Calculus of several variables

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User Generated

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Obpun

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Mathematics

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I need the solution for problems 2.1 2.2 and the problems in the paper

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1. Assume that A and B are contented subsets and ACB. Find the relationship (> or = or 5) between each pair. (a) 4A and GB (b) 4A and GASB (C) YB and GAYB 0, otherwise, 2. Let f: R+ R be defined by f(x) = { i sin x, 0 3 x < 27, (a) Find f* and calculate | F+ (b) Find f- and calculate 15 (c) Calculate / f and verify that | = +-15. The following two problems are bonus problems. You can earn a maximum of 10 extra points by completing the problem 3 and 4. . Note. Do not use computer system (Wolfram Alpha, Maple, Mathematica, etc.). Producing answer using the computer systems would result in -10 points for penalty. 3. Evaluate the integral / «(arctan a)? da (hint. Integration by parts) 4. Evaluate the integral / met die -dx (hint. x4 + 1 = x4 + 2x2 + 1 - 2x2) 3 Step Functions and Riemann Sums 223 Exercises 2.1 If f and g are admissible functions on R™ and CER, show that f+g and cfare admis- sible. 2.2 Show that the positive and negative parts of an admissible function are admissible. 2.3 If D is the set of discontinuities of f: RM → R, show that the set of discontinuities of fpA is contained in DUOA. 2.4 If A and B are contented sets with AC B, apply Proposition 2.5 with f=9A=%B9A, g=QB to show that V(A) SU(B). 2.5 Let A and B be contented sets with An B negligible. Apply Proposition 2.7 with f= MAU B to show that A UB is contented with U(A U B)= v(A) +v(B). 2.6 If fand g are integrable functions with f(x) < g(x) for all x, prove that $1$g without using Axioms I and II. Hint: 15g=ft
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Explanation & Answer

I finished. Please ask if I wrote too more "it is obvious"-es (or in any other need).

1. 𝜑𝐴 = 1 only on 𝐴, 𝜑𝐵 = 1 only on 𝐵.
a. Therefore for 𝐴 ⊂ 𝐵 𝜑𝐴 ≤ 𝜑𝐵 .

And 0 ≤ 𝜑𝐴 , 𝜑𝐵 ≤ 1, therefore
b. 𝜑𝐴 ≥ 𝜑𝐴 𝜑𝐵 (multiply true equality 1 ≥ 𝜑𝐵 by non-negative 𝜑𝐴 ).
c. 𝜑𝐵 ≥ 𝜑𝐴 𝜑𝐵 (multiply true equality 1 ≥ 𝜑𝐴 by non-negative 𝜑𝐵 ).

2.a.b.c. sin 𝑥 ≥ 0 for 𝑥 ∈ [0, 𝜋] and sin 𝑥 ≤ 0 for 𝑥 ∈ [𝜋, 2𝜋]. Therefore
1
[0, 𝜋],
𝑓 + (𝑥) = {2 sin 𝑥 , 𝑥 ∈
0,
otherwise,

1
[𝜋, 2𝜋],
𝑓 − (𝑥) = {− 2 sin 𝑥 , 𝑥 ∈
0,
otherwise.

Simply
𝜋

𝜋
1
1
1
∫ 𝑓 = ∫ sin 𝑥 𝑑𝑥 = (− cos 𝑥)
= − (−1 − 1) = 𝟏,
2
2
2
𝑥=0
ℝ
+

0

2𝜋

2𝜋
1
1
1
∫ 𝑓 − = ∫ (− sin 𝑥) 𝑑𝑥 = ( cos 𝑥)
= (1 − (−1)) = 𝟏,
2
2
2
𝑥=𝜋
ℝ
𝜋

2𝜋

∫ 𝑓=∫
ℝ

0

2𝜋
1
1
1
sin 𝑥 𝑑𝑥 = (− cos 𝑥)
= − (1 − 1) = 𝟎.
2
2
2
𝑥=0

3. Integration by parts:
∫ 𝑥(arctan 𝑥)2 𝑑𝑥 = |𝑢 = (arctan 𝑥)2 , 𝑑𝑢 =

2
1 2
arctan
𝑥
,
𝑑𝑣
=
𝑥𝑑𝑥,
𝑣
=
𝑥 |=
1 + 𝑥2
2

1
𝑥2
1
1
= 𝑥 2 (arctan 𝑥)2 − ∫
arctan 𝑥 𝑑𝑥 = 𝑥 2 (arctan 𝑥)2 − ∫ (1 −
) arctan 𝑥 𝑑𝑥 =
2
2
1+𝑥
2
1 + 𝑥2
1
1
= 𝑥 2 (arctan 𝑥)2 − ∫ arctan 𝑥 𝑑𝑥 + ∫
arctan 𝑥 𝑑𝑥 =
2
1 + 𝑥2
1
1
= 𝑥 2 (arctan 𝑥)2 − ∫ arctan 𝑥 𝑑𝑥 + ...