Mathematics

# Let $I = \int\limits_1^3 {\sqrt {{x^4} + {x^2}} dx,}$then I=

$2\sqrt 2 < 1 < 6\sqrt {10}$

##### SOLUTION
$\sum { = \int\limits_1^3 {\sqrt {{x^4} + {x^2}} dx} }$
$= \int\limits_1^3 {x\sqrt {{x^2} + 1} dx}$
$= \frac{1}{2}\int\limits_{\sqrt 2 }^{\sqrt {10} } {{7^2}dx}$
$= \frac{1}{2}{\left( {\frac{{{7^3}}}{3}} \right)_{\sqrt 2 }}^{\sqrt {10} }$
$= \frac{1}{6}\left( {10\sqrt {10} - 2\sqrt 2 } \right)$
$= \frac{1}{3}\left( {5\sqrt {10} - \sqrt 2 } \right)$
$= 4.8$
$\therefore 2\sqrt 2 < I < 6\sqrt {10}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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