Discussion Forum

 Let  $F(X)=\int_{c}^{x^{2}+\frac{\pi}{6}} 2\cos^{2}tdt$ for all xε R and  $f:\left[0,\frac{1}{2}\right]\rightarrow (0,\infty)$ be a continuous function, For $a \epsilon \left[ 0,\frac{1}{2}\right]$ , 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

Answer:

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, $f(x)=\int_{x}^{x^{2}+\frac{\pi}{6}} 2\cos^{2}t dt$

 On differentiating,

                  $F'(x)=2\cos^{2}\left(x^{2}+\frac{\pi}{6}\right).2x-2\cos^{2}x.1$

 Again differentiating,

            $F"(x)=4\left\{\cos ^{2}\left(x^{2}+\frac{\pi}{6}\right)-2xcos\left(x^{2}+\frac{\pi}{6}\right)\sin\left(x^{2}+\frac{\pi}{6}\right)2x\right\}+\left\{4\cos x.\sin x\right\}$

            $=4\left\{\cos ^{2}\left(x^{2}+\frac{\pi}{6}\right)-4x^{2}\cos \left( x^{2}+\frac{\pi}{6}\right) \sin\left(x^{2}+\frac{\pi}{6}\right)\right\}+2\sin 2x$

          $\therefore F"(0)=4\left\{\cos^{2}\left(\frac{\pi}{6}\right)\right\}=3$

                    $\therefore f(0)=3$