Let {I_1} = int_1^2 frac{dx}{sqrt{1 + x^2}} a | vedclass

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(B) We are given ${I_1} = \int_1^2 \frac{dx}{\sqrt{1 + x^2}}$ and ${I_2} = \int_1^2 \frac{dx}{x}$.For all $x \in (1, 2)$,we know that $1 + x^2 > x^2$.

Vedclass · Accepted Answer

${I_2} > {I_1}$

Vedclass · Answer

(B) We are given ${I_1} = \int_1^2 \frac{dx}{\sqrt{1 + x^2}}$ and ${I_2} = \int_1^2 \frac{dx}{x}$.For all $x \in (1, 2)$,we know that $1 + x^2 > x^2$.Taking the square root on both sides,we get $\sqrt{1 + x^2} > x$.Since both sides are positive for $x \in (1, 2)$,we can take the reciprocal,which reverses the inequality:$\frac{1}{\sqrt{1 + x^2}} Integrating both sides with respect to $x$ from $1$ to $2$:$\int_1^2 \frac{dx}{\sqrt{1 + x^2}} Therefore,${I_1}  {I_1}$.

Let ${I_1} = \int_1^2 \frac{dx}{\sqrt{1 + x^2}}$ and ${I_2} = \int_1^2 \frac{dx}{x}$,then:

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