(11-12) For each n ∈ N, define fn : [0, 1] → R byfn(x) =(xn cos(1/x) if x > 0,0 if x = 0.11. Prove that fn ∈ R[0, 1] for all n ∈ N.12. Prove thatlimn→∞Z 10fn = 0.Hint) Use the comparison principle

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11. To prove that fn ∈ R[0, 1] for all n ∈ N, we need to show that fn is Riemann integrable on the interval [0, 1] for all natural numbers n.

The function fn is defined on [0, 1] and is bounded since -1 ≤ cos(1/x) ≤ 1 for all x in (0, 1] and 0 ≤ xn ≤ 1 for all x in [0, 1] and all n in N.

The function fn is also continuous on (0, 1] since it is a product of continuous functions on this interval. At x = 0, we have fn(0) = 0 for all n in N.

Since the function is bounded and continuous on [0, 1], it is Riemann integrable on this interval. Hence, fn ∈ R[0, 1] for all n ∈ N.

12. To prove that lim (n→∞) ∫ (from 0 to 1) fn = 0, we can use the comparison principle.

The comparison principle states that if 0 ≤ g(x) ≤ f(x) for all x in [a, b] and if ∫ (from a to b) g and ∫ (from a to b) f both exist, then 0 ≤ ∫ (from a to b) g ≤ ∫ (from a to b) f.

In this case, we can take g(x) = 0 and f(x) = |fn(x)|. We have 0 ≤ |fn(x)| for all x in [0, 1] and all n in N.

Since fn is Riemann integrable on [0, 1] for all n in N, ∫ (from 0 to 1) |fn| exists for all n in N.

By the comparison principle, we have 0 ≤ ∫ (from 0 to 1) fn for all n in N.

As n→∞, xn cos(1/x) → 0 for all x in (0, 1] and fn(0) = 0. Hence, lim (n→∞) ∫ (from 0 to 1) fn = 0.