The value of `a(a >0)` for which the area bounded by the curves `y=x/6+1/(x^2),y=0,x=a ,a n dx=2a` has the least value is__

Answer

1

Explanation

`f(a)=overset(2a)underset(a)int((x)/(6)+(1)/(x^(2)))dx=((x^(2))/(12)-(1)/(x))_(a)^(2a)`
`=((4a^(2))/(12)-(1)/(2a)-(a^(2))/(12)+(1)/(a))=(a^(2))/(4)+(1)/(2a)`
`"Let "f'(a)=(2a)/(4)-(1)/(2a^(2))=0`
`rArr" "a=1` which is point of minima.