Sigma Percentile
JEE Main 2023 (08 April Shift 1)
LEVELJEE Advanced

Animated Solution for Mathematics - Definite Integration: Let denote the greatest integer . Then is equal to

Enter Numerical Value:

Visualized Solution

Splitting the Integral

  • Let
  • Using linearity of integrals, we can split this into two parts.
  • Let and

Analyzing

  • Consider the function on the interval .
  • At the boundaries, and .
  • The minimum value is at , where .
  • Thus, for all .

Greatest Integer of

  • Since strictly inside the interval, its greatest integer value is constant.
  • for .
  • The isolated points where it equals do not affect the definite integral.

Evaluating

Analyzing

  • Now consider on the interval .
  • is a strictly decreasing function in this domain.

Splitting Points for

  • The greatest integer function jumps at integer values.
  • We must find where equals .

Intervals for

  • On ,
  • On ,
  • On ,
  • On ,

Setting up

Evaluating

Final Substitution

  • Recall our original split:
  • Substitute and

Final Answer

  • Final Answer:

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

We are tasked with evaluating the integral:
By the linearity of the integral, we can decompose this into two distinct parts:
Let us define these as and respectively.

Phase 1

The Cosecant Mystery
Consider the function on the interval . At the boundaries, and .
However, between these points, the function dips, reaching its minimum value of at . Thus, for all in our open interval, .
Because the function never reaches inside this interval, the greatest integer is simply . The integral becomes:

Phase 2

The Cotangent Odyssey
Now, we turn to . The cotangent function is strictly decreasing on this interval.
At , . At , .
As it descends, it crosses the integers . We split the integral at the points where equals these integers: 1. 2. 3.
This divides our interval into four sub-intervals where the greatest integer is constant: - - - -

Phase 3

The Arithmetic Finale
We set up as the sum of these integrals:
Calculating these values:
Finally, we substitute and back into our original expression:
The final result is 14.

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