Problem 27
Question
Let \( A \) be the area under the graph of an increasing continuous function \( f \) from \( a \) to \( b \), and let \( L_n \) and \( R_n \) be the approximations to \( A \) with \( n \) subintervals using left and right endpoints, respectively. (a) How are \( A \), \( L_n \), and \( R_n \) related? (b) Show that $$ R_n - L_n = \frac{b - a}{n} [f(b) - f(a)] $$ Then draw a diagram to illustrate this equation by showing that the \( n \) rectangles representing \( R_n - L_n \) can be reassembled to form a single rectangle whose area is the right side of the equation. (c) Deduce that $$ R_n - A < \frac{b - a}{n} [f(b) - f(a)] $$
Step-by-Step Solution
Verified Answer
(a) \(L_n \leq A \leq R_n\).
(b) Proved as \( \large R_n - L_n = \frac{b-a}{n} [f(b) - f(a)] \) and visualized geometrically.
(c) \(R_n - A < \frac{b-a}{n} [f(b) - f(a)]\).
1Step 1: Understand the Relationship among A, L_n, and R_n
The area under the curve of an increasing continuous function from \( a \) to \( b \) is denoted by \( A \). The left endpoint approximation, \( L_n \), uses the left endpoints of subintervals, and since \( f \) is increasing, \( L_n \) underestimates the actual area \( A \). The right endpoint approximation, \( R_n \), uses the right endpoints of subintervals, which overestimates \( A \). Therefore, \( L_n \leq A \leq R_n \).
2Step 2: Prove R_n - L_n Formula
Consider the formula for \( R_n - L_n \) as the sum of the width of each subinterval times the difference of function values at their endpoints.The difference exposure can be expressed as:\[R_n - L_n = \sum_{i=1}^{n}\left[ f\left(x_{i}\right) - f\left(x_{i-1}\right) \right]\Delta x \]where \( \Delta x = \frac{b-a}{n} \) and \( x_{i} = a + i \Delta x \). Notice that \( R_n \) is based on right endpoints, \( f(x_i) \), and \( L_n \) is based on left endpoints, \( f(x_{i-1}) \). Substitute and simplify:\[R_n - L_n = \frac{b-a}{n} \sum_{i=1}^{n} \left[ f\left(a + i\Delta x\right) - f\left(a + (i-1)\Delta x\right) \right] = \frac{b-a}{n} [f(b) - f(a)] \]This shows that \( R_n - L_n = \frac{b-a}{n} [f(b) - f(a)] \).
3Step 3: Diagram and Explanation
Visualize \( R_n - L_n \) as a difference of areas formed by the actual rectangles: each subinterval \( i \) from \( a \) to \( b \) contributes a small strip of width \( \Delta x \) and height difference \( f(x_i) - f(x_{i-1}) \). Aggregating these strips effectively forms a single large rectangle with base \( b-a \) and height \( \frac{[f(b) - f(a)]}{n} \), which directly corresponds to the formula \( \frac{b-a}{n} [f(b) - f(a)] \).
4Step 4: Deduce the Inequality
Since \( f(x) \) is increasing, for any subinterval, the height \( f(x_i) \) used in \( R_n \) is always greater than \( f(x) \) anywhere in that subinterval, but less or equal for \( f(b) \). Thus, the aggregate difference \([f(b) - f(x)]\) for actual midpoints is less than \((f(b) - f(a))/n\) for the entire range from \( a \) to \( b \).Mathematically:\[0 < \int_a^b f(x)\,dx - L_n < \frac{b-a}{n} [f(b) - f(a)]\]Therefore, \( R_n - A < \frac{b-a}{n} [f(b) - f(a)] \) in the limit of the improper summation representation implies this directly.
Key Concepts
Definite IntegralsApproximation MethodsContinuous FunctionsArea Under a Curve
Definite Integrals
Definite integrals represent the accumulated sum of a function over a particular interval. These are used to calculate the total area under the curve of a given function, from one point to another along the x-axis. When we talk about the definite integral of a continuous function from point \( a \) to \( b \), we denote it as \( \int_a^b f(x) \, dx \). This essentially quantifies the total "mass" beneath the curve between these two points.
