Question

A continuous real function $f$ satisfies $f\left(2x\right)=3f\left(x\right)\forall x\in R$. If $\underset{0}{\stackrel{1}{\int }}f\left(x\right)dx=1$, then the value of definite integral $\underset{1}{\stackrel{2}{\int }}f\left(x\right)dx$ is

Answer 1

(5) We have $f\left(2x\right)=3f\left(x\right)$ and ${\int }_0^{1}f\left(x\right)dx=1$

From equations (1) and $\left(2\right),\frac{1}{3}{\int }_0^{1}f\left(2x\right)dx=1$

Put $2x=t,\frac{1}{6}{\int }_0^{2}f\left(t\right)dt=1$

$\Rightarrow {\int }_0^{2}f\left(t\right)dt=6$

$\Rightarrow {\int }_0^{1}f\left(t\right)dt+{\int }_1^{2}f\left(t\right)dt=6$

Hence, ${\int }_1^{2}f\left(t\right)dt=6-{\int }_0^{1}f\left(t\right)dt=6-1=5$