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Answer

3

Explanation

Since, `F'(a) +2` is the area bounded by x = 0, y = 0, y =f(x) and x= a.
`therefore int_(0)^(a) f(x)dx=F'(a)+2`
Using Newton-Leibnitz formula,
`f(a)=F''(a)`
`and f(0)=f''(0) " …(i)" `
Given, `F(x) =int_(x)^(x^(2)+(pi)/(6)) 2cos^(2)t dt`
On differentiating,
`F'(x)=2cos^(2)(x^(2)+(pi)/(6))*2x-2cos^(2)x*1`
Again differentiating,
`F''(x)=4{cos^(2)(x^(2)+(pi)/(6))*2xcos(x^(2)+(pi)/(6))sin(x^(2)+(pi)/(6))2x}+{4cos x*sinx}`
`=4{cos^(2)(x^(2)+(pi)/(6))-4x^(2)cos(x^(2)+(pi)/(6))sin(x^(2)+(pi)/(6))}+2sin 2x`
`therefore F''(0)=4{cos^(2)((pi)/(6))}=3`
`therefore f(0)=3`