Question
Let $$T>0$$ be a fixed real number. Suppose $$f$$ is a continuous
function such that for all $$x \in R,\,f\left( {x + T} \right) = f\left( x \right).$$
If $$I = \int\limits_0^T {f\left( x \right)dx} $$ then the value of $$\int\limits_3^{3 + 3T} {f\left( {2x} \right)dx} $$ is-
A.
$$\frac{3}{{2I}}$$
B.
$$2I$$
C.
$$3I$$
D.
$$6I$$
Answer :
$$3I$$
Solution :
Given that $$T>0$$ is a fixed real number. $$f$$ is continuous $$\forall \,x \in R$$ such that $$f\left( {x + T} \right) = f\left( x \right)$$
$$ \Rightarrow f$$ is a periodic function of period $$T$$
Also given $$I = \int_0^T {f\left( x \right)dx} $$
Then, let $${I_1} = \int_3^{3 + 3T} {f\left( {2x} \right)dx} $$
$$\eqalign{
& {\text{Put }}2x = z\,\, \Rightarrow dx = \frac{{dz}}{2} \cr
& {\text{Also as }}x \to 3,\,z \to 6;\, \cr
& {\text{As }}x \to 3 + 3T,\,z \to 6 + 6T \cr
& {I_1} = \frac{1}{2}\int_6^{6 + 6T} {f\left( z \right)dz} \cr
& = \frac{1}{2}\left[ {\int_6^T {f\left( z \right)dz} + \sum\limits_{n = 1}^5 {\int_{nT}^{\left( {n + 1} \right)T} {f\left( z \right)dz} } + \int_{6T}^{6T + 6} {f\left( z \right)dz} } \right] \cr
& {\text{Now, }}\int_{nT}^{\left( {n + 1} \right)T} {f\left( z \right)dz} = \int_0^T {f\left( {nT + u} \right)du,} \cr
& {\text{where }}z = nT + u \cr
& = \int_0^T {f\left( u \right)du = 1\,\,\,\,\,\,\,\,\,\,\left[ {\because f\left( {nT + u} \right) = f\left( u \right)} \right]} \cr} $$
Similarly, we can show that
$$\eqalign{
& \int_{6T}^{6T + 6} {f\left( z \right)dz} = \int_0^6 {f\left( z \right)dz} \cr
& \therefore {I_1} = \frac{1}{2}\left[ {\int\limits_6^T {f\left( z \right)dz} + 5I + \int\limits_0^6 {f\left( z \right)dz} } \right] \cr
& = \frac{1}{2}\left[ {\int\limits_6^T {f\left( z \right)dz} + 5I} \right] \cr
& = \frac{1}{2}\left( {6I} \right) \cr
& = 3I \cr} $$