Problem 23 - Calculus for Biology and Medicine

Question

In Problems 1-30, use integration by parts to evaluate each integral. $$ \int_{0}^{1} x e^{-x} d x $$

Step-by-Step Solution

Answer

The value of the integral is \( 1 - 2/e \).

Step 1: Identify Functions for Integration by Parts

The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). For the integral \( \int x e^{-x} \, dx \), we'll choose \( u = x \) (a polynomial) and \( dv = e^{-x} \, dx \) (an exponential function).

Step 2: Differentiate and Integrate

Differentiate \( u = x \), thus \( du = dx \). Integrate \( dv = e^{-x} \, dx \) to get \( v = \int e^{-x} \, dx = -e^{-x} \).

Step 3: Apply Integration by Parts Formula

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

Step 4: Simplify and Evaluate Remaining Integral

Simplify to: \( -xe^{-x} + \int e^{-x} \, dx = -xe^{-x} - e^{-x} + C \). Evaluate from 0 to 1: \([-(1)e^{-1} - e^{-1}] - [-(0)e^{0} - e^{0}] = \) \(-2e^{-1} + 1\).

Step 5: Compute the Definite Integral

Calculate the final result for the definite integral from 0 to 1. Substitute and simplify: \(-2/e + 1 = 1 - 2/e\).

Key Concepts

Definite Integral

In calculus, a definite integral is used to calculate the area under a curve between two points on the x-axis. It represents the accumulation of quantities, providing a numerical value that signifies the total area.
When solving \[\int_{0}^{1} x e^{-x} \, dx, \]we compute the definite integral from 0 to 1. This means evaluating the construct of function accumulation over a specific interval — in this case, the interval [0, 1].
The application of the definite integral here involves first obtaining an antiderivative function and then applying the limits of integration to find the net area. The evaluation of this integral involves taking the difference between the function values at these bounds. This is crucial in determining net quantities, such as area, volume, or total change in value over an interval.
The calculated solution\[1 - \frac{2}{e},\]as the result of this exercise, shows the exact value of the area under the curve from 0 to 1.

Exponential Function

An exponential function is characterized by a constant base raised to a variable exponent. In the context of this integral\[\int x e^{-x} \, dx,\]we see the function \(e^{-x}\), which is a decaying exponential.
Exponential functions are often recognized for their unique growth or decay properties, depending on whether the exponent is positive or negative. Here, the negative exponent indicates decay, meaning the function decreases as \(x\) increases.
When applying integration techniques, such as by parts in this exercise, exponential functions often simplify the process because they have straightforward derivatives and integrals. For instance, integrating \(e^{-x} \, dx\) results in \(-e^{-x}\), as shown in the steps.
Exponential functions are omnipresent in various fields because they model growth and decay processes, such as population growth, radioactive decay, and interest calculations, making them highly practical in real-world applications.

Polynomial Differentiation

Polynomial functions, like the function \(u = x\) in our integration by parts exercise, are among the simplest to differentiate.
Differentiation is the process of finding the derivative of a function, which represents an instantaneous rate of change. The derivative of a polynomial function adheres to a simple rule: multiply by the power and reduce the power by one.
For example, differentiating \(u = x\) leads to \(du = dx\). This expresses that the rate of change of \(x\) with respect to itself is constant, which is a fundamental characteristic of polynomial functions.
Polynomials are a common component in calculus due to their simplicity and ease of manipulation. When applying techniques like integration by parts, polynomials allow quick setup and calculation steps. Their intuitive differentiation makes them excellent candidates for being the "\(u\)" part in such formulas, simplifying the overall problem-solving process.