1. We are given $p'(x)=p'(1-x), \, \forall x\in[0,1],$ with $p(0)=1,\qquad p(1)=41.$

2. Define a new function $q(x)=p(x)+p(1-x).$ Differentiate it: $q'(x)=p'(x)-p'(1-x).$ Using the given condition $p'(x)=p'(1-x)$, we get $q'(x)=0.$ Hence $q(x)$ is constant on $[0,1]$.

3. So, $p(x)+p(1-x)=\text{constant}.$ Find the constant by putting $x=0$: $p(0)+p(1)=1+41=42.$ Therefore, $p(x)+p(1-x)=42 \quad \forall x\in[0,1].$

4. Now let $I=\int_0^1 p(x)\,dx.$ Using the substitution $x\mapsto 1-x$, $\int_0^1 p(1-x)\,dx=\int_0^1 p(x)\,dx=I.$

5. Add the two integrals: $2I=\int_0^1 \bigl(p(x)+p(1-x)\bigr)\,dx.$ Since $p(x)+p(1-x)=42$, $2I=\int_0^1 42\,dx=42.$ Thus, $I=21.$

6. Therefore, $\int_0^1 p(x)\,dx=21.$

7. Checking options:

   • A: $21$ ✅
   • B: $41$ ❌
   • C: $42$ ❌
   • D: $\sqrt{41}$ ❌

So the correct option is A.