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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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$I_1 = \int \limits _1^2 |x| dx ; I_2=\int \limits_1^2 \large\frac{1}{\sqrt {1+x^2}}$$ dx $

\[\begin {array} {1 1} (a)\;I_1 > I_2 \\ (b)\;I_1=I_2 \\ (c)\;I_1 < I_2 \\ (d)\;None \end {array}\]

Can you answer this question?
 
 

1 Answer

0 votes
By using property
$f(x) > g(x); \int \limits_a^b f(x)dx > \int \limits _a^b g(x) dx $
=> $|x| < \sqrt {1+(x)^2}$
=> $ \sqrt{1+x^2} > \sqrt {x^2}$
$\large\frac{1}{\sqrt {1+x^2}} < \frac{1}{\sqrt {x^2}}$
$\int \limits _a ^b \large\frac{1}{\sqrt {1+x^2}}$$ dx < \int \limits _a^b \large\frac{1}{\sqrt {x^2}}$
=> $I_2 < I_1$
Hence a is the correct answer.
answered Dec 16, 2013 by meena.p
 

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