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(A) Given $p'(x) = p'(1 - x)$.Integrating both sides with respect to $x$,we get $p(x) = -p(1 - x) + C$.At $x = 0$,$p(0) = -p(1) + C$.Substituting the

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$21$

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(A) Given $p'(x) = p'(1 - x)$.Integrating both sides with respect to $x$,we get $p(x) = -p(1 - x) + C$.At $x = 0$,$p(0) = -p(1) + C$.Substituting the given values $p(0) = 1$ and $p(1) = 41$,we have $1 = -41 + C$,which implies $C = 42$.Thus,$p(x) + p(1 - x) = 42$.Let $I = \int_{0}^{1} p(x) dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we have $I = \int_{0}^{1} p(1 - x) dx$.Adding the two expressions for $I$,we get $2I = \int_{0}^{1} (p(x) + p(1 - x)) dx$.Substituting $p(x) + p(1 - x) = 42$,we get $2I = \int_{0}^{1} 42 dx = 42[x]_{0}^{1} = 42$.Therefore,$I = 21$.

Let $p(x)$ be a function defined on $R$ such that $p'(x) = p'(1 - x)$ for all $x \in [0, 1]$,$p(0) = 1$,and $p(1) = 41$. Then $\int_{0}^{1} p(x) dx = $

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