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

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) \(\frac{3}{2} 1\) (B) \(2 I\) (C) \(3 I\) (D) \(6 I\)

Step-by-Step Solution

Answer

\(3I\)

Step 1: Understanding the periodicity of the function

The given function \(f\) satisfies the condition \(f(x+T) = f(x)\) for all \(x \in \mathbb{R}\). This means that \(f\) is a periodic function with period \(T\).

Step 2: Change the variable for integration

We need to evaluate the integral \(\int_{3}^{3+3T} f(2x) \, dx\). Let's change the variable of integration. Set \(u = 2x\), then \(du = 2 \, dx\) or \(dx = \frac{1}{2} \, du\). When \(x = 3\), \(u = 6\). When \(x = 3+3T\), \(u = 6 + 6T\).

Step 3: Substitute and transform the integral

The integral \(\int_{3}^{3+3T} f(2x) \, dx\) becomes \(\frac{1}{2} \int_{6}^{6+6T} f(u) \, du\) after substituting \(u = 2x\).

Step 4: Exploit function periodicity for new bounds

Using the periodicity \(f(u+T) = f(u)\), the integral \(\int_{6}^{6+6T} f(u) \, du\) can be broken down into smaller periodic intervals. This is equivalent to \(6\) full periods of \(f\), from 0 to \(T\).

Step 5: Calculate using the period value

Because each period from \(0\) to \(T\) contributes equally, \(\int_{6}^{6+6T} f(u) \, du = 6 \times I\), where \(I = \int_{0}^{T} f(x) \, dx\). Thus, \(\frac{1}{2} \int_{6}^{6+6T} f(u) \, du = \frac{1}{2} \times 6I = 3I\).

Key Concepts

Periodic Function

A periodic function is one that repeats its values in regular intervals or periods. For a function to be periodic, there must exist a non-zero constant, referred to as the period, such that repeating this interval will yield the same function values. In our exercise, the function \( f \) is defined to have a period \( T \), as given by the property \( f(x+T) = f(x) \).
This means that the entire behavior of the function \( f \) from \( x \) to \( x+T \) is identical in all subsequent intervals of length \( T \).
The periodicity allows us to transform larger intervals with the same repetitive subintervals, essentially converting a complex integral into a simpler one with repeated base integrals.

• This property is crucial when calculating definite integrals over multiples of the period, since the value of the integral over one period can be used to determine the integral over any multiple of the period.
• Periodic functions are often encountered in trigonometry, where sine and cosine functions are classic examples, repeating every \( 2\pi \).

Change of Variables

The change of variables is a mathematical technique used to simplify the integration process. It involves substituting a new variable in place of the original one to make the integral more manageable.
In the given problem, we have the original integral \( \int_{3}^{3+3T} f(2x) \, dx \). By setting \( u = 2x \), we transform the variable \( x \) into \( u \).
This method allows us to modify the limits of integration to suit the new variable:

• Initially, when \( x = 3 \), \( u \) becomes \( 6 \).
• When \( x = 3 + 3T \), \( u \) becomes \( 6 + 6T \).

Thus, the differential \( dx \) changes to \( \frac{1}{2} \, du \).
This is particularly useful when the new integral expression adheres better to the periodic properties of \( f \). Changing variables can simplify computation and reveal properties of the function that were less apparent in the original form.

Integration by Substitution

Integration by substitution is a method closely tied to the change of variables. It's often used to transform complex integrals into simpler forms that are easier to tackle, much like how substitution in algebra simplifies equations.
In our problem, after substituting \( u = 2x \), the derivative \( dx \) transforms to \( \frac{1}{2} \, du \), and our bounds become \( 6 \) to \( 6+6T \).
This approach allows us to leverage the periodic nature of \( f \), as it aids in breaking down the integral into smaller, more manageable parts:

• The integral \( \int_{6}^{6+6T} f(u) \, du \) is simplified using the substitution, and accounts for the periodic nature of \( f \).
• Each full period contribution is repeated, allowing us to express the entire integral as \( 6 \times I \), greatly simplifying computation.

The trick with substitution is finding the right function to substitute so that the transformed integral becomes elementary or maintains the properties needed, like periodicity in this case. This method is powerful and widely applicable in integration problems to simplify the process or make evaluation possible.