Lower Riemann Integral (I)

To find the lower Riemann integral, we need to find the infimum (greatest lower bound) of the function f(x) on each subinterval of the partition.

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

For each subinterval [x_i-1, x_i], where x_i = iΔx, we need to find the infimum of f(x) on that subinterval.

If x_i is rational, then f(x_i) = x_i * x_i^2 = x_i^3.
If x_i is irrational, then f(x_i) = x_i^2 * x_i^3 = x_i^5.

Since both x_i^3 and x_i^5 are non-negative for x_i in [0,1], the infimum of f(x) on each subinterval is 0.

Therefore, the lower Riemann sum is given by the sum of the infimums multiplied by the width of each subinterval:

I = Σ(0 * Δx) = 0.

Therefore, the value of I is 0.

Upper Riemann Integral (J)

To find the upper Riemann integral, we need to find the supremum (least upper bound) of the function f(x) on each subinterval of the partition.

For each subinterval [x_i-1, x_i], where x_i = iΔx, we need to find the supremum of f(x) on that subinterval.

If x_i is rational, then f(x_i) = x_i * x_i^2 = x_i^3.
If x_i is irrational, then f(x_i) = x_i^2 * x_i^3 = x_i^5.

Since x_i^5 is always greater than or equal to x_i^3 for x_i in [0,1], the supremum of f(x) on each subinterval is x_i^5.

Therefore, the upper Riemann sum is given by the sum of the supremums multiplied by the width of each subinterval:

J = Σ(x_i^5 * Δx).

To evaluate this sum, we can use the fact that the sum of the nth powers of the first n positive integers is given by the formula:

1^5 + 2^5 + ... + n^5 = (n^2 * (n+1)^2 * (2n^2 + 2n - 1))/12.

In our case, n = 1/Δx = n, so the upper Riemann sum becomes:

J = (n^2 * (n+1)^2 * (2n^2 + 2n - 1))/12 * Δx.

Taking the limit as n approaches infinity (or Δx approaches 0), we get:

J = (1^2 * (1+1)^2 * (2*1^2 + 2*1 - 1))/12 * 0 = (4/12) * 0 = 0.

Therefore, the value of J is 0.

The correct answer is neither (a), (b), (c), nor (d). Both the lower R