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Let p(x) be a function on the real line satisfying p'(x)=p'(1-x) for every x in the interval [0,1], with p(0)=1 and p(1)=41. Then the value of ∫₀¹ p(x) dx is
- 21
- 41
- 42
- √41
Correct answer: 21
Solution
The condition p'(x) = p'(1-x) implies that the derivative of p is symmetric about x = 0.5, which suggests that p(x) can be expressed in a form that reflects this symmetry. Given the boundary conditions p(0) = 1 and p(1) = 41, we can deduce that the average value of p over the interval [0, 1] is 21, leading to the conclusion that the integral of p from 0 to 1 equals 21.
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