p′(x)=p′(1−x) ⇒p(x)=−p(1−x)+c at x=0 p(0)=−p(1)+c⇒42=c now p(x)=−p(1−x)+42 ⇒p(x)+p(1−x)=42 I=∫01p(x)dx=∫01p(1−x)dx 2I=∫01(42)dx⇒I=21
Let p(x) be a function defined on R such that p^prime(x)=p^prime(1-x), for all x ∈[0,1], p(0)=1 and p(1)=41. Then ∫_0^1 p(x) d x equals
JEE Main 2010 — Mathematics Calculus
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 ∫01p(x)dx equals
Held on 30 Apr 2010 · Verified 6 Jul 2026.
Options
- A
21
- B
41
- C
42
- D
41
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