can we conclude that $f$ is a constant? Suppose $f(x) \in C[a,b]$ and $\varphi(x)$ is Riemann integrable and satisfy : $\int_a^b\varphi(x)\mathscr{dx}=0$.   $\int_a^bf(x)\varphi(x)\mathscr{dx}=0$       can we conclude that $f(x)\equiv c$?

My approach:

for $f$ is continuous on $[a,b]$, then there exists $M$ and $m$ in $\Bbb{R}$,satisfy: $m \leq f(x) \leq M$,so we have $M-f(x)\geq0$,so I can use the mean value theorem:
$$\int_a^b(M-f(x))\varphi(x)\mathscr{dx}=\mu\int_a^b(M-f(x))\mathscr{dx}=0 \Rightarrow \int_a^b f(x)\mathscr{dx} =M(b-a)$$
where $\inf\varphi(x)\leq\mu \leq \sup \varphi(x)$
use MVT to : $$\int_a^b(f(x)-m)\varphi(x)\mathscr{dx}$$we get $\int_a^bf(x)\mathscr{dx}=m(b-a)$ this implies $m=M$,then $f$ is constant on $[a,b]$
Am I right?


No. For example, you could have $\varphi=0$, in which case you will have
$$\int_a^b f(x)\varphi(x)\,dx=0$$
for any $f\in C[a,b]$ whatsoever.
In fact, any function $\varphi$ which is $0$ on some open interval in $[a,b]$ will admit non-constant functions $f\in C[a,b]$ such that $\int_a^b f(x)\varphi(x)\,dx=0$.
A slightly less trivial example: for any $t>0$, we could choose $\varphi(x)=x$ on $[-t,t]$, which satisfies $\int_{-t}^t\varphi(x)\,dx=0$, and any even function $f\in C[-t,t]$ will have
$$\begin{align}\int_{-t}^t f(x)\varphi(x)\,dx&=\int_{-t}^0xf(x)\,dx+\int_0^t xf(x)\,dx\\\\
&=-\int_0^{-t}xf(x)\,dx+\int_0^txf(x)\,dx\\\\
&=-\int_0^{t}xf(x)\,dx+\int_0^txf(x)\,dx\\\\
&=0.\end{align}$$
