Question

Let \(T\gt 0\) be a fixed number. If \(f:\mathbb{R}\to\mathbb{R}\) is a continuous function such that \[ f(x+T)=f(x),\qquad x\in \mathbb{R}. \] If \[ I=\int_0^T f(x)\,dx, \] then \[ \int_0^{5T} f(2x)\,dx= \]

• A. \(10I\)
• B. \(\frac{5}{2}I\)
• C. \(5I\)
• D. \(2I\)

Hint:

If \(f(x)\) is periodic with period \(T\), then \[ \int_0^{nT}f(x)\,dx=n\int_0^T f(x)\,dx. \] For \(f(kx)\), first use substitution and then apply periodicity.

Answer 1

The Correct Option is C

Step 1: Use substitution.
Consider \[ \int_0^{5T}f(2x)\,dx \] Let \[ u=2x \] Then, \[ du=2dx \] So, \[ dx=\frac{du}{2} \] When \[ x=0, \] we get \[ u=0 \] When \[ x=5T, \] we get \[ u=10T \] Therefore, \[ \int_0^{5T}f(2x)\,dx = \frac{1}{2}\int_0^{10T}f(u)\,du \]

Step 2: Use periodicity of \(f\).
Given, \[ f(x+T)=f(x) \] So, \(f\) is periodic with period \(T\).
Hence, \[ \int_0^{10T}f(u)\,du = 10\int_0^T f(u)\,du \] But, \[ I=\int_0^T f(x)\,dx \] Therefore, \[ \int_0^{10T}f(u)\,du=10I \]

Step 3: Substitute back.
Thus, \[ \int_0^{5T}f(2x)\,dx = \frac{1}{2}\cdot 10I \] \[ =5I \]

Step 4: Final conclusion.
Therefore, \[ \boxed{5I} \]