1. Key Concepts and Formulas

• Integration of Derivatives: If $F'(x) = f(x)$, then $\int f(x) dx = F(x) + C$. Specifically, if $p'(x)$ is given, integrating it yields $p(x) + C$.
• Substitution Rule for Integration: For an integral of the form $\int g(u(x)) u'(x) dx$, the substitution $u = u(x)$ simplifies it to $\int g(u) du$. For $\int p'(ax+b) dx$, let $u = ax+b$, so $du = a dx$, which means $dx = \frac{1}{a} du$. Thus, $\int p'(ax+b) dx = \frac{1}{a} \int p'(u) du = \frac{1}{a} p(u) + C = \frac{1}{a} p(ax+b) + C$.
• Property of Definite Integrals: For any continuous function $f(x)$ and constants $a$ and $b$: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ For the interval $[0, 1]$, this property becomes: $\int_0^1 f(x) dx = \int_0^1 f(1-x) dx$ This property is particularly useful when the integrand has symmetry or when a functional relationship involving $f(x)$ and $f(a+b-x)$ can be established.

2. Step-by-Step Solution

We are given the differential equation $p'(x) = p'(1-x)$ for $x \in [0,1]$, with initial conditions $p(0) = 1$ and $p(1) = 41$. We need to find the value of $\int_0^1 p(x) dx$.

Step 1: Derive a functional relation for $p(x)$ We are given $p'(x) = p'(1-x)$. To find a relation for $p(x)$, we integrate both sides with respect to $x$. Integrating the left side: $\int p'(x) dx = p(x) + C_1$ Integrating the right side requires a substitution. Let $u = 1-x$. Then $du = -dx$, so $dx = -du$. $\int p'(1-x) dx = \int p'(u) (-du) = -\int p'(u) du = -p(u) + C_2 = -p(1-x) + C_2$ Equating the two integrated forms: $p(x) + C_1 = -p(1-x) + C_2$ Rearranging the constants, we get a functional relation: $p(x) + p(1-x) = C_2 - C_1$ Let $C = C_2 - C_1$. Then the functional relation is: $p(x) + p(1-x) = C$

Step 2: Determine the constant of integration $C$ We use the given boundary conditions $p(0) = 1$ and $p(1) = 41$ to find the value of $C$. Substitute $x=0$ into the functional relation $p(x) + p(1-x) = C$: $p(0) + p(1-0) = C$ $p(0) + p(1) = C$ Now, substitute the given values $p(0)=1$ and $p(1)=41$: $1 + 41 = C$ $C = 42$ So, the specific functional relation for $p(x)$ is: $p(x) + p(1-x) = 42$

Step 3: Evaluate the definite integral using the property of definite integrals Let the integral we want to evaluate be $I$. $I = \int_0^1 p(x) dx$ We apply the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. For the interval $[0,1]$, this means: $I = \int_0^1 p(1-x) dx$ Now, we add the original integral expression for $I$ and this new expression: $I + I = \int_0^1 p(x) dx + \int_0^1 p(1-x) dx$ $2I = \int_0^1 [p(x) + p(1-x)] dx$ From Step 2, we know that $p(x) + p(1-x) = 42$. Substitute this into the integral: $2I = \int_0^1 42 dx$ Now, we evaluate this simple definite integral: $2I = \left[ 42x \right]_0^1$ $2I = 42(1) - 42(0)$ $2I = 42$ Solving for $I$: $I = \frac{42}{2}$ $I = 21$

3. Common Mistakes & Tips

• Incorrectly integrating $p'(1-x)$: A common mistake is to forget the negative sign that arises from the substitution $u=1-x$. The integral of $p'(1-x)$ is $-p(1-x)$, not $p(1-x)$.
• Forgetting the constant of integration: When integrating $p'(x)$ and $p'(1-x)$, the constants of integration are crucial for establishing the correct functional relationship between $p(x)$ and $p(1-x)$. These constants are resolved using the given boundary conditions.
• Not recognizing the symmetry property: The integral property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ is a powerful tool for integrals over symmetric intervals. Recognizing when to apply it, especially in conjunction with a functional relation like $f(x) + f(a+b-x) = K$, can significantly simplify the problem.

4. Summary

The problem required us to evaluate a definite integral of a function $p(x)$ for which a property of its derivative was given. By integrating the derivative relation $p'(x) = p'(1-x)$, we derived a functional equation $p(x) + p(1-x) = C$. Using the boundary conditions $p(0)=1$ and $p(1)=41$, we found the constant $C$ to be $42$, leading to the specific relation $p(x) + p(1-x) = 42$. Finally, we employed the property of definite integrals $\int_0^1 p(x) dx = \int_0^1 p(1-x) dx$. By adding the integral to itself and substituting the functional relation, we simplified the problem to integrating a constant, yielding the result $\int_0^1 p(x) dx = 21$.

The final answer is $\boxed{21}$ which corresponds to option (A).