Mathematics

If  $$\mathrm { I } _ { 1 } = \int _ { 1 } ^ { 2 } x [ \sqrt { x } + \sqrt { 3 - x } ] d x$$  and  $$\mathrm { I } _ { 2 } = \int _ { 1 } ^ { 2 } ( \sqrt { x } + \sqrt { 3 - x } ) d x$$  then  $$\dfrac { I _ { 1 } } { I _ { 2 } } =$$


ANSWER

$$3/2$$


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Single Correct Medium Published on 17th 09, 2020
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