limit, \( \lim_{a \to -\infty} \), which evaluates the expression as \( a \) approaches the infinity.

This limit evaluates the behavior of the function as it nears the boundary of the defined region. For our integral, we considered what happens to \( \int_{a}^{0} x \cdot 2^{x} \, dx \) as \( a \) tends towards \(-\infty\). This setup allowed us to properly evaluate the integral's convergence, leading us swiftly to the conclusion that it indeed converges and equals \( \frac{1}{(\ln(2))^2} \). Using limits of integration properly allows us to safely navigate and solve improper integrals efficiently.