Analyzing the Setup
Imagine you are standing on the x-axis, looking at the graph of f(x)=∣sinx∣. 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 x=0 to x=10π.
The Power of Periodicity
To solve this, we must first identify the 'heartbeat' of the function. A standard sine wave repeats every 2π.
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 π interval. Therefore, the fundamental period T of ∣sinx∣ is π.
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 10π, and our period T is π. This means we have exactly n=10 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 10.
The Elegant Calculation
Now, let us focus on just one hump, the integral from 0 to π. In this interval, sinx is non-negative, so the modulus is redundant: ∣sinx∣=sinx.
Our integral simplifies to:
The anti-derivative of sinx is −cosx. 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 cosπ=−1 and cos0=1, 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 ∣sinx∣ is exactly 2 square units. With 10 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.