For [x ( 0,5 2 ),] define [f( x ) = _0^x t t,dt. ] Then f has

Question

For $$x \in \left( {0,{{5\pi } \over 2}} \right),$$ define $$f\left( x \right) = \int\limits_0^x {\sqrt t \sin t\,dt.} $$  Then $f$ has

Accepted Answer

$$f'\left( x \right) = \sqrt x \sin x$$

At local maxima or minima, $$f'\left( x \right) = 0$$

$$ \Rightarrow x = 0$$ or $sin$ $x=0$

$$ \Rightarrow x = 2\pi ,\,\,\pi  \in \left( {0,{{5\pi } \over 2}} \right)$$

$$f''\left( x \right) = \sqrt x \cos \,x + {1 \over {2\sqrt x }}\sin \,x$$

$$ = {1 \over {2\sqrt x }}\left( {2x\,\cos \,x + \sin \,x} \right)$$

At $$x = \pi ,$$  $$f''\left( x \right) < 0$$

Hence, local maxima at $$x = \pi $$

At $$x = 2\pi ,\,\,\,f''\left( x \right) > 0$$

Hence local minima at $$x = 2\pi $$

Solutions · 1