Given f(x)=x∫0xf(t)dt+x3 note that ∫0xf(t)dt=xf(x)−x2, also f(0)=0 then we will have the following: f′(x)=∫0xf(t)dt+x(f(x)−f(0))+3x2=xf(x)−x2+xf(x)−xf(0)+3x2=xf(x)+2x2+xf(x)−xf(0)=xf(x)+2x2+xf(x)−x(0)=xf(x)+2x2+xf(x)⟹f′(a)=af(a)+2a2+af(a)=a1+2a2+a(1)=2a2+a+a1
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