(10) 3. Evaluate the improper integral of \( f(x)=x^{1 / 3} \) on the interval \( [0,8] \) by taking the limit of a monotone sequence of functions defined on \( [0,8] \). Specifically, define the sequence \( f_{n}(x) \) on \( [0,8] \), compute \( \int f_{n}(x) d x \), and then evaluate the limit. Can \( f(x) \) be shown to be 2-summable by this method?

To evaluate the improper integral of \( f(x) = x^{1/3} \) on the interval \( [0, 8] \) using a monotone sequence of functions, we can define a sequence \( f_n(x) \) that approximates \( f(x) \) as \( n \) approaches infinity. ### Step 1: Define the Sequence We can define the sequence \( f_n(x) \) as follows: \[ f_n(x) = \begin{cases} x^{1/3} & \text{if } x \leq n \\ 0 & \text{if } x > n \end{cases} \] This means that \( f_n(x) \) is equal to \( f(x) \) for \( x \) in the interval \( [0, n] \) and is zero for \( x > n \). As \( n \) increases, \( f_n(x) \) approaches \( f(x) \) for all \( x \in [0, 8] \). ### Step 2: Compute the Integral Next, we compute the integral of \( f_n(x) \) over the interval \( [0, 8] \): \[ \int_0^8 f_n(x) \, dx = \int_0^8 x^{1/3} \, dx \] Since \( n \) is greater than or equal to 8, we can compute the integral directly without worrying about the upper limit. ### Step 3: Evaluate the Integral To evaluate the integral, we use the power rule: \[ \int x^{1/3} \, dx = \frac{x^{1/3 + 1}}{1/3 + 1} = \frac{x^{4/3}}{4/3} = \frac{3}{4} x^{4/3} \] Now we evaluate this from 0 to 8: \[ \int_0^8 x^{1/3} \, dx = \left[ \frac{3}{4} x^{4/3} \right]_0^8 = \frac{3}{4} (8^{4/3}) - \frac{3}{4} (0^{4/3}) \] Calculating \( 8^{4/3} \): \[ 8^{4/3} = (2^3)^{4/3} = 2^4 = 16 \] Thus, we have: \[ \int_0^8 x^{1/3} \, dx = \frac{3}{4} \cdot 16 = 12 \] ### Step 4: Evaluate the Limit Since \( f_n(x) \) converges to \( f(x) \) as \( n \to \infty \), we can conclude that: \[ \lim_{n \to \infty} \int_0^8 f_n(x) \, dx = \int_0^8 f(x) \, dx = 12 \] ### Step 5: 2-Summability To determine if \( f(x) \) is 2-summable, we check if the integral converges: \[ \int_0^8 f(x) \, dx = 12 \] Since this integral converges, we can conclude that \( f(x) \) is indeed 2-summable. ### Final Answer The value of the improper integral is \( 12 \), and \( f(x) \) can be shown to be 2-summable by this method.