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Step-by-Step Solution
Step 1: Understand the Given Condition on the Derivative
We are given that for all $x$ in $[0,1]$,
$$p'(x) = p'(1-x).$$
This tells us that the derivative of $p$ at $x$ is the same as its derivative at $1 - x$.
Step 2: Relate $p(x)$ and $p(1-x)$
If $p'(x) = p'(1-x)$ for all $x$, then consider the function
$$g(x) = p(x) + p(1 - x).$$
Taking the derivative of $g(x)$ with respect to $x$ gives
$$g'(x) = p'(x) - p'(1 - x) \quad (\text{because } \frac{d}{dx}\,p(1-x) = -p'(1-x)).$$
Since $p'(x) = p'(1-x)$, it follows that
$$g'(x) = p'(x) - p'(x) = 0.$$
Hence, $g(x) = p(x) + p(1-x)$ is a constant function on $[0,1].$
Step 3: Determine the Constant
To find the constant, use the values given at the endpoints:
$$p(0) = 1 \quad \text{and} \quad p(1) = 41.$$
Then
$$p(0) + p(1 - 0) = p(0) + p(1) = 1 + 41 = 42.$$
Therefore,
$$p(x) + p(1 - x) = 42 \quad \text{for all } x \in [0,1].$$
Step 4: Evaluate the Definite Integral $I = \int_{0}^{1} p(x)\,dx$
Let
$$I = \int_{0}^{1} p(x)\,dx.$$
By the change of variable $u = 1 - x$, observe that when $x$ goes from $0$ to $1,$ $u$ also goes from $1$ to $0.$ Reversing the limits changes the sign, so we carefully write
$$\int_{0}^{1} p(1-x)\,dx = \int_{1}^{0} p(u)\,(-du) = \int_{0}^{1} p(u)\,du = I.$$
Thus,
$$\int_{0}^{1} p(1 - x)\,dx = I.$$
Now, because $p(x) + p(1-x) = 42,$ adding the two integrals gives
\[
\int_{0}^{1} p(x)\,dx \;+\; \int_{0}^{1} p(1-x)\,dx
= \int_{0}^{1} \bigl[p(x) + p(1-x)\bigr]\,dx
= \int_{0}^{1} 42 \,dx = 42.
\]
Hence,
\[
I + I = 42 \quad \Longrightarrow \quad 2I = 42 \quad \Longrightarrow \quad I = 21.
\]
Step 5: Conclude the Final Answer
Therefore, the value of the definite integral is
$$\int_{0}^{1} p(x)\,dx = 21.$$
