Animated Solution for Mathematics - Definite Integration: 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 $I={\int }_0^{T}f(x)dx$ then the value of ${\int }_3^{3+3T}f(2x)dx$ is

Imagine you are standing on the shore, watching a rhythmic, repeating wave crash against the sand. This is the physical essence of a periodic function. In mathematics, we define this as $f(x+T)=f(x)$, where $T$ is the period.

It means that no matter where you are on the $x$-axis, if you jump forward by $T$, you land on the exact same value. Today, we are going to solve a problem that tests your ability to manipulate these waves using the power of calculus.

We are given that the area under one such wave, from $0$ to $T$, is:

Our mission is to evaluate a new, slightly more complex integral:

When you see $f(2x)$ inside an integral, your first instinct should be to simplify the argument. The $2x$ is a transformation—a horizontal compression of the function.

To handle this, we use the substitution method. Let $z=2x$. This immediately tells us that $dz=2dx$, or more usefully:

This is our first step in taming the expression. We are essentially changing our coordinate system to make the function look like the original $f(z)$ we know and love.

Here is where many students stumble. When you change the variable from $x$ to $z$, you absolutely must change the limits of integration.

The original integral runs from $x=3$ to $x=3+3T$. Using our substitution $z=2x$:

For the lower limit, when $x=3$, $z=2(3)=6$. For the upper limit, when $x=3+3T$, $z=2(3+3T)=6+6T$.

Now, our integral $J$ looks like this:

By pulling the constant $\frac{1}{2}$ outside, we get:

This is much cleaner, but we still have that integral from $6$ to $6+6T$.

Now, we invoke the most powerful tool in our arsenal: the property of periodic functions. The property states that for any periodic function with period $T$, the integral over any interval of length $nT$ is equal to $n$ times the integral over one period.

Mathematically:

Look at our integral: ${\int }_6^{6+6T}f(z)dz$. Here, our starting point $a$ is $6$, and the length of the interval is $6T$. This means $n=6$.

The property tells us that the starting point $6$ is irrelevant! The integral is simply $6$ times the integral over one period:

Since we know ${\int }_0^{T}f(z)dz=I$, our integral becomes $6I$.

We are almost at the finish line. We have $J=\frac{1}{2}(6I)$.

A quick division gives us the final result:

It is elegant, it is precise, and it is a perfect example of how understanding the geometric nature of a function can turn a daunting integral into a simple algebraic exercise. You have successfully navigated the substitution, avoided the limit trap, and utilized the periodic property.