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Evaluate the terms of \(\sum_{i=1}^{4} f\left(x_{i}\right) \Delta x,\) with \(x_{1}=0, x_{2}=2, x_{3}=4, x_{4}=6,\) and \(\Delta x=0.5,\) for each function. $$f(x)=6+2 x$$

Short Answer

Expert verified
The evaluated sum is 24.

Step by step solution

01

Understand the Problem

We have a summation of the form \(\sum_{i=1}^{4} f(x_i) \Delta x\) where we need to evaluate \(f(x)\) at specific values of \(x\) and multiply each by \(\Delta x\). The function given is \(f(x) = 6 + 2x\).
02

Evaluate f(x) at x1

Substitute \(x_1 = 0\) into \(f(x) = 6 + 2x\). Thus, \(f(0) = 6 + 2(0) = 6\).
03

Evaluate f(x) at x2

Substitute \(x_2 = 2\) into \(f(x) = 6 + 2x\). Thus, \(f(2) = 6 + 2(2) = 10\).
04

Evaluate f(x) at x3

Substitute \(x_3 = 4\) into \(f(x) = 6 + 2x\). Thus, \(f(4) = 6 + 2(4) = 14\).
05

Evaluate f(x) at x4

Substitute \(x_4 = 6\) into \(f(x) = 6 + 2x\). Thus, \(f(6) = 6 + 2(6) = 18\).
06

Multiply Each Term by Δx

Multiply each evaluated \(f(x_i)\) by \(\Delta x = 0.5\). Thus, \(f(0)\Delta x = 6 \times 0.5 = 3\), \(f(2)\Delta x = 10 \times 0.5 = 5\), \(f(4)\Delta x = 14 \times 0.5 = 7\), \(f(6)\Delta x = 18 \times 0.5 = 9\).
07

Sum the Results

Add up all the results from Step 6: \(3 + 5 + 7 + 9 = 24\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Evaluation
Function evaluation is a fundamental step when working with Riemann sums. It helps you understand how the function behaves at specific points within the interval. In our exercise, the function \( f(x) = 6 + 2x \) needs to be evaluated at several specific values of \( x \). This means we substitute the given \( x_i \) values into the function to find \( f(x_i) \). Let's look at how this is done:
  • For \( x_1 = 0 \), substitute into the function to get \( f(0) = 6 + 2(0) = 6 \).
  • Similarly, for \( x_2 = 2 \), substitute to find \( f(2) = 6 + 2(2) = 10 \).
  • Proceed in this manner for \( x_3 = 4 \) and \( x_4 = 6 \) to get \( f(4) = 14 \) and \( f(6) = 18 \) respectively.
This step helps in preparing each term that will be later multiplied by \( \Delta x \). Each function evaluation gives you a piece of the puzzle needed to form the Riemann sum.
Delta x
\( \Delta x \) is a small change in \( x \) that represents the width of each subinterval in a partitioned interval when performing a Riemann sum. In the context of calculating an integral with Riemann sums, \( \Delta x \) plays a crucial role in determining the approximation of the area under a curve. Initially, choose your points for \( x \) values, for instance, \( x_1, x_2, x_3, \) and \( x_4 \) from the interval:
  • In our problem, \( \Delta x \) is given as 0.5, which means each segment between consecutive \( x_i \) values (e.g., from \( x_1 \) to \( x_2 \)) will be 0.5 units wide.
  • Every function evaluation from earlier must now be multiplied by this \( \Delta x \).
This width \( \Delta x \) is a significant feature in forming each term of the Riemann sum, allowing us to approximate the area efficiently.
Summation
Summation is the process of adding a sequence of numbers, which in this case, are the function evaluations multiplied by \( \Delta x \). It's the final and crucial step in calculating a Riemann sum. Once each of the \( f(x_i) \) values has been appropriately evaluated and adjusted for \( \Delta x \), the resulting numbers are summed up:
  • Our calculations reveal these values: \( f(0) \times \Delta x = 3 \), \( f(2) \times \Delta x = 5 \), \( f(4) \times \Delta x = 7 \), and \( f(6) \times \Delta x = 9 \).
  • Adding these gives us the total: \( 3 + 5 + 7 + 9 = 24 \).
Summation in Riemann sums brings together all evaluated pieces over the interval to approximate the integral of a function.

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