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

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Aicharya · Accepted Answer

$\mathrm{I}=\int_1^3 \sqrt{3+\mathrm{x}+\mathrm{x}^2} \mathrm{dx}=\int_1^3 \sqrt{\left(\mathrm{x}+\frac{1}{2}ight)^2+\left(\frac{\sqrt{11}}{2}ight)^2} \mathrm{dx}$ $ \begin{gathered} =\left[\frac{\left(x+\frac{1}{2}ight)}{2} \sqrt{x^2+x+3}+\frac{11}{4 	imes 2} \log \left(x+\frac{1}{2}ight)+\sqrt{x^2+x+3} \midight]_1^3 \ =\left[\frac{7}{4} \sqrt{15}+\frac{11}{8} \log \left|\frac{7}{2}+\sqrt{15}ight|-\frac{3}{4} \sqrt{5}-\frac{11}{8} \log \left|\frac{3}{2}+\sqrt{5}ight|ight] \ =\frac{1}{4}[7 \sqrt{15}-3 \sqrt{5}]+\frac{11}{8}\left[\log \left(\frac{7+2 \sqrt{15}}{3+2 \sqrt{5}}ight)ight] \ =\frac{\sqrt{5}}{4}[7 \sqrt{3}-3]+\frac{11}{8} \log \left[\frac{7+2 \sqrt{15}}{3+2 \sqrt{5}}ight] \ =5.100066+0.9344 \Rightarrow I-6.0344 \in(2 \sqrt{5}, 2 \sqrt{15}) \end{gathered} $

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

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