f′(x)=f′(2−x)
f(x)=−f(2−x)+c
put x=0
f(0)=−f(2)+c
c=f(0)+f(2)=1+e2
so, f(x)+f(2−x)=1+e2
I=∫02f(x)dx
I=∫02f(2−x)dx
2I=∫02(f(x)+f(2−x))dx
2I=(1+e2)∫02dx
I=1+e2
Let f(x) be a differentiable function defined on [0,2] such that f′(x)=f′(2−x) for all x∈(0,2),f(0)=1 and f(2)=e2. Then the value of ∫02f(x)dx is
Held on 24 Feb 2021 · Verified 6 Jul 2026.
2(1+e2)
1+e2
1−e2
2(1−e2)
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