Problem 8 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(T>0\) be a fixed real number. Suppose \(f\) is a continuous function such that for all \(x \in R, f(x+T)=f(x)\). If \(I=\int_{0}^{T} f(x) d x\) then the value of \(\int_{3}^{3+3 T} f(2 x) d x\) is (A) \(\frac{3}{2} P\) (B) \(2 I\) (C) \(3 I\) (D) \(6 I\)

Step-by-Step Solution

Answer

The value is \(3I\), which corresponds to option (C).

Step 1: Substitution for Simplification

Consider the integral \( \int_{3}^{3+3T} f(2x) dx \). Let \( u = 2x \), which gives \( du = 2dx \) or \( dx = \frac{1}{2}du \). The limits of integration also change: when \( x = 3 \), \( u = 6 \), and when \( x = 3+3T \), \( u = 6 + 6T \).

Step 2: Adjust Integral for Substitution

Using the substitution from Step 1, the original integral becomes \[ \int_{6}^{6+6T} f(u) \frac{1}{2} du = \frac{1}{2} \int_{6}^{6+6T} f(u) du. \]

Step 3: Use Periodicity of Function

Since \( f(x+T) = f(x) \) for all \( x \), \( f \) is periodic with a period of \( T \). Thus, the interval from \( 6 \) to \( 6 + 6T \) can be broken into 6 periods each of length \( T \).

Step 4: Evaluate the Integral Using Periodicity

The integral \( \int_{6}^{6+6T} f(u) du \) can be evaluated as 6 times the integral over one period \( T \), since the periodic function repeats exactly. Therefore, \( \int_{6}^{6+6T} f(u) du = 6I \).

Step 5: Multiply by Substitution Factor

Finally, multiply the result from Step 4 by \( \frac{1}{2} \) from the substitution. This gives \[ \frac{1}{2} \times 6I = 3I. \]

Key Concepts

Definite Integrals

A definite integral is a brilliant mathematical tool used to determine the area under a curve between two specified points on the x-axis.
Think of this as a way of summing up infinitely many small quantities to get a whole. This is expressed as \( \int_{a}^{b} f(x) dx \), where \( f(x) \) is the function and \( a \) and \( b \) are the limits of integration.
Here's why it's helpful:

• Calculates the exact area under a curve
• Utilized in finding volumes, and more complex scenarios like work done by a force
• Particularly useful with periodic functions to simplify repetitive calculations

In our exercise, we're dealing with the definite integral \( \int_{3}^{3+3T} f(2x) dx \), where the periodic nature of the function \( f(x) \) simplifies the calculation significantly.
Understanding the definite integral helps in grasping how the overall repetition within periodic functions integrates into a whole.

Substitution Method

The substitution method is a technique used to simplify integrals.
It's like changing the variable to transform a complex integral into an easier one. Imagine it as changing a lens to get a clearer view.
Here's how substitution works:

• Choose a new variable, often \( u \), to replace part of the integrand
• Find the differential of this substitution to replace \( dx \)
• Translate the integration limits to match the new variable

In our problem, we set \( u = 2x \), which simplifies the original tough integral \( \int_{3}^{3+3T} f(2x) \).
After calculating, substituting \( dx = \frac{1}{2} du \) makes the integral more straightforward, allowing us to focus on just \( f(u) \) instead.
This process not only simplifies the integration but also aligns with our limits of integration based on the function's attribute of periodicity.

Continuous Functions

Continuous functions are essential in calculus because they have no breaks, gaps, or jumps.
They're called 'continuous' as they flow smoothly without any interruptions across their domain. This is represented mathematically such that for any point \( x \) in its domain, a small change in \( x \) results in a small change in \( f(x) \).
Key characteristics include:

• No abrupt changes or 'jumps' in the function's graph
• All definite integrals defined and manageable due to the absence of gaps
• Important for applying theorems like the Intermediate Value Theorem

In our problem, the function \( f(x) \) is continuous and periodic, which makes integrating it over a defined interval very structured and predictable.
This continuity, coupled with the periodicity condition \( f(x+T) = f(x) \), means true consistency across every periodic segment.
Thus, every segment contributes equally when calculating the necessary integrals, making them simpler and straightforward to determine.