$\text{ A continuous real function f satisfies }f\left(2x\right)=3f\left(x\right)\mathrm{\forall }x\in R\text{. If }{\int }_0^1 f\left(x\right)dx=1,\text{ then the value of definite integral }{\int }_1^2 f\left(x\right)dx\text{ is }$

$\begin{array}{l}\text{ We have }f\left(2x\right)=3f\left(x\right)\\ \text{ and }{\int }_0^1 f\left(x\right)dx=1\\ \text{ From the above equations }\\ \frac{1}{3}{\int }_0^1 f\left(2x\right)dx=1\\ \text{ Put }2x=t,\frac{1}{6}{\int }_0^2 f\left(t\right)dt=1\end{array}$

$\begin{array}{l}\Rightarrow {\int }_0^2 f\left(t\right)dt=6\\ \Rightarrow {\int }_0^1 f\left(t\right)dt+{\int }_1^2 f\left(t\right)dt=6\\ \text{ Hence, }{\int }_1^2 f\left(t\right)dt=6-{\int }_0^1 f\left(t\right)dt=6-1=5\end{array}$