Problem 41 - Precalculus

Question

For a new machine shop employee, the rate of production for a specialize part is modeled by the function $$p(x)=\frac{1}{2} x^{2}+3$$ (production increases quickly with experience), where \(p(x)\) represents the number of parts completed per day after \(x\) days on the job. The area under this curve in the interval \(x \in[0,4]\) then represents the total number of parts produced in the first 4 days. Using the rectangle method results in the expression \(\frac{4}{n} \sum_{i=1}^{n}\left[\frac{1}{2}\left(\frac{4}{n} i\right)^{2}+3\right] .\) Find the total number of parts produced by applying the summation properties/formulas and taking the limit as \(n \rightarrow \infty\).

Step-by-Step Solution

Answer

Approximately 22.67 parts are produced in the first 4 days.

Step 1: Understand the Problem Context

We need to find the total number of parts produced over the first four days. The production rate is modeled by the function \(p(x)=\frac{1}{2} x^{2}+3\). We're finding the definite integral of this rate function from \(x=0\) to \(x=4\) using the limit of a Riemann sum as \(n\to\infty\).

Step 2: Setup the Riemann Sum

The Riemann sum approximates the integral by dividing the interval into \(n\) rectangles. The width of each rectangle is \(\frac{4}{n}\), and the height is determined by \(p\left(\frac{4}{n}i\right)\). The sum is \[\frac{4}{n} \sum_{i=1}^{n}\left[\frac{1}{2}\left(\frac{4}{n} i\right)^{2}+3\right].\]

Step 3: Substitute and Simplify the Expression in the Summation

Substitute \(\frac{4}{n}i\) into the function \(p(x)\). This gives \(\left[\frac{1}{2}\left(\frac{4}{n} i\right)^{2}+3\right]=\left[\frac{8}{n^2}i^2+3\right]\).

Step 4: Separate the Summation into Two Parts

The sum can be split as follows: \[\frac{4}{n}\sum_{i=1}^{n}\left(\frac{8}{n^2}i^2+3\right) = \frac{4}{n}\left(\sum_{i=1}^{n}\frac{8}{n^2}i^2 + \sum_{i=1}^{n} 3\right).\]

Step 5: Evaluate Each Part Using Summation Formulas

Use the summation formula \(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\) and \(\sum_{i=1}^{n} 1 = n\). This transforms the sum into: \[\frac{4}{n} \left(\frac{8}{n^2}\frac{n(n+1)(2n+1)}{6} + 3n\right).\]

Step 6: Simplify the Expression

Simplify each term: \[= \frac{32}{6n^3} n(n+1)(2n+1) + \frac{12}{n}n.\]

Step 7: Evaluate the Limit as \(n \to \infty\)

Simplifying further gives: \(\frac{32}{6n} (n+1)(2n+1) \to \frac{32}{6} \times 2= \frac{32}{3}\) and \(12\). Adding these results as \(n \to \infty\), the total is: \(\frac{32}{3} + 12 = \frac{32 + 36}{3} = \frac{68}{3}.\)

Step 8: Write the Final Answer

The total number of parts produced in the first 4 days is thus \(\frac{68}{3}\), which simplifies to about 22.67 parts.

Key Concepts

Riemann sum

In mathematics, a Riemann sum is a method used to approximate the total area under a curve, which in calculus is known as the definite integral. The idea is to break the interval over which you want to integrate into many small subintervals.
Then, calculate the area of rectangles that approximate the area under the curve. These rectangles have a width determined by the difference between the limits of integration divided by the number of rectangles, often denoted as \( \frac{b-a}{n} \), where \( b \) and \( a \) are the endpoints of the interval.
In our example, we have the function \( p(x) = \frac{1}{2}x^2 + 3 \) which describes production over time. As \( n \to \infty \), the Riemann sum approaches the exact area under the curve from \( x = 0 \) to \( x = 4 \).
To form the Riemann sum, substitute \( x_i = \frac{4}{n}i \) into the function to find the height of each rectangle. Then multiply by the width \( \frac{4}{n} \) to find the Riemann sum:

• \[ \frac{4}{n} \sum_{i=1}^{n} \left[ \frac{1}{2} \left( \frac{4}{n} i \right)^2 + 3 \right] \]

This sum represents the approximate total number of parts produced over four days.

summation formulas

Summation formulas are incredibly useful when dealing with Riemann sums as they allow you to simplify the sums of certain sequences. In this scenario, two basic summation formulas are particularly helpful: the sum of integers and the sum of squares.
For our function \( p(x) \), we apply these formulas to simplify the Riemann sum. Consider this as breaking the task into manageable pieces, where

• The sum of the first \( n \) integers: \( \sum_{i=1}^{n} 3 = 3n \)
• The sum of the squares of the first \( n \) integers: \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \)

When calculating the Riemann sum for \( p(x) \), these formulas allow us to separate and handle each part of the function \( \left( \frac{8}{n^2}i^2 + 3 \right) \). By applying these formulas, we get:

• \( \frac{8}{n^2} \sum_{i=1}^{n} i^2 + 3 \sum_{i=1}^{n} 1 \)
• Transforming it into: \( \frac{32n(n+1)(2n+1)}{6n^3} + \frac{12n}{n} \)

Using these formulas simplifies the process of moving towards calculating the definite integral.

limit evaluation

In calculus, evaluating limits is crucial for transitioning from an approximation to an exact value. When working with Riemann sums, taking the limit as \( n \to \infty \) transforms the sum into a definite integral.
For the given exercise, once the Riemann sum is expressed using summation formulas, the next step is to evaluate its limit. This involves identifying terms that simplify or cancel out as \( n \) grows very large.In our specific example, when evaluating:

• \( \frac{32(n+1)(2n+1)}{6n^3} \to \frac{32}{6} \times 2 \)
• The constant sum simplifies as \( \frac{12n}{n} \to 12 \)

Combining these results we find the total number of parts by taking the sum of the constants. As \( n \to \infty \), we see the Riemann sum converge to \( \frac{68}{3} \), confirming that \( \frac{68}{3} \approx 22.67 \) parts are produced in the first four days. This highlights the power of limits in establishing precise results from approximations.