Problem 208 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of \(\int_{1}^{a}[x] f^{\prime}(x) d x, a>1\) where \([x]\) denotes the greatest integer not exceeding \(x\) is (A) \(a f(\mathrm{~A})-\\{f(1)+f(2)+\ldots+f([a])\\}\) (B) \([a] f(\mathrm{~A})-\\{f(1)+f(2)+\ldots+f([a])\\}\) (C) \([a] f([a])-\\{f(1)+f(2)+\ldots+f(a)\\}\) (D) \(a f([a])-\\{f(1)+f(2)+\ldots+f(a)\\}\)

Step-by-Step Solution

Answer

The answer is (B): \([a] f(a) - {f(1)+f(2)+\ldots+f([a])}\).

Step 1: Understand the Problem

We are asked to evaluate the integral \(\int_{1}^{a}[x] f^{\prime}(x) \, dx\), where \([x]\) is the greatest integer function, also known as the floor function. It assigns an integer value to \(x\) which is the largest integer less than or equal to \(x\). The choices involve using floor function \[a\].

Step 2: Break the Integral into Intervals

The greatest integer function \([x]\) is constant over intervals \([n, n+1)\) where \(n\) is an integer. Thus, break the integral from 1 to \(a\) into segments between consecutive integers: \[ \int_{1}^{a}[x] f^{\prime}(x) \, dx = \sum_{n=1}^{[a]} \int_{n}^{n+1} n f^{\prime}(x) \, dx \] and if \(a\) is not an integer, we will separately consider the part from \([a]\) to \(a\).

Step 3: Evaluate Each Segment

Since \([x] = n\) is constant over each interval \([n, n+1)\), we can write: \[ \int_{n}^{n+1} n f^{\prime}(x) \, dx = n \int_{n}^{n+1} f^{\prime}(x) \, dx = n \left[ f(x) \right]_{n}^{n+1} = n(f(n+1) - f(n)) \].

Step 4: Sum Over all Intervals

For integer intervals, sum the result over all segments: \[ \sum_{n=1}^{[a]} n (f(n+1) - f(n)) \]This simplifies to: \[ f([a]+1) ([a]) - \sum_{n=1}^{[a]} f(n) \] using telescopic cancellation.

Step 5: Consider the Remaining Part if \(a\) is Not an Integer

If \(a\) is not an integer, add the term from \([a]\) to \(a\):\[ \int_{[a]}^{a} [x] f^{\prime}(x) \, dx = [a] \int_{[a]}^{a} f^{\prime}(x) \, dx = [a] (f(a) - f([a])) \].

Step 6: Combine Results

The entire integral, therefore, is:\[ [a] f(a) - \sum_{n=1}^{[a]} f(n) \].

Step 7: Identify Correct Answer Option

Looking at the options, option B: \[ [a] f(a) - \{ f(1) + f(2) + \ldots + f([a]) \} \] matches our derived result.

Key Concepts

Greatest Integer Function

The greatest integer function, denoted as \([x]\), assigns to any real number \(x\) the largest integer less than or equal to \(x\). This function is widely used in integral calculus and throughout mathematics to simplify calculations involving integer values. For example, when \(x = 3.7\), the greatest integer function will return \([3.7] = 3\). This is particularly useful because it allows us to handle non-integer input in a manner that is consistent and mathematically rigorous. When you encounter an integral with the greatest integer function, it typically involves segmenting the function into intervals where the value is constant.

Floor Function

The floor function is another name for the greatest integer function, represented by the same notation \([x]\). It is one of the standard ways of rounding down a real number to an integer.

Some important points to remember about the floor function include:

• It always rounds down to the nearest integer.
• If \(x\) itself is an integer, then \([x] = x\).
• It is a step function, meaning it jumps at integer values.

These properties help us evaluate integrals, especially when calculating with piecewise or discontinuous sections such as in a telescopic series.

Integral Calculus

Integral calculus is a branch of calculus that deals with integrals, encompassing processes of finding antiderivatives and calculating areas under curves. In essence, it allows us to sum an infinite series of infinitesimally small quantities.

The definite integral is one type of integral that calculates the net area between the curve and the horizontal axis, from one point to another. It is written as \(\int_{a}^{b} f(x) \, dx\) and evaluated as the limit of a sum. In the given problem, evaluating pieces over intervals defined by the floor function transforms the problem into a manageable form. The goal is to sum discrete contributions over defined intervals, leading to a comprehensive result.

Telescopic Series

A telescopic series is a sequence of terms designed so that most terms cancel out when summed, leaving behind a significantly reduced expression.

These series are invaluable in integral calculus for simplifying complex sums. When you encounter a sum like \(\sum (f(n+1) - f(n))\), the intermediate terms cancel, collapsing the expression to \(f([a]+1) - f(1)\) or similar, depending on the limits.

This method is employed to address problems involving piecewise functions like the floor function because it distills sums down to their essential components. The telescopic property is crucial in reaching a simplified, tractable solution as demonstrated in the solution where segments cancel each other out, leading to easier calculation of \(\int_{1}^{a}[x] f^{\prime}(x) \, dx\).