Problem 5 - Applied Calculus

Question

Use the following table to estimate \(\int_{0}^{15} f(x) d x\). $$ \begin{array}{c|c|c|c|c|c|c} \hline x & 0 & 3 & 6 & 9 & 12 & 15 \\ \hline f(x) & 50 & 48 & 44 & 36 & 24 & 8 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The estimated value of the integral is 543.

Step 1: Understand the Setup

We need to estimate the integral of the function \( f(x) \) over the interval from \( x = 0 \) to \( x = 15 \). We will use the given table of values for \( f(x) \) to approximate this integral.

Step 2: Choose the Method

We will use the trapezoidal rule to approximate the integral based on the table values. This rule is suited for estimating definite integrals when only discrete data points are available.

Step 3: Apply the Trapezoidal Rule Formula

The trapezoidal rule is given by: \[\int_{a}^{b} f(x) \, dx \approx \frac{1}{2} \sum_{i=1}^{n}(x_i - x_{i-1})(f(x_{i-1}) + f(x_i))\] where \(n\) is the number of subintervals. In our case, the subintervals are \([0, 3]\), \([3, 6]\), \([6, 9]\), \([9, 12]\), \([12, 15]\), with five subintervals.

Step 4: Calculate Each Subinterval Contribution

- For \([0, 3]\): \( \frac{1}{2} \cdot (3 - 0) \cdot (50 + 48) = \frac{1}{2} \cdot 3 \cdot 98 = 147 \)- For \([3, 6]\): \( \frac{1}{2} \cdot (6 - 3) \cdot (48 + 44) = \frac{1}{2} \cdot 3 \cdot 92 = 138 \)- For \([6, 9]\): \( \frac{1}{2} \cdot (9 - 6) \cdot (44 + 36) = \frac{1}{2} \cdot 3 \cdot 80 = 120 \)- For \([9, 12]\): \( \frac{1}{2} \cdot (12 - 9) \cdot (36 + 24) = \frac{1}{2} \cdot 3 \cdot 60 = 90 \)- For \([12, 15]\): \( \frac{1}{2} \cdot (15 - 12) \cdot (24 + 8) = \frac{1}{2} \cdot 3 \cdot 32 = 48 \)

Step 5: Sum Up the Contributions

Add all the contributions together to get the approximate value of the integral: \[ 147 + 138 + 120 + 90 + 48 = 543 \]

Key Concepts

Understanding Numerical Integration

Numerical integration is a powerful mathematical technique that helps us calculate approximate values of definite integrals.
It is especially useful when the function under investigation is complex or we only have function values at discrete data points.
Unlike traditional integration methods that require an explicit function, numerical integration focuses on estimate techniques using available data.

• It's ideal for situations where a function is challenging to integrate symbolically.
• Instead of trying to find an antiderivative, approximations build over discrete segments of the domain.
• This approach is vital in applied sciences, where exact function forms aren't always known.

In our example, we are presented with a set of points, and our goal is to find the definite integral of these points over a defined range using established numerical methods.

Exploring Definite Integral and its Importance

A definite integral represents the area under a curve over a specified range. Simply put, it sums up infinite tiny segments from a function's graph to find the total accumulated value over an interval.
This visualization aids in understanding real-world problems like finding distances, probabilities, and accumulated changes. In our scenario,

• We aim to calculate the integral from 0 to 15, signifying the total area between the curve defined by our data points and the x-axis.
• The resulting value gives us insights into quantity or change accumulated across these points.
• This measure is independent of how scattered or irregular the function may appear.

The task utilizes numerical methods like the trapezoidal rule to derive the definite integral from available discrete data.

Making Use of Discrete Data Points

In numerical integration, discrete data points serve as the primary building blocks for our calculations. These points may come from experimental data, computed algorithms, or measurements.
With discrete data, we have values for a specific set of inputs rather than a continuous function formula.

• The trapezoidal rule, used in this example, leverages these discrete points to draw many tiny adjacent trapezoids under the curve and sum their areas.
• Given our table, the data points at specific x-values are crucial in forming these approximations.
• These data-driven methods are foundational for computations in many scientific fields, where continuous data curves aren't accessible.

By understanding how discrete data points fit into the broader context of numerical analysis, learners can better appreciate how approximations are made without explicit function equations.