Sigma Percentile
JEE Main 2002
LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: is

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Visualized Solution

Visualizing the Function

  • Let the given integral be .
  • The function is a periodic function.
  • Observe the graph: it consists of repeating "humps" above the x-axis.

Identifying the Period

  • The standard period of is .
  • Due to the modulus, the negative half-cycle is reflected.
  • Therefore, the pattern repeats every units.
  • The period of is .

Extending to

  • The upper limit of our integral is .
  • This means there are exactly identical humps from to .
  • Calculating the total area is equivalent to finding the area of one hump and multiplying by .

The Periodicity Property of Definite Integrals

  • We use the definite integral property for periodic functions:
  • Here, is the period and is an integer.

Applying the Property

  • In our problem, the function is .
  • The period is .
  • The upper limit is , so .
  • Substituting these into the property:

Simplifying the Modulus

  • We are now integrating only over the interval .
  • In the first and second quadrants (), the sine function is positive or zero.
  • Therefore, .
  • The integral becomes: .

Integrating

  • The anti-derivative of is .
  • Applying this to our integral:
  • We can pull the negative sign outside to avoid confusion:

Substituting the Limits

  • Now, substitute the upper limit () and the lower limit ().
  • Recall the trigonometric values:

Final Arithmetic

  • Substitute the values into the expression:
  • Simplify the terms inside the bracket:

Final Conclusion

  • The value of the integral is .
  • Key Takeaway: Always check for periodicity when dealing with integrals of modulus trigonometric functions.
  • The area of one hump of is exactly square units.

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

Imagine you are standing on the -axis, looking at the graph of . Usually, the sine function is a wave that dances above and below the axis, a perfect balance of positive and negative.
But the modulus operator—those two vertical bars—acts like a mirror. It takes every negative 'valley' and flips it upwards, transforming the sine wave into a continuous, rhythmic series of identical humps.
This is the geometric reality of our problem: we are calculating the total area under these humps from to .

The Power of Periodicity

To solve this, we must first identify the 'heartbeat' of the function. A standard sine wave repeats every .
But because we have reflected the negative parts, the pattern repeats much faster. Look closely at the graph: the function returns to its starting shape after every interval. Therefore, the fundamental period of is .
In the world of JEE Advanced, recognizing periodicity is often the key to unlocking a problem. We use the property:
Here, our upper limit is , and our period is . This means we have exactly identical humps. Instead of wrestling with a complex integral over a large range, we can simply find the area of one single hump and multiply it by .

The Elegant Calculation

Now, let us focus on just one hump, the integral from to . In this interval, is non-negative, so the modulus is redundant: .
Our integral simplifies to:
The anti-derivative of is . Applying the Fundamental Theorem of Calculus:
To avoid any sign errors, let us pull the negative sign outside:
Substituting the limits, we get:
Knowing that and , the expression becomes:
The negative signs cancel out, leaving us with a clean, positive result of 20.

Final Reflections

We started with a daunting integral, but by visualizing the geometry and applying the property of periodicity, we reduced it to a simple arithmetic task.
The area of a single hump of is exactly square units. With such humps, the total area is 20.
Always remember: when you see modulus signs in a trigonometric integral, stop and look for the symmetry. It is often the shortest path to the answer.

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