Question:
Consider the function f, defined by f(x) = x^2 for rational x and f(x) = x^3 for irrational x. Let I be the Lower Riemann integral and J be the Upper Riemann integral. Find the values of I and J.

Solution:
To find the values of I and J, we need to calculate the Lower Riemann sum and the Upper Riemann sum for the function f over a given interval.

Lower Riemann Sum (I):
The Lower Riemann sum is obtained by taking the infimum of the function values in each subinterval.

Let's consider a partition P of the interval [0, 1] into n subintervals. The width of each subinterval is Δx = (1 - 0)/n = 1/n.

For each subinterval [x_i-1, x_i], we choose a point c_i in the interval such that x_i-1 ≤ c_i ≤ x_i. Since the function f(x) = x^2 for rational x and f(x) = x^3 for irrational x, we have two cases:

1. If c_i is rational, then f(c_i) = c_i^2.
2. If c_i is irrational, then f(c_i) = c_i^3.

Therefore, the Lower Riemann sum I is given by:

I = Σ f(c_i) Δx

To simplify the calculation, let's choose the points c_i such that c_i = x_i-1, i.e., we choose the left endpoints of each subinterval.

I = Σ f(x_i-1) Δx
= Σ (x_i-1)^2 Δx (for rational c_i)
+ Σ (x_i-1)^3 Δx (for irrational c_i)

Now, let's evaluate each sum separately:

1. Σ (x_i-1)^2 Δx (for rational c_i):
Since the function f(x) = x^2 for rational x, we can replace (x_i-1)^2 with f(x_i-1):

Σ (x_i-1)^2 Δx = Σ f(x_i-1) Δx

This sum represents the Lower Riemann sum for the function f(x) = x^2 over the interval [0, 1]. We can evaluate it using the formula for the sum of squares of consecutive integers:

Σ (x_i-1)^2 Δx = (0^2 + 1^2 + 2^2 + ... + (n-1)^2) Δx
= [(n-1)(n)(2n-1)/6] Δx

Now, substitute Δx = 1/n:

Σ (x_i-1)^2 Δx = [(n-1)(n)(2n-1)/6] (1/n)
= (n^2 - n)(2n-1)/6n
= (2n^3 - 3n^2 + n)/6n

2. Σ (x_i-1)^3 Δx (for irrational c_i):
Since the function f(x) = x^3 for irrational x, we can replace (x_i-1)^3 with