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BITSAT 2011

Exam: BITSAT
Year: 2011
Subject: Mathematics
Chapter: Definite Integration

BITSAT 2011 · Mathematics · Definite Integration

Let ( I _ 1 =∫_1^2 1 1+x^2 dx) and (I _ 2 =∫_1^2 (1)/(x) d x), then

Let (

\mathrm{I}}_{1}=\int_1^2 \frac{1}{\sqrt{1+\mathrm{x}^2}} \mathrm{dx}) and (\mathrm{I}

}_{2}=\int_1^2

\frac{1}{x}

d x), then

Options

A $\mathrm{I}_{1} >$ $\mathrm{I}_{2}$ $)$
B $\mathrm{I}_{2} >$ $\mathrm{I}_{1}$ $)$
C( $\mathrm{I}}_{1}= \mathrm{I}$ }_{2})
DNone of these

Correct answer

B. $\mathrm{I}_{2} >$ $\mathrm{I}_{1}$ $)$

Step-by-step solution

If (I_1=\int_1^2

\frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}^2}},

I_2=\int_1^2

\frac{\mathrm{dx}}{\mathrm{x}}

) (I_1=\ell n \left(\frac{2+\sqrt{5}}{1+\sqrt{2}}\right), \quadI_2=\ell n 2 \Rightarrow I_1 < I_2)

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