Sigma Percentile
JEE Main 2022 (28 June Shift 1)
LEVELJEE Advanced

Animated Solution for Mathematics - Definite Integration: Let denote the greatest integer less than or equal to . Then, the value of the integral is equal to

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Visualized Solution

Understanding the Problem

  • Let
  • The integral is , where is the greatest integer function.
  • Goal: Identify intervals where is constant.

Analyzing the Function

  • Vertex:
  • Max value:
  • Endpoints: and
  • Range of for is

Finding Roots:

  • Set
  • Roots: and
  • For , , so

Finding Points:

  • Set
  • Points: and
  • For and ,

Finding Points:

  • Set
  • For ,

Finding Points:

  • The remaining interval is up to
  • At ,
  • For ,

Splitting the Integral

Calculating the First Three Parts

  • Sum of first three

Calculating the Fourth Part

Calculating the Fifth Part

Final Summation and Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

Imagine you are standing before a complex, shifting landscape. You have a quadratic function, , and you are asked to find the area under its 'staircase' version, the greatest integer function .
When we see , we shouldn't panic. Instead, we should see a staircase where is the smooth, continuous path and the greatest integer function is the rigid, stepped structure that follows it.
Our goal is to calculate the area under this staircase from to .

Mapping the Terrain

First, let us understand the behavior of our quadratic function . This is an inverted parabola.
To understand its journey, we find its vertex using:
At this peak, the value is . At the boundaries, and . This tells us that as moves from to , the function climbs from to and then descends to .

Finding the Jump Points

The greatest integer function only changes its value when crosses an integer. We need to find exactly where this happens.
We set , which gives us , yielding roots at and . In this interval, is between and , so .
Next, we look for . Solving gives and .
Finally, for , we solve , or . Using the quadratic formula, we find:
We take the positive root for the right side of the parabola.

The Sum of Rectangles

Now, the integral transforms into a sum of simple rectangular areas:
Each integral is just the height of the step multiplied by the width of the interval. The first three parts are straightforward: , , and . Summing these gives .
For the fourth part:
For the final part:

Final Synthesis

Adding these together:
Converting to , we get:
This is the beauty of calculus. We took a complex, discontinuous function and, by understanding its geometric soul, broke it down into simple, manageable pieces. You have successfully navigated the staircase!

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