If I=int_1^3 sqrt{3+x+x^2} dx,then I lies in | vedclass

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(A) Let $f(x) = \sqrt{3+x+x^2}$. Since $f(x)$ is an increasing function on $[1, 3]$,we have $f(1) \le f(x) \le f(3)$ for all $x \in [1, 3]$.$f(1) = \s

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$(2 \sqrt{5}, 2 \sqrt{15})$

Vedclass · Answer

(A) Let $f(x) = \sqrt{3+x+x^2}$. Since $f(x)$ is an increasing function on $[1, 3]$,we have $f(1) \le f(x) \le f(3)$ for all $x \in [1, 3]$.$f(1) = \sqrt{3+1+1} = \sqrt{5}$.$f(3) = \sqrt{3+3+9} = \sqrt{15}$.By the property of definite integrals,$\int_a^b f(x) dx \le (b-a) 	imes \max(f(x))$ and $\int_a^b f(x) dx \ge (b-a) 	imes \min(f(x))$.Here,$a=1, b=3$,so $b-a = 2$.Thus,$2 	imes \sqrt{5} \le I \le 2 	imes \sqrt{15}$.Therefore,$I \in [2 \sqrt{5}, 2 \sqrt{15}]$.Comparing this with the given options,the correct interval is $(2 \sqrt{5}, 2 \sqrt{15})$.

If $I=\int_1^3 \sqrt{3+x+x^2} dx$,then $I$ lies in the interval

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