Discussion
The Lebesgue integral of the function \(f : [0, 1] \to \mathbf{R}\) defined by :
\[ f(x) = \begin{cases} 1/\sqrt[3]{x} & \text{if } 0 \lt x \le 1 \\ 0 & \text{if } x = 0 \end{cases} \]
is :
Answer: C
The integral is 3/2.
\(f(x) = x^{-1/3}\) is non-negative and the single point \(x = 0\) has measure zero, so the Lebesgue integral equals the improper Riemann integral:
\[ \begin{aligned} \int_{0}^{1} x^{-1/3}\,dx &= \left[\frac{x^{2/3}}{2/3}\right]_{0}^{1} \\ &= \frac{3}{2} \end{aligned} \]
The integral is finite because the exponent \(1/3\) is less than 1. The value 3/4 is \(\int_{0}^{1} x^{1/3}dx\), the wrong sign of exponent; 2/3 inverts the answer; and 4/3 does not come from either.
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