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Let f (x) =integral upper limit is x lower limit is 0 (2u+1/u^2-2u+2)du then find the least value of f (x) in [-1,1].

Narasimha Reddy

12 Years agoGrade Upto college level

1 Answer

Sher Mohammad

12 Years ago

2u+-2+2+1=2(u-1)+3

u^2-2u+2=(u-1)^2+1

2u+1/u^2-2u+2

$\int_0^x \left(\frac{2 (u-1)}{(u-1)^2+1}+\frac{3}{(u-1)^2+1}\right) \, du=\log \left(x^2-2 x+2\right)-3 \tan ^{-1}(1-x)+\frac{3 \pi }{4}-\log (2)$

f'(x)=0, imply x=-1/2

$\frac{3 \pi }{4}-\log (2)+\log \left(\frac{13}{4}\right)-3 \tan ^{-1}\left(\frac{3}{2}\right)$

Sher Mohammad

B.Tech, IIT Delhi

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