Problem 65 - Calculus Single Variable

Question

Let \(c\) be the midpoint of \([a, b]\). Let \(m\) and \(h\) be any constants. Show that $$ \int_{a}^{b}(m(x-c)+h) d x=h(b-a) $$

Step-by-Step Solution

Answer

The integral \(\int_{a}^{b} (m(x-c)+h) dx\) evaluates to \(h(b-a)\).

Step 1: Understanding the Problem

We have the integral of the function \(m(x-c) + h\) over the interval \([a, b]\). We know that \(c\) is the midpoint, which means \(c = \frac{a+b}{2}\). We need to show that the integral equals \(h(b-a)\).

Step 2: Substituting the Midpoint

First, identify the midpoint \(c\) of \([a, b]\) as \(c = \frac{a+b}{2}\). Substitute this expression for \(c\) into the integral, resulting in \(\int_{a}^{b} \left(m \left(x - \frac{a+b}{2}\right) + h\right) dx\).

Step 3: Breaking Down the Integral

Separate the integral into two parts: \(\int_{a}^{b} m\left(x - \frac{a+b}{2}\right) dx + \int_{a}^{b} h dx\). This allows us to handle the linear and constant components independently.

Step 4: Evaluating the First Integral

The first part, \(\int_{a}^{b} m\left(x - \frac{a+b}{2}\right) dx\), can be solved by recognizing it is a linear function shifted by the constant \(m\). The integral simplifies to \([mx - m\frac{a+b}{2}]\) evaluated from \(a\) to \(b\). Specifically, it simplifies to zero because the symmetrical bounds about the midpoint cancel its contribution.

Step 5: Evaluating the Second Integral

The second integral \(\int_{a}^{b} h dx\) is straightforward. Since \(h\) is constant, the integral becomes \(h(x)\) evaluated from \(a\) to \(b\), resulting in \(h(b-a)\).

Step 6: Combining Results

Add the results from the two integrals: the first integral contributes nothing as it evaluates to zero, and the second yields \(h(b-a)\). Thus, the total integral \(\int_{a}^{b} (m(x-c)+h) dx\) equals \(h(b-a)\).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, representing the accumulation of a quantity over a given interval. Unlike indefinite integrals, which represent a family of functions, definite integrals yield a specific numerical value. This value quantifies the net area under the curve of a function over a specified interval.

For a function \(f(x)\), the definite integral from \(a\) to \(b\) is denoted as \(\int_{a}^{b} f(x) \, dx\). In essence, it calculates the area between the function \(f(x)\) and the x-axis, from \(x = a\) to \(x = b\). A key property of definite integrals is that they account for the direction: areas above the x-axis contribute positively, while those below contribute negatively.

Definite integrals are crucial for analyzing real-world problems where we need to determine the accumulated change or total quantity represented by a continuous function. They provide valuable insights in physics, engineering, and various fields of science.

Midpoint Rule

The Midpoint Rule is a numerical method for approximating the value of a definite integral. It is particularly useful when an integral is difficult to evaluate analytically. The concept involves using rectangles to estimate the area under a curve, taking the midpoint of each interval to determine the height of each rectangle.

To apply the midpoint rule, you begin by subdividing the integral's interval \([a,b]\) into smaller subintervals. The width of each subinterval is typically denoted as \(\Delta x\). The midpoint \(c_i\) of each subinterval \([x_{i-1}, x_i]\) is calculated, which then helps to estimate the integral by summing up the areas of rectangles of height \(f(c_i)\) and width \(\Delta x\).

• The formula for the midpoint rule is given by: \(\int_{a}^{b} f(x) \, dx \approx \sum_{i=1}^{n} f(c_i) \Delta x\)
• Each \(c_i = \frac{x_{i-1} + x_i}{2}\)

The accuracy of the midpoint rule depends on the number of subintervals; more subdivisions generally lead to a better approximation. This rule is particularly advantageous when the function is approximately linear over each subinterval.

Linear Functions

Linear functions are the simplest type of function characterized by a constant rate of change. Their general form is \(f(x) = mx + b\), where \(m\) represents the slope of the line, and \(b\) is the y-intercept, indicating where the line crosses the y-axis.

These functions are called linear because their graph is a straight line. The slope \(m\) denotes the steepness and direction of the line. If \(m\) is positive, the line slopes upwards, while a negative \(m\) indicates a downward slope. The y-intercept \(b\) provides the initial value of the function when \(x = 0\).

• Key properties of linear functions include their constant slope and the fact that any two points on the line have the same rate of change.
• Linear functions are unique because they have no maximum or minimum points within their domain; they extend infinitely in both directions.

Linear functions are crucial in mathematics and real-world applications because they often describe relationships with constant rates, such as speed, cost, or resource consumption. Understanding these functions is essential for grasping more complex calculus concepts, as seen in the integration of linear terms in the original exercise.