Problem 18 - Applied Calculus

Question

Both exactly [e.g. \(\ln (3 \pi)]\) and numerically [e.g. \(\ln (3 \pi) \approx 2.243]\). $$ \int_{0}^{10} z e^{-z} d z $$

Step-by-Step Solution

Answer

The exact answer is \(1 - 11e^{-10}\) and approximately 0.9995.

Step 1: Identify the Integral Form

The given integral is \( \int_{0}^{10} z e^{-z} \, dz\). We recognize that this is an integration by parts problem, where we can solve for a product of two functions.

Step 2: Choose Functions for Integration by Parts

For integration by parts, we choose \(u = z\) and \(dv = e^{-z} \, dz\). This implies \(du = dz\) and \(v = -e^{-z}\), since the integral of \(dv = e^{-z} \), is \(v = -e^{-z}\).

Step 3: Apply the Integration by Parts Formula

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we compute: \[ uv = -z e^{-z} \bigg|_{0}^{10} = -10 e^{-10} - (0) \] and \[ \int v \, du = \int -e^{-z} \, dz = e^{-z} \bigg|_{0}^{10} = 1 - e^{-10} \].

Step 4: Substitute and Simplify

Substituting these values in the integration by parts formula, we get: \[ \int_{0}^{10} z e^{-z} \, dz = \left[-z e^{-z} \right]_{0}^{10} + \left[1 - e^{-10} \right] = -10 e^{-10} + 1 - e^{-10} = 1 - 11e^{-10} \].

Step 5: Numerical Approximation

Calculating the numerical approximation, we find \(e^{-10} \approx 4.54 \times 10^{-5}\). Thus, \(1 - 11 \times 4.54 \times 10^{-5} \approx 0.9995\).

Key Concepts

Integration by Parts

Integration by parts is a technique often used to solve integrals involving the product of two functions. It's based on the product rule for differentiation and is especially useful when integrals cannot be solved through basic algebraic manipulations.
To use this method, you need to choose your functions strategically. Given an integral of the form \( \int u \, dv \), you select one part of the integrand to be \( u \) and the other to be \( dv \). Here’s how:

• Identify parts: Set \( u = z \) (a function that simplifies upon differentiation) and \( dv = e^{-z} \, dz \) (a function that easily integrates).
• Differentiate and integrate: Compute \( du = dz \) and find \( v = -e^{-z} \), since the integral of \( e^{-z} \) is \(-e^{-z}\).

This leads us to the integration by parts formula: \( \int u \, dv = uv - \int v \, du \).
Applying this strategy simplifies complex integrals into more manageable computations, as shown in our example problem.

Definite Integrals

Definite integrals are used to calculate the net area under a curve between two points, giving us a numerical value rather than a function. In our problem, we deal with the definite integral \( \int_{0}^{10} z e^{-z} \, dz \).
The procedure involves evaluating the antiderivative at a specific upper limit and subtracting the antiderivative evaluated at the lower limit.

• Compute boundary terms: Find \( uv \) from earlier, and evaluate from \( 0 \) to \( 10 \).
• Substitute limits: In our example, \( \left[-z e^{-z} \right]_{0}^{10} = -10 e^{-10} - 0 \).

This results in a number representing the accumulated total between these two limits. Definite integrals offer precise answers, unlike indefinite integrals that contain arbitrary constants.

Exponential Functions

Exponential functions, like \( e^{-z} \), frequently occur in integration problems due to their continuous and smooth nature. These functions are essential as they challenge integration techniques but often have straightforward derivatives and integrals.
Key properties of exponential functions include:

• The base \( e \), known as Euler's number, approximates to 2.718 and is used extensively in calculus.
• The function \( e^{-z} \) decreases as \( z \) increases, which is particularly useful in our integral problem as it helps limit the scale of the function’s value.

The ease of integrating and differentiating such functions (\( \frac{d}{dz} e^{-z} = -e^{-z} \)) makes problems like our given exercise approachable when combined with other integration techniques like integration by parts.