Problem 55 - A Graphical Approach to College Algebra

Question

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. $$f(x)=2 x^{2}$$

Step-by-Step Solution

Answer

The value of the summation is 56.

Step 1: Understanding the Summation

The given expression is a Riemann sum, which calculates the sum of function values at specific points multiplied by the width of each interval, \( \Delta x = 0.5 \). We need to evaluate \( f(x_i) \) for each \( i \) and then multiply by \( \Delta x \).

Step 2: Evaluate \( f(x_1) \)

First, substitute \( x_1 = 0 \) into the function \( f(x) = 2x^2 \). Therefore, \( f(x_1) = 2(0)^2 = 0 \).

Step 3: Evaluate \( f(x_2) \)

Next, substitute \( x_2 = 2 \) into the function. Thus, \( f(x_2) = 2(2)^2 = 2 \times 4 = 8 \).

Step 4: Evaluate \( f(x_3) \)

Substitute \( x_3 = 4 \) into \( f(x) \). So, \( f(x_3) = 2(4)^2 = 2 \times 16 = 32 \).

Step 5: Evaluate \( f(x_4) \)

Finally, substitute \( x_4 = 6 \) into the function. Thus, \( f(x_4) = 2(6)^2 = 2 \times 36 = 72 \).

Step 6: Sum the Products

Calculate the total sum of all the evaluated terms multiplied by \( \Delta x \). So, the sum becomes:\[(0 + 8 + 32 + 72) \times 0.5\]

Step 7: Simplify the Sum

Perform the addition first to get the sum of function evaluations:\(0 + 8 + 32 + 72 = 112\).

Step 8: Multiply by \( \Delta x \)

Finally, multiply the result by \( \Delta x = 0.5 \):\[112 \times 0.5 = 56\].

Key Concepts

Definite Integral

In calculus, a definite integral is a fundamental concept used to find the area under a curve on a given interval. It represents the accumulation of quantities, such as distance, area, or volume, over a continuous interval. The notation for a definite integral typically involves an integral sign with upper and lower bounds, such as \[\int_{a}^{b} f(x) \, dx,\]where \( a \) and \( b \) are the limits of integration, and \( f(x) \) is the integrand, the function being integrated.

Definite integrals can be thought of as the limit of Riemann sums. As the number of intervals increases and the width of each interval decreases, the Riemann sum approaches the exact area. This understanding bridges the gap between discrete sums and continuous integration, providing a basis for applying calculus to solve real-world problems, such as calculating total accumulated change over any interval.

Function Evaluation

When evaluating a function, you're essentially substituting specific values into the function's formula to get results. This operation is crucial in many mathematical contexts, particularly when dealing with Riemann sums.

For example, consider the function \( f(x) = 2x^2 \). If you need to evaluate this function at specific points, you replace \( x \) with a desired value and simplify. Let's evaluate it:

• At \( x_1 = 0 \), \( f(x_1) = 2(0)^2 = 0 \).
• At \( x_2 = 2 \), \( f(x_2) = 2(2)^2 = 8 \).
• At \( x_3 = 4 \), \( f(x_3) = 2(4)^2 = 32 \).
• At \( x_4 = 6 \), \( f(x_4) = 2(6)^2 = 72 \).

Function evaluation is a step-by-step substitution and calculation process that provides the values needed for further analysis or computation in mathematics.

Interval Width

Interval width, often denoted as \( \Delta x \), plays a key role in calculating Riemann sums and definite integrals. It represents how wide each subdivision or segment of your interval is, giving context to the scope of each function evaluation.

For example, suppose you're working within the interval \([0, 6]\) and you want to break it into equal parts, each with a width of \(0.5\). The formula for determining the number of intervals is\[n = \frac{{b-a}}{{\Delta x}},\]where \( b \) is the upper bound, \( a \) is the lower bound, and \( \Delta x \) is the interval width.

In practical terms, \( \Delta x \) helps determine the precision of the Riemann sum. The smaller the interval width, the more accurate the approximation to the definite integral. Interval width sets the granularity of the subdivision, influencing the calculation and understanding of accumulated values.

Summation Notation

Summation notation, often represented by the Greek letter sigma (\( \Sigma \)), is a concise way to express the sum of a sequence of numbers or expressions. This notation is widely used in calculus, particularly with Riemann sums.

The form\[\sum_{i=1}^{n} a_i\]indicates that you are summing the terms \( a_i \) as \( i \) ranges from 1 to \( n \). In Riemann sums, each term often consists of the product of a function evaluation at a specific point and the interval width, formally written as \( f(x_i)\Delta x \).

Understanding how to use summation notation is vital for handling complex calculations in calculus efficiently. It allows you to structure and simplify the representation of aggregate calculations, ultimately tying them back to concepts like definite integrals. By using summation notation, mathematicians and students alike can more easily grasp the complexity of continuous sums.