JEE Main 2022 (27 June Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Definite Integration: The integral $\int_{0}^{1} \frac{1}{[ \frac{1}{x} ]} d x$ , where $[.]$ denotes the greatest integer function is equal to

Q: Why do we split the integral at $1/n$? Is there a simpler way?

The greatest integer function $[1/x]$ is inherently discontinuous. It jumps at every integer value of $1/x$. Since $x$ ranges from $0$ to $1$, $1/x$ ranges from $1$ to $\infty$. The function $[1/x]$ remains constant between these integer values, which is why we must partition the domain into intervals like $(1/2, 1], (1/3, 1/2]$, and so on. There is no shortcut; this partitioning is the fundamental definition of integrating a step function.

Q: Wait, why does the sum $\sum_{n=1}^{\infty} \frac{1}{n^2}$ equal $\frac{\pi^2}{6}$?

That is the famous Basel Problem, first solved by Leonhard Euler. While it is a standard result in advanced calculus, for JEE purposes, it is a powerful tool to keep in your arsenal. Whenever you encounter the sum of the reciprocals of squares, you can confidently use $\frac{\pi^2}{6}$.

Q: The options provided in the question don't match my answer. Did I make a mistake?

It is common in competitive exams to occasionally encounter questions with errors in the options. Trust your derivation. If you have correctly identified the intervals, simplified the integrand to $1/n$, performed the partial fraction decomposition, and correctly evaluated the telescoping series, your result of $\frac{\pi^2}{6} - 1$ is mathematically rigorous and correct.

Select Answer:

Visualized Solution

Understanding the Integrand

We need to evaluate $I = \int_{0}^{1} \frac{1}{[ \frac{1}{x} ]} d x$ .
The function involves the Greatest Integer Function (GIF), denoted by $[\cdot]$ .
The GIF changes its value whenever its argument, $\frac{1}{x}$ , is an integer.

Finding the Critical Points

Let's find where $\frac{1}{x} = n$ for integers $n = 1, 2, 3, \dots$
This occurs at $x = 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots$
These points divide our integration domain $(0, 1]$ into infinite sub-intervals: $(\frac{1}{2}, 1], (\frac{1}{3}, \frac{1}{2}], \dots$

Analyzing a General Interval

Consider a general sub-interval $x \in (\frac{1}{n + 1}, \frac{1}{n}]$ .
In this interval, taking reciprocals gives $n \leq \frac{1}{x} < n + 1$ .
Therefore, the GIF evaluates to $[\frac{1}{x}] = n$ .
The integrand simplifies to a constant: $\frac{1}{[ \frac{1}{x} ]} = \frac{1}{n}$ .

Setting Up the Infinite Series

The integral is the sum of areas of rectangles over these intervals.
$I = \sum_{n = 1}^{\infty} \int_{\frac{1}{n + 1}}^{\frac{1}{n}} \frac{1}{n} d x$
For $n = 1$ , area is $A_{1} = \int_{1/2}^{1} 1 d x$ .
For $n = 2$ , area is $A_{2} = \int_{1/3}^{1/2} \frac{1}{2} d x$ .

Evaluating the General Term

Let's evaluate the integral for the $n$ -th term:
$T_{n} = \frac{1}{n} [x]_{\frac{1}{n + 1}}^{\frac{1}{n}}$
$T_{n} = \frac{1}{n} (\frac{1}{n} - \frac{1}{n + 1})$
$T_{n} = \frac{1}{n} (\frac{n + 1 - n}{n ( n + 1 )}) = \frac{1}{n ^{2} ( n + 1 )}$

Partial Fraction Decomposition

We need to sum $T_{n} = \frac{1}{n ^{2} ( n + 1 )}$ from $n = 1$ to $\infty$ .
We can decompose this using partial fractions:
$\frac{1}{n ^{2} ( n + 1 )} = \frac{A}{n} + \frac{B}{n ^{2}} + \frac{C}{n + 1}$
$1 = A n (n + 1) + B (n + 1) + C n^{2}$

Solving for Constants

Substitute $n = 0 ⟹ 1 = B (1) ⟹ B = 1$
Substitute $n = - 1 ⟹ 1 = C (- 1)^{2} ⟹ C = 1$
Compare coefficients of $n^{2}$ : $0 = A + C ⟹ A = - 1$
So, $T_{n} = \frac{1}{n ^{2}} - \frac{1}{n} + \frac{1}{n + 1}$

Summing the Infinite Series

Now, sum the terms: $I = \sum_{n = 1}^{\infty} (\frac{1}{n ^{2}} - (\frac{1}{n} - \frac{1}{n + 1}))$
Split into two sums: $I = \sum_{n = 1}^{\infty} \frac{1}{n ^{2}} - \sum_{n = 1}^{\infty} (\frac{1}{n} - \frac{1}{n + 1})$
The first sum is the famous Basel problem: $\sum_{n = 1}^{\infty} \frac{1}{n ^{2}} = \frac{π ^{2}}{6}$

Evaluating the Telescoping Series

The second sum is a telescoping series:
$S_{2} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots$
All intermediate terms cancel out!
$S_{2} = 1 - lim_{n \to \infty} \frac{1}{n + 1} = 1 - 0 = 1$

Final Answer

Combining the two results:
$I = \frac{π ^{2}}{6} - 1$
Note: The options provided in the original problem text appear to be incorrect. The mathematically rigorous answer is $\frac{π ^{2}}{6} - 1$ .

