Let F(x)=x2+π6∫x2cos2tdt for all x∈R and f:[0,12]→[0,∞) be a continuous function. For a∈[0,12], if F′(a)+2 is the area of the region bounded by x=0,y=0,y=f(x) and x=a, then f(0) is
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Solution
F(x)=x2+π6∫x2cos2tdtUsing Newton-Leibnitz's theorem,F′(x)=2cos2(x2+π6)×2x−2cos2x×1=4xcos2(x2+π6)−2cos2xGiven,a∫0f(x)dx=F′(a)+2⇒a∫0f(x)dx=4acos2(a2+π6)−2cos2a+2
Differentiating w.r.t. 'a', we get f(a)=4a×2cos(a2+π6)[−sin(a2+π6)]×2a +4cos2(a2+π6)−4cosa×(−sina)⇒f(0)=0+4cos2(π6)−0⇒f(0)=4×(√32)2=3