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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 ` isA. `(2)/(3)`B. 2C. `(10)/(3)`D. 6 |
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Answer» Correct Answer - 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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