Description
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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Explanation & Answer
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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 �...
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