Math235 – Fall 2022
Integration & Series
12/20/2022
Evaluate the following integrals:
1.
2𝑥 2 − 𝑥 + 4 𝐴𝑥 + 𝐵 𝐶
𝑥−1 1
= 2
+ = 2
+
3
𝑥 + 4𝑥
𝑥 +4 𝑥 𝑥 +4 𝑥
=∫
𝑥−1
𝑥 2 +4
𝑑𝑥 + ∫
𝑑𝑥
𝑥
𝑙𝑒𝑡 𝑥 = 2 𝑡𝑎𝑛 𝜃
∴ 𝑑𝑥 = 2𝑠𝑒𝑐 2 𝜃𝑑𝜃
∫
2𝑡𝑎𝑛𝜃 − 1
2𝑡𝑎𝑛𝜃 − 1
𝑠𝑖𝑛𝜃
𝜃
2
2𝑠𝑒𝑐
𝜃𝑑𝜃
+
𝑙𝑛𝑥
=
∫
𝑑𝜃
+
𝑙𝑛𝑥
=
∫
𝑑𝜃
−
+ 𝑙𝑛𝑥
4(𝑡𝑎𝑛2 𝜃 + 1)
2
𝑐𝑜𝑠𝜃
2
𝑙𝑒𝑡 𝑡 = 𝑐𝑜𝑠𝜃
∴ 𝑑𝑡 = −𝑠𝑖𝑛𝜃𝑑𝜃
𝑥
𝑥
−1 𝑥
𝑡𝑎𝑛−1
𝑡𝑎𝑛−1
−𝑑𝑡 𝑡𝑎𝑛 2
𝑥
2
2 + 𝑙𝑛𝑥 + 𝐶
=∫
−
+ 𝑙𝑛𝑥 = −𝑙𝑛𝑡 −
+ 𝑙𝑛𝑥 + 𝐶 = −𝑙𝑛|
|−
2
𝑡
2
2
2
√𝑥 + 4
2.
1 − 𝑥 + 2𝑥 2 − 𝑥 3 𝐴 𝐵𝑥 + 𝐶
𝐷𝑥 + 𝐸
1 𝑥+1
𝑥
= + 2
+ 2
= − 2
+ 2
2
2
2
(𝑥
(𝑥
𝑥(𝑥 + 1)
𝑥 𝑥 +1
+ 1)
𝑥 𝑥 +1
+ 1)2
=∫
1
𝑥
−
𝑥+1
𝑥2 + 1
+
𝑥
( 𝑥 2 + 1)
𝑑𝑥 = ln|𝑥| − ∫
2
𝑥
1
𝑥
∫
∫
𝑑𝑥
−
𝑑𝑥
+
𝑑𝑥
( 𝑥 2 + 1) 2
𝑥2 + 1
𝑥2 + 1
𝑠𝑒𝑡 𝑡 = 𝑥 2 + 1 ; 𝑑𝑡 = 2𝑥 𝑑𝑥
ln|𝑥| − ∫
𝑥
1
𝑥
1 1
1 1
−1
|
|
∫
∫
∫
∫ 𝑑𝑡
𝑑𝑥
−
𝑑𝑥
+
𝑑𝑥
=
ln
𝑥
−
𝑑𝑡
−
tan
𝑥
+
( 𝑥 2 + 1) 2
𝑥2 + 1
𝑥2 + 1
2 𝑡
2 𝑡2
1
1
1
1
= ln|𝑥| − ln|𝑡 | − tan−1 𝑥 − + 𝐶 = ln|𝑥| − ln(𝑥 2 + 1) − tan−1 𝑥 −
+𝐶
2
2𝑡
2
2(𝑥 2 + 1)
3.
𝑙𝑒𝑡 (𝑥 5 + 5𝑥 3 + 5𝑥) = 𝑡
∴ 5(𝑥 4 + 3𝑥 2 + 1) 𝑑𝑥 = 𝑑𝑡
=∫
𝑑𝑡 1
1
= 𝑙𝑛(𝑡 ) + 𝐶 = 𝑙𝑛(𝑥 5 + 5𝑥 3 + 5𝑥) + 𝐶
5𝑡 5
5
4.
𝑙𝑒𝑡 𝑒 𝑥 = 𝑡
∴ 𝑒 𝑥 𝑑𝑥 = 𝑑𝑡
=∫
𝑡𝑑𝑡
−1
2
−1
2
=∫
𝑑𝑡 + ∫
𝑑𝑡 = ∫
𝑑 (𝑡 + 1) + ∫
𝑑 (𝑡 + 2)
(𝑡 + 1)(𝑡 + 2)
𝑡+1
𝑡+2
𝑡+1
𝑡+2
= − 𝑙𝑛(𝑡 + 1) + 2 𝑙𝑛(𝑡 + 2) + 𝐶 = − 𝑙𝑛(𝑒 𝑥 + 1) + 2 𝑙𝑛(𝑒 𝑥 + 2) + 𝐶
5.
