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If `f(x)={{:(sqrt({x}),"for",xcancelinZ),(1,"for",x in Z):} and g(x)={x}^(2)` where {.} denotes fractional part of x then area bounded by f(x) and g(x) for `x in 0,6 ` is

A

`(2)/(3)`

B

2

C

`(10)/(3)`

D

6

Text Solution

Verified by Experts

The correct Answer is:
B
f(x) and g(x) are periodic with period 1.
`therefore"Required area is "int_(0)^(6)[f(x)-g(x)]dx=6int_(0)^(1)(sqrtx-x^(2))dx=2`
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