Problem 23 - Calculus for Biology and Medicine

Question

Use integration by parts to evaluate the integrals. $$ \int_{0}^{1} x e^{-x} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(-1\).

Step 1: Identify Functions for Integration by Parts

To apply integration by parts, we need to identify functions to use as \( u \) and \( dv \). For the integral \( \int x e^{-x} dx \), choose \( u = x \) and \( dv = e^{-x} dx \). This allows us to differentiate and integrate these choices easily.

Step 2: Differentiate and Integrate the Functions

Differentiate \( u = x \) to get \( du = dx \). Integrate \( dv = e^{-x} dx \) to find \( v = -e^{-x} \). Now, we have \( du \) and \( v \) that allow us to use the integration by parts formula.

Step 3: Apply the Integration by Parts Formula

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Substitute the values we identified: \( \int x e^{-x} dx = x(-e^{-x}) - \int (-e^{-x}) dx \).

Step 4: Simplify the Expression

Simplify the expression: \( -xe^{-x} + \int e^{-x} dx \). We can compute \( \int e^{-x} dx \), which is \( -e^{-x} \). Thus, the expression becomes \( -xe^{-x} - (-e^{-x}) \).

Step 5: Combine and Simplify Further

Combine the terms: \( -xe^{-x} + e^{-x} \), which simplifies to \( e^{-x}(1 - x) \). This represents the indefinite integral \( \int x e^{-x} dx = e^{-x}(1 - x) + C \), where \( C \) is the constant of integration.

Step 6: Evaluate the Definite Integral from 0 to 1

Now substitute the limits of integration: \[ \int_{0}^{1} x e^{-x} \, dx = \left[ e^{-x}(1-x) \right]_{0}^{1} \]. Calculate it as \( e^{-1}(1-1) - e^{0}(1-0) = 0 - 1 = -1 \).

Key Concepts

Integration Techniques

One of the key methods to integrate complex functions is Integration by Parts. It is particularly useful when the standard integration approaches, like substitution, become challenging. This technique is based on the product rule of differentiation and is expressed through the formula:

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

Where \( u \) and \( dv \) are parts of the original integrand that we choose strategically for easier computation.
Here’s a simple way to decide what parts of the integrand will become \( u \) and \( dv \):

• Choose \( u \) to be the function that simplifies upon differentiation. Logarithmic, algebraic, and trigonometric functions are great candidates.
• Choose \( dv \) for the part that is easily integrable. Exponential and basic trigonometric functions often work well here.

In our given exercise \( \int x e^{-x} \, dx \), we picked \( u = x \) because it simplifies to \( dx \) when differentiated, and \( dv = e^{-x} \, dx \), as it integrates smoothly to \( -e^{-x} \).
Integration by parts transforms a complex integral into simpler parts that are more straightforward to solve.

Definite Integrals

Definite integrals help us find the area under a curve between two specific points. This is represented as \( \int_{a}^{b} \), where \( a \) and \( b \) are the limits of the integration.
After computing the indefinite integral, which provides the general form of the antiderivative, we evaluate it at both upper and lower limits. The result gives us the definite value of the integral over the interval \([a,b]\).
In our problem \( \int_{0}^{1} x e^{-x} \, dx \), the integral was first solved without bounds to find \( \int x e^{-x} \, dx = e^{-x}(1 - x) + C \), with \( C \) as the constant of integration. Then, by substituting the upper limit (1) and the lower limit (0) into the expression and computing, we obtained:

• \( \left[ e^{-x}(1-x) \right]_{0}^{1} = 0 - 1 \)
• This simplifies to \(-1\), which is the definitive area under the curve from 0 to 1, albeit on the negative side because the function dips below the x-axis.

Understanding definite integrals is essential not just in pure math but across fields such as physics and engineering where they are applied to calculate quantities like displacement and pressure.

Calculus Applications

Understanding how integrals can be applied in real-world context makes calculus a powerful tool in various disciplines.**Utilizing Integration for Problem Solving**
Integration, whether definite or indefinite, is essential in solving a variety of practical problems:

• Calculate areas under curves, as seen in our example, which is fundamental in physics to find concepts like displacement over time with velocity functions.
• Modeling population dynamics in biology where growth rates and carrying capacity can be determined through integrative techniques.
• Engineers use integrals to compute electrical currents (as seen in the discipline of circuit analysis) or to determine the amount of material required for a build.

In the context of the given exercise, integration demonstrates how a function representing continuous processes like decay (exponential decay, expressed with terms similar to \( e^{-x} \)) can be analyzed over a specific range.
By understanding both the 'how' and 'why' of integration, we can appreciate its versatility and application in solving tangible problems across various industries.