Grade upto college level Algebra

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

kumar

12 Years agoGrade upto college level

2 Answers

SADDAM NAFEES

12 Years ago

wrong category

Thanks & Regards

SADDAM NAFEES,

askIITians Faculty

Sher Mohammad

12 Years ago

$(u-2)^{2}-u^{2}-1=4u$

$\frac{1}2{}\frac{4u+2}{u^2-2u+2}=\frac{1}2\frac{(u-2)^{2}-u^{2}-1+2}{(u-2)^{2}-2}=\frac{1}2\frac{(u-2)^{2}-u^{2}+1}{(u-2)^{2}-2}$

$\frac{1}2{}\frac{4u+2}{u^2-2u+2}=\frac{1}2\frac{(u-2)^{2}-2-u^{2}+3}{(u-2)^{2}-2}=\frac{1}{2}+\frac{3-u^{2}}{(u-2)^{2}-2}$

$\frac{1}2{}\frac{4u+2}{u^2-2u+2}=\frac{1}{2}+\frac{3}{(u-2)^{2}-2}-\frac{u^{2}}{(u-2)^{2}-2}$

$\int \frac{3}{(u-2)^{2}-2}=\frac{3 \left(\log \left(-u+\sqrt{2}+2\right)-\log \left(u+\sqrt{2}-2\right)\right)}{2 \sqrt{2}}$

$\int -\frac{u^{2}}{(u-2)^{2}-2}=-u-\left(2+\frac{3}{\sqrt{2}}\right) \log \left(-u+\sqrt{2}+2\right)-\left(2-\frac{3}{\sqrt{2}}\right) \log \left(u+\sqrt{2}-2\right)$

$\int \frac{1}{2}=\frac{1}{2}u$

Sher Mohammad,

B.Tech, IIT Delhi

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

kumar

2 Answers

SADDAM NAFEES

Sher Mohammad

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