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Question:

If for a continuous function f(x), ∫t-πt(f(x)+x)dx=π2-t2, for all t≥-π, then f(-π/3) is equal to :

π

π2

π3

π6

Solution:

Given ∫t-πt(f(x)+x)dx=π2-t2
Differentiate with respect to t
f(t)+t=0
⇒ f(t)=-t
Put t=-π/3
f(-π/3)=-(- π/3) = π/3