Let p(x) be a function defined on R such that p'(x) = p'(1 - x) , for all x [0, 1] , p(0) = 1 and p(1) = 41 . Then _0^1 p(x) dx equals: - Mathematics MCQ AIEEE 2026 | ExamTest.live

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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) d x equals:

Solution & Step-by-step Explanation

Given p^{'} (x) = p^{'} (1 - x) .Integrating both sides: p (x) = - p (1 - x) + C .At x = 0 : p (0) = - p (1) + C ⟹ 1 = - 41 + C ⟹ C = 42 .So p (x) + p (1 - x) = 42 .Let I = \int_{0}^{1} p (x) d x .Using property \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x : I = \int_{0}^{1} p (1 - x) d x . 2 I = \int_{0}^{1} [p (x) + p (1 - x)] d x = \int_{0}^{1} 42 d x = 42 . I = 21 .

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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) d x equals:

21

41

42

20

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