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Question

Let F(x)=x2+π6x2cos2t dt for all xR 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  


Solution

F(x)=x2+π6x2cos2t dtUsing Newton-Leibnitz's theorem,F(x)=2cos2(x2+π6)×2x2cos2x×1         =4xcos2(x2+π6)2cos2xGiven, a0f(x)dx=F(a)+2a0f(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)0f(0)=4×(32)2=3

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