Question

Let $f$ is a continuous function for all $x\in R$ and $f\left(x\right)=f(x+T),T>0.$ If $I=\underset{0}{\overset{T}{\int }}f\left(x\right)dx,$ then $\underset{2}{\overset{2+10T}{\int }}f\left(\frac{x}{2}\right)dx$ is equal to

• A. $5I$
• B. $10I$
• C. $20I$
• D. $40I$

Answer 1

The correct option is B $10I$

As, $f\left(x\right)=f(x+T),f\left(x\right)$ has period $T.$
Now, $I_1=\underset{2}{\overset{2+10T}{\int }}f\left(\frac{x}{2}\right)dx$
Put $\frac{x}{2}=y,dx=2dy$
$\Rightarrow I_1=2\underset{1}{\overset{1+5T}{\int }}f\left(y\right)dy=2\times 5\underset{0}{\overset{T}{\int }}f\left(y\right)dy\because \underset{a}{\overset{a+nT}{\int }}f\left(y\right)dy=n\underset{0}{\overset{T}{\int }}f\left(y\right)dy$
$\therefore I_1=10I$