advanced calculus of several variables, homework help

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Mathematics

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I want you to answer 4 q . tow of them from book ( 1.3 and 1.7) and another two q posted in files. also I include the lrctures.

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Textbook, Page 213, 1.3, 1.7, plus, (b) Verify that a(0}(0) = 5* **dr by calculating the Riemann integral 1. Given function f(x) = 2 on the interval (1,3). Let op() = {(x,y) ER:13 as 3,0 Sy Sz} (a) Use the definition of area to show that the area of the set of() exists and determine the area a(O()). ſi zdr. 2. Given function f(x) = x2 - 4.0-5. (a) Determine /+() and f (x). Sketch the graphs of the two functions. (b) Calculate the Riemann integrals Ls+(x)da, [1+(x)dar, and LS(z)de. Verify that ſys(a)de = s+(a)de - L3(dr. The integration by parts formula takes the familiar and memorable form Ju dv = w - uv - sodu if we write u = f(x); 1=968), and further agree to the notation du = f'(x) dx, do = g'(x) dx in the integrands. Exercises ..., R.', then on 1.1 Calculate the area under y = x over the interval [0, 1] by making direct use of the defini- tion of area. 1.2 Apply the fundamental theorem to calculate the area in the previous exercise. 1.3 Prove Property A of area. Hint: Given SCT, suppose to the contrary that a(T) must contain [0, 1] x [0, 1] 2 -I alę,">, ¿ y=fexs A rexida , - of A = f fersche Sco 5 yo flas I Is e subit of A Ex Isdratinal ms in [ail]
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Explanation & Answer

The solutions are ready!

1. 𝑓(𝑥) = 𝑥 2 on [1, 3].
a. Divide the interval by 𝑛 equal parts and inscribe the set 𝑂 into the set
𝑘

′′
′′
𝑈𝑛′′ = ⋃2𝑛−1
𝑘=0 𝑅𝑘 where 𝑅𝑘 = [1 + 𝑛 , 1 +

𝑘+1
𝑘+1 2
×
(1
+
) ].
]
[0,
𝑛
𝑛

Similarly, inscribe the set
𝑘



𝑈𝑛′ = ⋃2𝑛−1
𝑘=0 𝑅𝑘 where 𝑅𝑘 = [1 + 𝑛 , 1 +

𝑘+1
𝑘 2
×
(1
+
) ]
]
[0,
𝑛
𝑛

into the set 𝑂.

Let’s find the areas of 𝑈𝑛′ and 𝑈𝑛′′ :
𝑎(𝑈𝑛′ )

2𝑛−1

2𝑛−1

𝑘=0

𝑘=0

1
𝑘 2 1
𝑘 𝑘2
1
2 2𝑛(2𝑛 − 1) 1 1
= ∑ ∙ (1 + ) = ∑ (1 + 2 + 2 ) = (2𝑛 + ∙
+ 2 ∙ (2𝑛 − 1)2𝑛(4𝑛 − 1))
𝑛
𝑛
𝑛
𝑛 𝑛
𝑛
𝑛
2
𝑛 6

(I believe you know these formulas of summations).
This, in turn, is equal to

1
(2𝑛
𝑛

=

1 1
𝑛 3

+ 2(2𝑛 − 1) + ∙ (2𝑛 − 1)(4𝑛 − 1)) =

1
(2𝑛2
𝑛2

1
3

+ 4𝑛2 − 2𝑛 + (2𝑛 − 1)(4𝑛 − 1)) =

1
1
(18𝑛2 − 6𝑛 + 8𝑛2 − 6𝑛 + 1) = 2 (26𝑛2 − 12𝑛 + 1).
2
3𝑛
3𝑛

The same way
𝑎(𝑈𝑛′′ )

2𝑛−1

2𝑛

2𝑛

𝑘=0

𝑘=1

𝑘=0

1
𝑘+1 2
1
𝑘 2
1
𝑘 2 1
= ∑ ∙ (1 +
) = ∑ ∙ (1 + ) = ∑ ∙ (1 + ) − =
𝑛
𝑛
𝑛
𝑛
𝑛
𝑛
𝑛

2𝑛

1
𝑘 𝑘2
1
2 2𝑛(2𝑛 + 1) 1 1
= (∑ (1 + 2 + 2 ) − 1) = (2𝑛 + 1 + ∙
+ 2 ∙ (2𝑛 + 1)2𝑛(4𝑛 + 1) − 1) =
𝑛
𝑛 𝑛
𝑛
𝑛
2
𝑛 6
𝑘=0

=

1
1 1
1
(2𝑛 + 1 + (4𝑛 + 2) + ∙ (2𝑛 + 1)(4𝑛 + 1) − 1) = 2 (6𝑛2 + 3𝑛 + 12𝑛2 + 6𝑛 + 8𝑛2 + 6𝑛 + 1 − 3𝑛) =
𝑛
𝑛 3
3𝑛
=

The limit of both quantities is
Therefore 𝑂 has area and it is

3

1

1

26
which
3
𝟐𝟔
.
𝟑

1
(26𝑛2 + 12𝑛 + 1).
3𝑛2

proves that their difference becomes infinitely small.

b. ∫1 𝑥 2 𝑑𝑥 = 3 (𝑥 3 )3𝑥=1 = 3 (27 − 1) =

𝟐𝟔
,
𝟑

the same.

2. 𝑓(𝑥) = 𝑥 2 − 4𝑥 − 5 = (𝑥 + 1)(𝑥 − 5), which means �...


Anonymous
Great content here. Definitely a returning customer.

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