dxdF(x)=xesinx or ∫14x3esinx3dx=∫14x33x2esinx3dx Let x3=t,3x2dx=dt when x=1,t=1&x−4,t=64 F(t)=∫164tesintdt=∫164 F(t)dt=F(64)−F(1)K=64
Let dxdF(x)=(xesinx )x>0. If ∫14x3esin 3dx=F(k)−F(1) then one of the possible values of k, is
Held on 30 Apr 2003 · Verified 6 Jul 2026.
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