Let f : R → R be a differentiable function such that f(0) = 0, f \left(\right. \frac{\pi}{2} \left.\right) = 3 and f '(0) = 1. If g(x) = \int_{x}^{\frac{\pi}{2}} \left[\right. f ' \left(\right. t \left.\right) \textrm{ } cosec \textrm{ } t - cot \textrm{ } t \textrm{ } cosec \textrm{ } t \textrm{ } f \left(\right. t \left.\right) \left]\right. \textrm{ } dt \textrm{ }\textrm{ }\textrm{ } for \textrm{ }\textrm{ } x \textrm{ }\textrm{ } \in \textrm{ }\textrm{ } \left(\right. 0 , \textrm{ } \frac{\pi}{2} \left]\right. \textrm{ }\textrm{ } then \textrm{ }\textrm{ } \underset{x \rightarrow 0}{Lim} \textrm{ } g \left(\right. x \left.\right) =

Engineering

Mathematics

Leibnitz Rule Derivative of Anti Derivatives

Question

Let f : R → R be a differentiable function such that f(0) = 0, $f (\frac{π}{2})$ = 3 and f '(0) = 1.

If g(x) = $\int_{x}^{\frac{π}{2}} [f' (t) cosec t - \cot t cosec t f (t)] dt for x \in (0, \frac{π}{2}] then \underset{x \to 0}{Lim} g (x) =$

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f (0) = 0, $f (\frac{π}{2})$ = 3 and f '(0) = 1

g(x) = $\int_{x}^{\frac{π}{2}} [\underset{II}{\underset{⏟}{f' (t)}} \underset{I}{\underset{⏟}{cosec t}} - \cot t cosec t f (t)] dt$

g(x) = ${(cosec t f (t))}_{x}^{\frac{π}{2}} + \int_{x}^{\frac{π}{2}} cosec t cost f (t) dt - \int_{x}^{\frac{π}{2}} \cot t cosec t f (t) dt$

= $f (\frac{π}{2})$ · 1 – f(x) cosec x = 3 – $\frac{f (x)}{sinx}$

$∴ \underset{x \to 0}{Lim} g (x) = 3 - \underset{x \to 0}{Lim} \frac{f (x)}{x}$ = 3 – f '(0) = 2