Problem 45 - Calculus Single Variable

Question

Evaluate each of the integrals. $$ \int\left(x / 2^{x}\right) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(-\frac{x + 1}{\ln 2} \cdot 2^{-x} + C\).

Step 1: Identify the Integration Method Needed

The integral \( \int \left( \frac{x}{2^x} \right) dx \) involves an expression in the form of \( x \) divided by \( 2^x \). This suggests the use of integration by parts, as it is a product of two functions: \( x \) and \( \frac{1}{2^x} \).

Step 2: Choose \( u \) and \( dv \) for Integration by Parts

Using integration by parts, we set \( u = x \), which means \( du = dx \). For \( dv \), we choose \( dv = \frac{1}{2^x} dx = 2^{-x} dx \) to be integrated. Let \( 2^{-x} = e^{-x \ln 2} \) so that its integral can be computed.

Step 3: Integrate \( dv \) to Find \( v \)

The integral of \( dv = e^{-x \ln 2} dx \) gives \( v \). This evaluates to \( v = -\frac{1}{\ln 2} e^{-x \ln 2} = -\frac{1}{\ln 2} \cdot 2^{-x} \).

Step 4: Apply Integration by Parts Formula

Use the formula \( \int u \, dv = uv - \int v \, du \). Substituting our expressions, we get: \[ \int \frac{x}{2^x} \, dx = x \left(-\frac{1}{\ln 2} \cdot 2^{-x}\right) - \int \left(-\frac{1}{\ln 2} \cdot 2^{-x}\right) \cdot 1 \, dx \].

Step 5: Simplify the Expression

Simplify the equation to \(-\frac{x}{\ln 2} \cdot 2^{-x} + \frac{1}{\ln 2} \int 2^{-x} \, dx \).

Step 6: Compute Remaining Integral

Compute \( \int 2^{-x} \, dx \). Rewrite \( 2^{-x} \) as \( e^{-x \ln 2} \) and integrate to obtain \( -\frac{1}{\ln 2} \cdot 2^{-x} \).

Step 7: Final Expression

Combine all parts to get the final answer: \[ \int \frac{x}{2^x} \, dx = -\frac{x}{\ln 2} \cdot 2^{-x} - \frac{1}{\ln 2} \left(-\frac{1}{\ln 2} \cdot 2^{-x}\right) + C \], which simplifies to \[ -\frac{x + 1}{\ln 2} \cdot 2^{-x} + C \].

Key Concepts

Definite Integral

Definite integrals are a powerful tool in calculus that allow us to calculate the accumulated quantity, or total, of a function over a specific interval. When evaluating definite integrals, we look for the net area under or above a curve:

• Definite integrals can help us calculate areas and can be found over a specific interval \(a\) to \(b\).
• The result of a definite integral is a specific numerical value, unlike indefinite integrals, which produce a family of functions with a constant \(+C\).

For example, if we had the definite integral \( \int_a^b \frac{x}{2^x} \, dx \), this would find the area under the curve from \(x = a\) to \(x = b\) for the function \( \frac{x}{2^x} \). The method of integration by parts, as applied in the previous solution, is often used in both definite and indefinite integrals to evaluate complex expressions.

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. These functions often have the form \(b^x\), where \(b\) is a positive constant.

• In our example, \(2^x\) is the exponential function where the base is \(2\).
• Exponential functions are notable for their rapid growth or decay, which sets them apart from polynomials or linear functions.
• When integrated, exponential functions can often result in expressions involving logarithms due to their natural properties.

One interesting property of exponential functions used in calculus, particularly in integration by parts, is the ability to rewrite them for easier manipulation. For instance, \(2^{-x}\) can be represented as \(e^{-x \ln 2}\), leveraging base changes to simplify calculations.

Integration Techniques

Integration techniques are essential for solving various integral problems, and one of the most versatile of these is integration by parts. This technique is especially useful when dealing with products of functions:

• The integration by parts formula is derived from the product rule of differentiation and is expressed as \( \int u \, dv = uv - \int v \, du \).
• To use integration by parts, you must choose appropriate \(u\) and \(dv\), derive \(du\) and integrate \(dv\) to find \(v\).

In the provided example, by selecting \(u = x\) and \(dv = \frac{1}{2^x} \, dx\), the task becomes more manageable. Taking derivatives and integrals of chosen parts transforms the original problem into simpler integral expressions. Using integration by parts not only simplifies complex integrals like \( \int \frac{x}{2^x} \, dx \) but also helps solve a variety of functions that are otherwise difficult to integrate directly.