Let Px be a function defined on R such that P'x=P'1-x, ∀x∈0, 1, P0=1, P1=41 then ∫01P(x)dx=

P1(x)=P1(1-x) ∫P1(x)dx=∫P1(1-x)dx P(x)=P(1-x)(-1)+C P(x)+P(1-x)=C put x=0 ⇒ P(0)+P(1)=C 1+41=C ∴ P(x)+P(1-x)=42 I=∫01P(x) dx I=∫01P(1-x)dx I+I=∫01 P(x)+P(1-x) dx=∫01 42dx =42(1-0) I=21

Let Px be a function defined on R such that P'x=P'1-x, ∀x∈0, 1, P0=1, P1=41 then ∫01P(x)dx=

P1(x)=P1(1-x) ∫P1(x)dx=∫P1(1-x)dx P(x)=P(1-x)(-1)+C P(x)+P(1-x)=C put x=0 ⇒ P(0)+P(1)=C 1+41=C ∴ P(x)+P(1-x)=42 I=∫01P(x) dx I=∫01P(1-x)dx I+I=∫01 P(x)+P(1-x) dx=∫01 42dx =42(1-0) I=21

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Let $P (x)$ be a function defined on R such that $P' (x) = P' (1 - x), \forall x \in [0, 1], P (0) = 1, P (1) = 41$ then $\int_{0}^{1} P (x) d x =$

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$\sqrt{41}$

answer is B.

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Detailed Solution

$P^{1} (x) = P^{1} (1 - x) \int P^{1} (x) dx = \int P^{1} (1 - x) dx P (x) = \frac{P (1 - x)}{(- 1)} + C P (x) + P (1 - x) = C put x = 0 \Rightarrow P (0) + P (1) = C 1 + 41 = C ∴ P (x) + P (1 - x) = 42 I = \int_{0}^{1} P (x) dx I = \int_{0}^{1} P (1 - x) dx I + I = \int_{0}^{1} (P (x) + P (1 - x)) d x = \int_{0}^{1} 42 dx = 42 (1 - 0) I = 21$