Let ∫extf(t)dt=sinx−xcosx−2x2 By using Leibnitz rule, we get dxd[∫extf(t)dt]=dxd[sinx−xcosx−2x2]⇒xf(x)−ef(e)⋅0=xsinx−x Now, put x=6π, we get 6π⋅f(6π)=6π⋅sin6π−6π⇒f(6π)=sin6π−1=21−1=−21
If ∫extf(t)dt=sinx−xcosx−2x2, for all x∈R−{0}, then the value of f(6π) is
Held on 7 May 2012 · Verified 6 Jul 2026.
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