In the context of our exercise, the area under the curve \( A \) is nothing but the definite integral of \( f \) from \( a \) to \( b \). Since \( f \) is continuous and increasing, the integration technique provides an exact value, unlike approximation methods which give estimates. The critical takeaway is that definite integrals provide precise measures of regions under graphs of functions, especially continuous ones.
In the context of our exercise, the area under the curve \( A \) is nothing but the definite integral of \( f \) from \( a \) to \( b \). Since \( f \) is continuous and increasing, the integration technique provides an exact value, unlike approximation methods which give estimates. The critical takeaway is that definite integrals provide precise measures of regions under graphs of functions, especially continuous ones.
Approximation Methods
Approximation methods are techniques used to estimate the value of definite integrals, especially when an exact solution is challenging to obtain. Riemann sums are a primary approach used for this purpose, utilizing either left endpoints, right endpoints, or midpoints.
In our specific exercise, two approximation methods are at work: the left endpoint method \( L_n \) and the right endpoint method \( R_n \).
In our specific exercise, two approximation methods are at work: the left endpoint method \( L_n \) and the right endpoint method \( R_n \).
- **Left Endpoint (\( L_n \))**: This method calculates the area using rectangular strips with heights determined by the function's value at the left edge of each subinterval. For an increasing function, \( L_n \) underestimates the area under the curve.
- **Right Endpoint (\( R_n \))**: Here, the heights of the rectangles are based on the function's value at the right edge of each subinterval. For increasing functions, \( R_n \) provides an overestimate of the area.
Continuous Functions
Continuous functions are those that, simply put, have no breaks, holes, or gaps in their graphs. They change smoothly over any interval, meaning you can draw the graph without lifting your pen off the paper. Mathematically, a function \( f \) is continuous at a point \( x = c \) if \( \lim_{x \to c} f(x) = f(c) \).
The significance of continuity in our exercise is substantial:
The significance of continuity in our exercise is substantial:
- Ensures the function has a well-defined area under its curve without jumps or breaks.
- Guarantees convergence of Riemann sums to the true area, which is the definite integral \( A \).
- Makes it possible to apply mathematical techniques for integration and approximation reliably.
Area Under a Curve
The area under a curve is a fundamental concept in calculus used to derive various physical and theoretical insights. It often represents quantities like distance, probability, or accumulation based on the context.
In mathematical terms, the area is calculated using integration and is formally written as \( \int_a^b f(x) \, dx \). In the context of our problem, this area, denoted by \( A \), is defined specifically for increasing continuous functions. The main points related to this are:
In mathematical terms, the area is calculated using integration and is formally written as \( \int_a^b f(x) \, dx \). In the context of our problem, this area, denoted by \( A \), is defined specifically for increasing continuous functions. The main points related to this are:
- The area under the curve from \( a \) to \( b \) is always defined and finite for continuous functions.
- Left and right approximations like \( L_n \) and \( R_n \) provide different estimates of this area, helping to better understand the integral at different approximation levels.
- A visualization of these methods, especially the difference \( R_n - L_n \), highlights how areas can add up to form a perfect rectangle, offering graphic insight into the integral's value.
Other exercises in this chapter
Problem 27
Evaluate the integral. \( \displaystyle \int^1_0 (u + 2) (u - 3) \,du \)
View solution Problem 27
Prove that \( \displaystyle \int^b_a x \, dx = \frac{b^2 - a^2}{2} \).
View solution Problem 28
Evaluate the indefinite integral. \( \displaystyle \int e^{\cos t} \sin t \, dt \)
View solution Problem 28
Evaluate the integral. \( \displaystyle \int^{2}_{1} \biggl( \frac{1}{x^2} - \frac{4}{x^3} \biggr) \,dx \)
View solution