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[Post New]15 Dec 2007 13:47:47 IST
    Subject: Re:least value
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iitkgp_bipin (6461)

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Olaaa!! Perrrfect answer. 1101  bad job dude!! I dont approve of this answer! 1  [1581 rates]
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f(x)= 5pi/4x (3sint+cost)dt

f '(x) = 3sinx + cosx

The interval [5pi/4,4pi/3] lies in 3rd quadrant where both sinx and cosx are less than zero.

so, f '(x) < 0 (f is decreasing in the given interval).

This means as x varies from 5pi/4 to 4pi/3, value of f(x) decreases and it attains minimum value at x=4pi/3.

so minimum f(x) = 5pi/44pi/3 (3sint+cost)dt

rest of integration is simple.
Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur

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