Question

It is known that \(\int_{a}^{b} f(x) d x>\int_{a}^{b} g(x) d x\). Does it follow that \(\mathrm{f}(\mathrm{x}) \geq \mathrm{g}(\mathrm{x}) \forall \mathrm{x} \in[\mathrm{a}, \mathrm{b}] ?\) Give examples.

Short answer

Answer: No, it is not necessarily true. The integral of a function represents the total area under the curve within a given interval, but it does not necessarily imply that the function must be greater than or equal to another function at every point within that interval. A counterexample has been provided above where the integral of function f(x) = x is greater than the integral of function g(x) = x^2 over the interval [0, 1], but f(x) is not greater than or equal to g(x) for every x in that interval.

Step-by-step solution

Understanding the integral inequality

Since the integral of f(x) is greater than the integral of g(x) over the interval [a, b], it implies that the total area under the curve of f(x), within this interval, is larger than the total area under the curve of g(x). However, it does not necessarily imply that f(x) is greater than or equal to g(x) for every x in [a, b].

Provide a counterexample

Let's consider two functions, f(x) = x and g(x) = x^2. If we integrate both of these functions over the interval [0, 1], we will get: $$\int_{0}^{1} x dx = \bigg[\frac{1}{2}x^2\bigg]_0^1 = \frac{1}{2}$$ and $$\int_{0}^{1} x^2 dx = \bigg[\frac{1}{3}x^3\bigg]_0^1 = \frac{1}{3}$$ It is clear that \(\int_{0}^{1} f(x) dx > \int_{0}^{1} g(x) dx\), as \(\frac{1}{2} > \frac{1}{3}\). However, if we compare the two functions f(x) and g(x) within this interval [0, 1], we can see that f(x) is not always greater than or equal to g(x). Particularly, when \(x = \frac{1}{2}\), we have: $$\frac{1}{2} = f(\frac{1}{2}) < g(\frac{1}{2}) = \frac{1}{4}$$ This contradicts the claim that f(x) ≥ g(x) for every x in [a, b]. Thus, knowing that the integral of f(x) is greater than the integral of g(x) over the interval [a, b] does not necessarily imply that f(x) ≥ g(x) for every x in the interval [a, b]. We have provided an example that contradicts this claim.

Key concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematics focused on the concept of an integral. This notion relates to the accumulation of quantities and the computation of areas under curves. An integral, in its simplest form, represents the area under the graph of a function within a certain range defined by an interval on the x-axis.

When we talk about the integral of a function with respect to a variable x, over an interval [a, b], we are essentially looking at the sum of the infinitely small areas under the curve between x = a and x = b. Calculating an integral, known as integration, can be done in various ways, including but not limited to geometric approximation, Riemann sums, or antiderivatives. The power of integral calculus lies in its ability to handle complex shapes and functions that are not easily quantifiable by basic geometry.

Dealing with integral inequalities (when we compare integrals of two functions over the same interval) is one of the many applications of integral calculus. Understanding this concept helps in a wide range of fields including physics, engineering, economics, and anywhere that requires measurement of total accumulation over a range.

Area Under the Curve

The 'area under the curve' is a term often used in the context of integral calculus to describe the region between the curve of a function and the x-axis. When we integrate a function over an interval, we are calculating the size of this region. The integral gives us a numerical value that, in a geometrical sense, represents this 'area.'

However, it's important to note that when the function takes on negative values, the integral may represent an 'area' with a negative value, as it lies below the x-axis. Therefore, when interpreting integral results, one must account for the direction of the curve relative to the x-axis.

To provide a visual example, consider plotting the graph of a function f(x) over an interval [a, b]. The curve created by plotting f(x) intercepts the x-axis at points a and b, and the space enclosed by this portion of the curve and the x-axis is what we refer to as the area under the curve. Whether this space is above or below the x-axis determines if its contribution is positive or negative.

Counterexample in Integrals

A counterexample in the context of integrals is a specific example used to show that a general statement concerning integrals is false. When comparing two functions through their integrals within the same interval, as seen in the original exercise, it's easy to mistakenly assume that if one function's integral is larger, then the function itself must always be greater than the other over that interval. As we've seen, this is not necessarily the case.

A counterexample demonstrates that while the total area (integral) for one function may exceed another's, at certain points within the interval, the function with the larger integral may actually be less than the other function. This phenomenon is due to the way integration sums up 'areas' without specifics about the function's behavior at every single point within the interval.

As in the step-by-step solution provided, where the function f(x) = x has a greater integral from 0 to 1 compared to the function g(x) = x^2, the integral gives us no information about the relative sizes of f(x) and g(x) at any particular point. The value f(1/2) is indeed less than g(1/2), proving that a greater integral over an interval does not imply pointwise superiority.