Mathematics

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


ANSWER

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


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
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