Let \text{F}\textrm{ } \left(\right. \text{x} \left.\right) \textrm{ } = \int_{x}^{x^{2} + \frac{\pi}{6}} \textrm{ }\textrm{ } 2 \left(cos\right)^{2} t \textrm{ }\textrm{ } dt for all x \in R and \textrm{ } \text{f}\textrm{ } : \textrm{ } \left[\right. 0 , \textrm{ }\textrm{ } \frac{1}{2} \left]\right. \rightarrow \left[\right. 0 , \infty \left.\right) be a continuous function. For \text{a} \textrm{ } \in \left[\right. 0 , \textrm{ }\textrm{ } \frac{1}{2} \left]\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

Engineering

Mathematics

Introduction to Area

Question

Let $F (x) = \int_{x}^{x^{2} + \frac{π}{6}} 2 \cos^{2} t dt$ for all x $\in$ R and $f : [0, \frac{1}{2}] \to [0, \infty)$ be a continuous function. For $a \in [0, \frac{1}{2}]$ , 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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$F' (x) = 2 \cos^{2} (x^{2} + \frac{π}{6}) \cdot 2 x - 2 \cos^{2} x$

$F " (x) = \cos^{2} (x^{2} + \frac{π}{6}) + (- 8 xcos (x^{2} + \frac{π}{6}) \sin (x^{2} + \frac{π}{6}) \cdot 2 x) - 2 \cos^{2} x$

$F " (0) = 4 \cdot (\frac{\sqrt{3}}{2}) = 3^{2}$

$\int_{0}^{α} f (x) dx = F' (α) + 2$

f(α) = F"(α)

f (0) = F"(0) = 3