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

The function

f (x) = \frac{1}{[ 1/ x ]}

involves the greatest integer function, which acts as a step function. To evaluate the integral of this function, we must decompose the domain into manageable intervals where the denominator remains constant.

The value of

[1/ x]

changes whenever

1/ x

is an integer. Specifically, for

x

in the interval

(\frac{1}{n + 1}, \frac{1}{n}]

, the value of

1/ x

lies in the range

[n, n + 1)

. Consequently, the greatest integer function is defined as:

[1/ x] = n

Slicing the Domain

By partitioning the interval

(0, 1]

into sub-intervals

(\frac{1}{n + 1}, \frac{1}{n}]

for

n = 1, 2, 3, \dots

, we transform the integral into an infinite sum of areas of rectangles. The total integral

I

is expressed as:

I = n = 1 \sum \infty \int_{\frac{1}{n + 1}}^{\frac{1}{n}} \frac{1}{n} d x

The height of each rectangle is

\frac{1}{n}

, and the width is the difference between the limits:

\frac{1}{n} - \frac{1}{n + 1}

. Thus, the area of the

n

-th rectangle is:

T_{n} = \frac{1}{n} (\frac{1}{n} - \frac{1}{n + 1}) = \frac{1}{n ^{2} ( n + 1 )}

The Algebraic Alchemy

To evaluate the infinite series

\sum_{n = 1}^{\infty} \frac{1}{n ^{2} ( n + 1 )}

, we apply partial fraction decomposition. We set:

\frac{1}{n ^{2} ( n + 1 )} = \frac{A}{n} + \frac{B}{n ^{2}} + \frac{C}{n + 1}

Solving for the constants, we find

B = 1

C = 1

, and

A = - 1

. This allows us to rewrite the general term as:

T_{n} = \frac{1}{n ^{2}} - (\frac{1}{n} - \frac{1}{n + 1})

The Grand Finale

The total integral is now the difference between two distinct series:

I = n = 1 \sum \infty \frac{1}{n ^{2}} - n = 1 \sum \infty (\frac{1}{n} - \frac{1}{n + 1})

The first part is the famous Basel Problem, which converges to

\frac{π ^{2}}{6}

. The second part is a telescoping series that simplifies as follows:

(1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + \dots = 1

Subtracting these results, we arrive at the final value of the integral:

$I = \frac{π ^{2}}{6} - 1$

Common Traps & FAQs

Why do we split the integral at

1/ n

? Is there a simpler way?

The greatest integer function

[1/ x]

is inherently discontinuous. It jumps at every integer value of

1/ x

. Since

x

ranges from

0

1

1/ x

ranges from

1

\infty

. The function

[1/ x]

remains constant between these integer values, which is why we must partition the domain into intervals like

(1/2, 1], (1/3, 1/2]

, and so on. There is no shortcut; this partitioning is the fundamental definition of integrating a step function.

Wait, why does the sum

\sum_{n = 1}^{\infty} \frac{1}{n ^{2}}

equal

\frac{π ^{2}}{6}

That is the famous Basel Problem, first solved by Leonhard Euler. While it is a standard result in advanced calculus, for JEE purposes, it is a powerful tool to keep in your arsenal. Whenever you encounter the sum of the reciprocals of squares, you can confidently use

\frac{π ^{2}}{6}

The options provided in the question don't match my answer. Did I make a mistake?

It is common in competitive exams to occasionally encounter questions with errors in the options. Trust your derivation. If you have correctly identified the intervals, simplified the integrand to

1/ n

, performed the partial fraction decomposition, and correctly evaluated the telescoping series, your result of

\frac{π ^{2}}{6} - 1

is mathematically rigorous and correct.

Animated Solution for Mathematics - Definite Integration: The integral $\int_{0}^{1} \frac{1}{[ \frac{1}{x} ]} d x$ , where $[.]$ denotes the greatest integer function is equal to

Select Answer:

Visualized Solution

Understanding the Integrand

Finding the Critical Points

Analyzing a General Interval

Setting Up the Infinite Series

Evaluating the General Term

Partial Fraction Decomposition

Solving for Constants

Summing the Infinite Series

Evaluating the Telescoping Series

Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Slicing the Domain

The Algebraic Alchemy

The Grand Finale

Similar Questions

Consider the integral $I = \int_{0}^{10} \frac{[ x ] e ^{[x]}}{e ^{x - 1}} d x$ where $[x]$ denotes the greatest integer less than or equal to $x$ . Then the value of $I$ is equal to:

The value of $\int_{- π /2}^{π /2} (\frac{1}{[ x ] + 4}) d x$ , where $[\cdot]$ denotes the greatest integer function, is

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫01​[x1​]1​dx, where [.] denotes the greatest integer function is equal to

Select Answer:

Visualized Solution

Understanding the Integrand

Finding the Critical Points

Analyzing a General Interval

Setting Up the Infinite Series

Evaluating the General Term

Partial Fraction Decomposition

Solving for Constants

Summing the Infinite Series

Evaluating the Telescoping Series

Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Slicing the Domain

The Algebraic Alchemy

The Grand Finale

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Animated Solution for Mathematics - Definite Integration: The integral $\int_{0}^{1} \frac{1}{[ \frac{1}{x} ]} d x$ , where $[.]$ denotes the greatest integer function is equal to

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If $[t]$ denotes the greatest integer $\leq t$ , then the value of $\frac{3 ( e - 1 )}{e} \int_{1}^{2} x^{2} e^{[x] + [x^{3}]} d x$ is :

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