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Let `f(x)=[x]+{x}^(3)` then the area of the figure bounded by `y=f^(-1)(x),y=0` between the ordinates `x=2` and `x=9/2` is `alpha`, then `alpha-3/(2^(10//3))+1/2` is equal to ________(where [.] denotes the greatest integer function) |
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Answer» Correct Answer - 9 `f^(-1)(x)=[x]+{x}^(1//3)` So, `int_(2)^(9//2)[x]dx+int_(2)^(9//2){x}^(1//3)=17/2+3/(2^(10//3))` |
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