Solution
If the function is constant, that implies that the derivative is 0
[tex]\begin{gathered} \Rightarrow\frac{dF(x)}{dx}=\frac{d}{dx}\int_x^{4x}\frac{1}{t}dt \\ \\ \text{ using fundamental theorem of calculus} \\ \\ \Rightarrow\frac{d}{dx}(\int_0^{4x}\frac{1}{t}dt-\int_0^x\frac{1}{t}dt) \end{gathered}[/tex][tex]\begin{gathered} \text{ using chain rule and fundamental theorem of calculus} \\ \\ \Rightarrow\frac{d}{dx}\int_0^{4x}\frac{1}{t}dt=\frac{d}{du}\int_0^u\frac{1}{t}dt\cdot\frac{d(4x)}{dx}=4\cdot\frac{1}{u}=4\cdot\frac{1}{4x}=\frac{1}{x} \end{gathered}[/tex]
Also,
[tex]\frac{d}{dx}\int_0^x\frac{1}{t}dt=\frac{1}{x}[/tex]
Then,
[tex]\operatorname{\Rightarrow}\frac{d}{dx}\int_x^{4x}\frac{1}{t}dt=\frac{1}{x}-\frac{1}{x}=0[/tex]
Since the derivative is 0, It is obvious that F(x) is a constant function.
