Problem 141 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of \(\int_{0}^{1}(\\{2 x\\}-1)(\\{3 x\\}-1) d x\), where \(\\{\cdot\\}\) denotes the fractional part is, (A) \(\frac{19}{72}\) (B) \(\frac{31}{9}\) (C) \(\frac{1}{8}\) (D) \(\frac{72}{19}\)

Step-by-Step Solution

Answer

The value of the integral is \( \frac{1}{8} \) (Option C).

Step 1: Understand the Fractional Part Function

The fractional part, denoted as \( \{x\} \), is defined as \( x - \lfloor x \rfloor \), where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \). It returns the "decimal" or "non-integer" part of \( x \). For instance, \( \{2x\} = 2x - \lfloor 2x \rfloor \) and \( \{3x\} = 3x - \lfloor 3x \rfloor \).

Step 2: Evaluate the fractional parts

For \( 0 \leq x < 1 \), the expression \( \{2x\} \) returns \( 2x \) itself because \( \lfloor 2x \rfloor = 0 \) for \( 0 \leq x < 0.5 \), and \( \{2x\} = 2x - 1 \) for \( 0.5 \leq x < 1 \). Similarly, for \( \{3x\} \), \( \lfloor 3x \rfloor \) determines if \( \{3x\} = 3x \) or \( 3x - n \) where \( n = 1 \) or \( 2 \) based on \( x \).

Step 3: Break the integration into intervals

Since the fractional parts \( \{2x\} \) and \( \{3x\} \) change behavior at each integer fraction multiple (i.e., at points where \( 2x \) or \( 3x \) hits a new integer), we need to consider these points: \( x = 0, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \) and \( 1 \). This results in the intervals: \([0, \frac{1}{3})\), \([\frac{1}{3}, \frac{1}{2})\), \([\frac{1}{2}, \frac{2}{3})\), and \([\frac{2}{3}, 1)\).

Step 4: Calculate the integral over each interval

For each interval, determine the forms of \( \{2x\} \) and \( \{3x\} \), and integrate \((\{2x\}-1)(\{3x\}-1)\) for that range. Specifically, this transforms our integral into a sum of simpler integrals over each interval. Calculate each sub-integral separately, taking care to substitute and evaluate the limits properly based on the behavior of \( \{2x\} \) and \( \{3x\} \) in that range.

Step 5: Sum the results of all intervals

Add the values of all computed integrals from Step 4 together to get the total area under the curve from \( x = 0 \) to \( 1 \). Make sure each calculation is correctly simplified to ensure the final result is accurate.

Key Concepts

Fractional Part Function

The fractional part function, denoted as \(\{x\}\), is a mathematical concept used to describe the "decimal" portion of a number. It is defined by the formula \(x - \lfloor x \rfloor\), where \(\lfloor x \rfloor\) represents the greatest integer less than or equal to \(x\). This function helps isolate the part of \(x\) that lies between successive integers, essentially ignoring the whole number component. For example, \(\{2x\}\) is calculated as \(2x - \lfloor 2x \rfloor\).
This function behaves differently based on the interval of \(x\). Within the bounds of

• \(0 \leq x < \frac{1}{3}\), \(\{3x\} = 3x\) as \(\lfloor 3x \rfloor = 0\)
• \(\frac{1}{3} \leq x < \frac{1}{2}\), \(\{3x\} = 3x - 1\) since \(\lfloor 3x \rfloor = 1\)
• Similarly for \(\{2x\}\) over different intervals like \(0 \leq x < \frac{1}{2}, \{2x\} = 2x\)

Understanding this behavior is crucial for calculating definite integrals involving fractional parts.

Integration by Parts

Integration by parts is a technique for finding integrals of products of functions. It is a useful tool particularly when dealing with integrals requiring a decomposition into simpler elements. The formula is described as: \[\int u \, dv = uv - \int v \, du\]Here, \(u\) and \(dv\) are chosen strategically:

• \(u\) should be differentiable to simplify the expression \(du\)
• \(dv\) needs to be integrable so \(v\) can be easily found

For the original problem, integration by parts helps in managing complex expressions into manageable segments. It's especially useful when dealing with polynomial or exponential components, as seen in fractional part integrals. By breaking apart the product of \((\{2x\}-1)(\{3x\}-1)\), you apply parts to integrate across specified intervals identified earlier. Remember, the key is selecting \(u\) and \(dv\) wisely to make the integration process smoother.

Piecewise Integration

Piecewise integration is a strategy used to evaluate integrals on sections of the interval where the function has different expressions. This is necessary when functions change their behavior or form at certain points. In the exercise, \(\{2x\}\) and \(\{3x\}\) alter at specific fractions of \(x\), prompting breaks at

• \(x = 0\)
• \(\frac{1}{3}\)
• \(\frac{1}{2}\)
• \(\frac{2}{3}\)
• \(1\)

At each interval, the behavior of the function is consistent and differs from adjacent segments. Therefore, the integration process involves calculating the definite integral for each piece, where each piece is characterized by its specific functional form. This leads to four separate integrals, each requiring evaluation and summing to obtain the total integral. Correctly understanding where the function changes is crucial to applying piecewise integration effectively. It ensures the integration captures all variations and singularities throughout the range \([0, 1)\).