𝑙𝑒𝑡 √𝑥 = 𝑡
∴
1
2 √𝑥
√3
∫
√3
3
𝑑𝑥 = 𝑑𝑡𝑎𝑛𝑑𝑡:
√3
→ √3
3
√3 2𝑑𝑡
𝑡
∗ 2𝑡𝑑𝑡 = ∫ 2
√3 𝑡 + 1
𝑡4 + 𝑡2
3
𝑙𝑒𝑡 𝑡 = 𝑡𝑎𝑛𝑢
∴ 𝑑𝑡 = 𝑠𝑒𝑐 2 𝑢𝑑𝑢
√3
∫
√3
3
−1
−1
𝑡𝑎𝑛 √3
𝑡𝑎𝑛 √3
2𝑑𝑡
1
3
2
−1
−1 √
∫
∫
=
2
∗
𝑠𝑒𝑐
𝑢𝑑𝑢
=
2𝑑𝑢
=
2
𝑡𝑎𝑛
−
2
𝑡𝑎𝑛
√3
√3
√3
𝑡2 + 1
𝑡𝑎𝑛2 𝑢 + 1
3
𝑡𝑎𝑛−1
𝑡𝑎𝑛−1
3
3
Determine whether the sequence converges or diverges. If it converges, find the limit.
6.
(converges)
3 𝑛
𝑎𝑛 = 9 × ( )
5
3
𝑙𝑖𝑚 9 × ( )𝑛 = 0
𝑛→∞
5
7.
(diverges)
8.
(converges)
∵ −1 ≤ 𝑠𝑖𝑛(2𝑛) ≤ 1𝑎𝑛𝑑1 + √𝑛 > 0
∴ 𝑙𝑖𝑚
−1
𝑛→∞ 1
+ √𝑛
≤ 𝑙𝑖𝑚
𝑠𝑖𝑛(2𝑛)
1 + √𝑛
𝑛→∞
∴ 0 ≤ 𝑙𝑖𝑚
𝑠𝑖𝑛(2𝑛)
𝑛→∞
∴ 𝑙𝑖𝑚
𝑠𝑖𝑛(2𝑛)
𝑛→∞
9.
1 + √𝑛
𝑛→∞ 1
1
+ √𝑛
≤0
=0
(converges)
𝑎𝑛 = 𝑙𝑛
2𝑛2 + 1
1
=
𝑙𝑛
2
−
(
)
𝑛2 + 1
𝑛2 + 1
𝑙𝑖𝑚 𝑙𝑛 (2 −
𝑛→∞
10.
1 + √𝑛
≤ 𝑙𝑖𝑚
1
) = 𝑙𝑛(2)
𝑛2 + 1
(converges)
∵ 0 ≤ 𝑐𝑜𝑠 2 𝑛 ≤ 1𝑎𝑛𝑑2𝑛 > 0
𝑐𝑜𝑠 2 𝑛
1
∴ 0 ≤ 𝑙𝑖𝑚
≤ 𝑙𝑖𝑚 𝑛
𝑛
𝑛→∞ 2
𝑛→∞ 2
𝑐𝑜𝑠 2 𝑛
≤0
𝑛→∞ 2𝑛
∴ 0 ≤ 𝑙𝑖𝑚
𝑐𝑜𝑠 2 𝑛
=0
𝑛→∞ 2𝑛
∴ 𝑙𝑖𝑚
Find the sum of the following series:
11.
𝐴𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑒𝑟𝑖𝑒𝑠,
𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚 𝑖𝑠 1; 𝑡ℎ𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑟𝑎𝑡𝑖𝑜 𝑖𝑠
1
4
∞
1 𝑘
1
4
∑( ) =
=
1
4
1−4 3
𝑘=0
12.
∞
=∑
𝑛=2
2
(𝑛 − 1)(𝑛 + 1)
𝑁
1
1
1 1
1
𝑠𝑒𝑡 𝑆𝑁 = ∑ (
−
) =1+ − −
𝑛−1 𝑛+1
2 𝑁 𝑁+1
𝑛=2
1 1
1
3
𝑙𝑖𝑚 𝑆𝑁 = 𝑙𝑖𝑚 {1 + − −
}=
𝑁→∞
2 𝑁 𝑁+1
2
𝑁→∞
13.
∞
=∑
𝑘=1
1
1
−
𝑛 𝑛+3
𝑁
1
1
1 1
1
1
1
𝑆𝑒𝑡 𝑆𝑁 = ∑ ( −
) =1 + + −
−
−
𝑛 𝑛+3
2 3 𝑁+1 𝑁+2 𝑁+3
𝑛=1
1 1
1
1
1
11
𝑙𝑖𝑚 𝑆𝑁 = 𝑙𝑖𝑚 {1 + + −
−
−
}=
𝑁→∞
2 3 𝑁+1 𝑁+2 𝑁+3
6
𝑁→∞
14. Find the values of x for which the series
converges. Find the sum of the series for those values of x.
𝐴𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 𝑡ℎ𝑒 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑒𝑟𝑖𝑒𝑠,
𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 |
∴ −1 < 𝑥 < 5
∞
∑(
𝑛=0
𝑥−2 𝑛
𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚
1
3
) =
=
=
3
1 − 𝑐𝑜𝑚𝑚𝑜𝑛 𝑟𝑎𝑡𝑖𝑜 1 − 𝑥 − 2 5 − 𝑥
3
𝑥−2
|<1
3