Question

Given that, when \(x>0\), the function \(f(x)\) is positive and is strictly decreasing, prove that \(f(n+1) \leq \int_{n}^{n+1} f(x) d x \leq f(n)\)

Short answer

Question: Prove that for a function \(f(x)\) which is positive and strictly decreasing when \(x>0\), the inequality \(f(n+1) \leq \int_{n}^{n+1} f(x) dx \leq f(n)\) holds for all n. Answer: We proved that for a strictly decreasing function \(f(x)\) that is positive for \(x>0\), the inequality \(f(n+1) \leq \int_{n}^{n+1} f(x) dx \leq f(n)\) holds for all n, using the properties of definite integrals and the fact that the function is strictly decreasing.

Step-by-step solution

Establish the inequality for the integral

Given that \(f(x)\) is strictly decreasing, we have: $$f(n+1) \leq f(x) \leq f(n)$$ for all \(x \in [n, n+1]\).

Integrate the inequality with respect to x

Integrate the inequality over the interval [n, n+1] with respect to x: $$\int_{n}^{n+1} f(n+1) dx \leq \int_{n}^{n+1} f(x) dx \leq \int_{n}^{n+1} f(n) dx$$

Evaluate the integrals and simplify

Evaluate the definite integrals and simplify the expressions: $$f(n+1) \int_{n}^{n+1} dx \leq \int_{n}^{n+1} f(x) dx \leq f(n) \int_{n}^{n+1} dx$$ $$f(n+1) (n+1 - n) \leq \int_{n}^{n+1} f(x) dx \leq f(n) (n+1 - n)$$ $$f(n+1) \leq \int_{n}^{n+1} f(x) dx \leq f(n)$$ Thus, we have proved the desired inequality \(f(n+1) \leq \int_{n}^{n+1} f(x) dx \leq f(n)\).

Key concepts

Definite Integrals

Definite integrals are a cornerstone of integral calculus, with a wide range of applications in science and engineering. They can be thought of as a method to calculate the total accumulation of a quantity. For instance, if you want to determine the area under a curve between two points on the x-axis, a definite integral is the tool you'll use.
In mathematical terms, a definite integral is represented as \( \int_{a}^{b} f(x) \, dx \) where \(f(x)\) is the function to be integrated over the interval \[a,b\]. It represents the sum of the infinite number of infinitesimal areas of rectangles with widths \(dx\) and heights \(f(x)\). But it's not just about areas; this concept is used to describe physical quantities such as displacement, volume, and total work done, among others.
In the problem provided, the definite integral \( \int_{n}^{n+1} f(x) \, dx \) was used to find the total value of function \(f(x)\) between \(n\) and \(n+1\). The solution shows that this integral value is bounded by the function values at the endpoints of the interval, understandable when considering the nature of the function's decrease over the interval.

Function Properties

A function can have various properties that define its behavior and how it interacts with operations such as integration. Functions can be increasing or decreasing, concave up or concave down, and continuous or discontinuous, among other characteristics.
Functions that are strictly decreasing, like the one in our example, have the property that for any two points \(x_1\) and \(x_2\) where \(x_1 < x_2\), the value of \(f(x_1)\) is greater than \(f(x_2)\). This is crucial for understanding the relationship between the function values and the definite integral. Since the function is positively decreasing, one can infer that the area under the curve from \(n\) to \(n+1\) will be trapped between the values of the function at these two points, consistent with the inequality \(f(n+1) \leq \int_{n}^{n+1} f(x) \, dx \leq f(n)\).
Intuitively, the property of being strictly decreasing means that as you move along the x-axis, the function value is consistently going down. This insight helps visually understand why the inequality holds true and is vital to the proof provided in the example.

Monotonically Decreasing Functions

Monotonically decreasing functions are a specific category of functions that have the property of never increasing as the input increases. More formally, a function \(f\) is monotonically decreasing if, for every \(x_1\) and \(x_2\) such that \(x_1 < x_2\), the inequality \(f(x_1) \geq f(x_2)\) holds.
This concept is a significant player in the inequality proof involving integrals. If \(f(x)\) is monotonically decreasing, it ensures that the integral over a set interval will not surpass the function's value at the beginning of the interval and will not fall below the function's value at the end of the interval. This characteristic gives rise to the nature of the inequality in the homework problem, as it guarantees that as we progress along the interval from \(n\) to \(n+1\), the total accumulation captured by the definite integral reflects the function's consistent decrease.
Understanding monotonically decreasing functions illuminates why the step-by-step solution confidently asserts the inequality involving \(f(n+1)\), \(f(n)\), and the integral over the interval. It also highlights the role monotonicity plays in determining bounds on definite integrals and hints at the deep connection between the shapes of functions and the values of their integrals.