Question

Area under the curve `y=sqrt(4x+1)` between x = 0 and x = 2 is ______.

पर्याय

• `56/3` sq. units

• `64/3` sq. units

• 8 sq. units

• `13/3` sq. units

Answer 1

Area under the curve `y=sqrt(4x+1)` between x = 0 and x = 2 is `underline(13/3 "sq. units")`.

Explanation:

Required area `=int_0^2sqrt(4x+1)  dx`

`=[(4x+1)^(3/2)/(4(3/2))]_0^2`

`=26/6`

`=13/3 "sq. units"`