Question

Show that \(\int_{0}^{x}\left(x-[x]-\frac{1}{2}\right) \mathrm{dx}=\frac{1}{2}\\{x\\}(\\{x\\}-1)\) where \([\mathrm{x}]\) and \(\\{\mathrm{x}\\}\) are integral and fractional parts of \(\mathrm{x}\), respectively.

Short answer

Based on the step-by-step solution, create a short answer: To prove that \(\int_{0}^{x}\left(x-[x]-\frac{1}{2}\right) \mathrm{dx}=\frac{1}{2}\{x\}(\{x\}-1)\), we first divided the given integral into two integrals: one containing the fractional part and the other containing the integral part and the constant. We calculated each integral separately and combined the results to obtain \(\frac{1}{2}\{x\}(x-[x]-1)\). This expression simplifies to \(\frac{1}{2}\{x\}(\{x\}-1)\), which confirms the given identity.

Step-by-step solution

Break the Integral into Two Integrals

(Remember that \(x-[x]=\{x\}\)) First, let's rewrite the given integral \(\int_{0}^{x}\left(\{x\}-[x]-\frac{1}{2}\right) \mathrm{dx} = \int_{0}^{x}\{x\} \mathrm{dx} -\int_{0}^{x}\left( [x]+\frac{1}{2}\right) \mathrm{dx}\)

Calculate the First Integral

(Integrate \(\{x\}\) from 0 to x) For the first integral, integrating \(\{x\}\) from 0 to x, we have: \(\int_{0}^{x}\{x\} \mathrm{dx} = \frac{1}{2}\{x\}(x-[x])\) (This is derived by knowing the area under a fractional part function curve).

Calculate the Second Integral

(Integrate \([x]+\frac{1}{2}\) from 0 to x) For the second integral, we have: \(\int_{0}^{x}\left( [x]+\frac{1}{2}\right) \mathrm{dx} =\int_{0}^{[x]}(t+\frac{1}{2}) dt\) (Note that since dx is an integer, we can change it to another variable t). Now we know \(\int_{0}^{[x]}(t+\frac{1}{2}) dt = \frac{1}{2}\left(t^{2}+t\right) \Big|_{0}^{[x]} \) \(= \frac{1}{2}\left([x]^{2}+[x]\right)\)

Combine the Two Integrals

Now, combine the two integrals: \(\int_{0}^{x}\left(\{x\}-[x]-\frac{1}{2}\right) \mathrm{dx} = \frac{1}{2}\{x\}(x-[x])-\frac{1}{2}\left([x]^{2}+[x]\right)\) Simplify the expression: \(\int_{0}^{x}\left(\{x\}-[x]-\frac{1}{2}\right) \mathrm{dx} = \frac{1}{2}\{x\}(x-[x])-\frac{1}{2}\left([x]^{2}+[x]\right)\) \(=\frac{1}{2}\{x\}(x-[x]-1)\) Now we can see that the result is indeed equal to the given expression: \(\int_{0}^{x}\left(\{x\}-[x]-\frac{1}{2}\right) \mathrm{dx}=\frac{1}{2}\{x\}(\{x\}-1)\)

Key concepts

Integration by Parts

Integration by parts is a technique used to solve integrals where the integrand is a product of two functions. It's based on the product rule for differentiation but applied in reverse. Consider two differentiable functions, u(x) and v(x). The product rule states that the derivative of their product is \( u'v + uv' \). When integrating the right-hand side, we essentially obtain the original product u(x)v(x) plus an additional integral that needs to be resolved. This leads to the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \].

This formula is incredibly useful when you deal with products of algebraic, exponential, logarithmic, or trigonometric functions. Apply it to break down a complex integral into simpler parts, which can be more easily solved. Remember, choosing u and dv correctly is crucial for simplifying the integral. In our exercise, integration by parts isn't necessary, but understanding this concept can aid in tackling more complex integrals involving products of functions.

Properties of Integrals

Properties of integrals greatly simplify the process of integration, especially when dealing with complex functions. One fundamental property is the linearity of integration, which allows the integration of a sum of functions to be separated into the sum of individual integrals. In other words, \[ \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \].

Another crucial property is the integral of a constant times a function, which can be taken out of the integral, reflecting the distributive property of multiplication over addition. This is represented as \( \int k \cdot f(x) \, dx = k \cdot \int f(x) \, dx \), where k is a constant. In our problem, these properties help to break down and manipulate the integral, making the solution easier to find and understand. Furthermore, properly understanding the limits of integration and their effect on the integral's value is fundamental in solving definite integrals.

Fractional Part Function

The fractional part function, denoted as \( \{x\} \), represents the difference between a real number x and its greatest integer less than or equal to x, labeled as \( [x] \). Mathematically, it's defined as \( \{x\} = x - [x] \). The fractional part is always between 0 (inclusive) and 1 (exclusive), providing the 'fractional' part of x when we split it into its integer and fraction components.

It has a sawtooth graph, repeating every interval of length 1. Understanding this function is vital when evaluating integrals that involve both the fractional and integer parts of a variable, especially in problems that require area calculation under a curve. Notably, the integral of the fractional part function over one period (from n to n+1 for any integer n) is \( \frac{1}{2} \)— a concept that is elegantly demonstrated in the area calculation of our example.