Final answer:
The function f(x)=x²+1/x does not belong to the Lebesgue space L1(0,∞) because the integral of 1/x diverges as x approaches 0, causing the integral of the function over the interval (0,∞) to be infinite.
Step-by-step explanation:
The question asks whether the function f(x)=x²+⅛/x belongs to the space Lᵗʳ(0,∞). In mathematical analysis, the space referred to here is typically the Lebesgue space of integrable functions, usually denoted by Lp with p representing a positive real number indicating the power to which the absolute value of the function is raised before integrating. The question seems to be asking if the function is Lebesgue integrable on the interval from 0 to infinity.
To determine if a function belongs to Lᵗʳ, it must meet the criteria where its absolute value raised to the p-th power is integrable on the given interval. That is to say, the integral of |f(x)|p (where in this case p would be 1 since we're talking about L1) over the interval from 0 to infinity must converge to a finite number.
For the function f(x)=x²+⅛/x, the portion x² is clearly Lebesgue integrable over (0, ∞) as it grows without bound. However, the term 1/x poses an issue at the lower limit, since the integral of 1/x from 0 to any positive value is divergent (i.e., the area under the curve approaches infinity). Hence, this function does not belong to the space L1 over the interval (0, ∞) because the integral of the function over the entire interval is not finite.