Sigma Percentile
JEE Main 2012
LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: If , then equals

Select Answer:

* Multiple Correct

Visualized Solution

Understanding

  • Given function:
  • This represents the net area under the curve from to .

Integration Formula

  • Standard integral:

Applying Limits

  • Integrate:
  • Apply limits:

Evaluating Upper Limit

  • Substitute upper limit :

Evaluating Lower Limit

  • Substitute lower limit :
  • Result:

Finding

  • We need to find the value of .
  • Substitute in our simplified .

Substitution

Expanding the Argument

  • Expand the bracket:

Periodicity of Sine

  • Recall trigonometric property:
  • The sine function repeats its values every radians.

Applying Periodicity

  • Here, is an even multiple of (i.e., ).
  • Therefore, .

Evaluating

  • To check the given options, let's find .
  • Substitute into .

Calculating

  • Since , we get .

Visualizing

  • Visually, the integral of over the interval covers exactly two full periods.
  • The positive and negative areas cancel out perfectly, yielding a net area of .

Checking Option 2

  • We know and .
  • Option 2:
  • Substitute :
  • This matches . Option 2 is correct.

Checking Option 3

  • Option 3:
  • Substitute :
  • This also matches . Option 3 is correct.
  • Final Answer: Both Option 2 and Option 3 are correct.

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

We are tasked with analyzing the function defined by the integral:
Think of this as an area function. The graph of is a wave oscillating with a frequency four times higher than the standard cosine wave. As increases, we accumulate the signed area under this wave.

The Integration

A Step of Precision
To find the explicit form of , we apply the fundamental theorem of calculus. Using the rule , where , we perform the integration:
Applying the limits from to , we obtain:
Since , the function simplifies to the elegant expression:

The Shift

Exploring
Next, we investigate the behavior of by substituting the shifted value into our derived function:
Recall that the sine function is periodic with a period of . Because represents exactly two full cycles, the sine function remains unchanged:
Therefore, we conclude that:

The Final Revelation

To finalize our understanding, we evaluate by substituting into our formula:
This result reveals a profound geometric truth: the integral of over the interval covers exactly two full periods of the wave. Consequently, the positive and negative areas cancel out perfectly.
Given that , we observe that:
Both expressions are equivalent to , demonstrating the inherent symmetry and periodic nature of the function.

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