Chapter 5: Problem 65
Let \(c\) be the midpoint of \([a, b]\). Let \(m\) and \(h\) be any constants. Show that $$ \int_{a}^{b}(m(x-c)+h) d x=h(b-a) $$
Short Answer
Expert verified
The integral \(\int_{a}^{b} (m(x-c)+h) dx\) evaluates to \(h(b-a)\).
Step by step solution
01
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 \(h(b-a)\).
02
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 \(\int_{a}^{b} \left(m \left(x - \frac{a+b}{2}\right) + h\right) dx\).
03
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.
04
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.
05
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 \(h(b-a)\).
06
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 \(h(b-a)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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 \(\Delta x\).
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 \(c_i = \frac{x_{i-1} + x_i}{2}\)
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 \(x = 0\).
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.