Determine if the following improper integrals converge or diverge. If an integral converges, find the value it converges to. (a) | 2+1) dx 2-9 (x-2)(x2+1) (Hint: The initial resulting limit here will be an indeterminate form so be careful) (b) 6x2 (b) j √64-3 dx
Determine if the following improper integrals converge or diverge. If an integral converges, find the value it converges to. (a) | 2+1) dx 2-9 (x-2)(x2+1) (Hint: The initial resulting limit here will be an indeterminate form so be careful) (b) 6x2 (b) j √64-3 dx
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![Determine if the following improper integrals converge or diverge. If an integral converges, find the
value it converges to.
(a)√(x-2) (x²+1) dx
(Hint: The initial resulting limit here will be an indeterminate form so be careful)
4
(b) f
6x2
0
√64-23dx
Q4)
-5
tim
x²-9
X2-9
x²+x-3
N
6-700√3
dx
хро
x²-9
dy
Xh-2x+4x-3
x²+x-2x²-3
3x²-4x+1
S
6x2
dx
57-7791
24 6х2
(64-x³½2 dx
4-64-X3
du=-3x²dx
-3x²du = dx
CH
3x²-41-+1
LH.
2
6x-4
증
lim
2
dx
6x-4
2
dx
+-700 6x-4
f
=
3
14
lim po
U=6x-4
du- bdx
du-dy
+ 4 = • du
14
làm 186
и
Tim
=>]
23
€300
du
34
bt-4
lim
t700
In 16x-41
61-4
- Tim In (6 (6+-4)) - 3 In (6 (14)-4))
t->00
= im 1/3 In (36-24) - 1/3 (84-24)
= lim / In (366-24) - In (60)
=
=
12
Vo
64
6x2
64
ه داد
X
D
64
4
I du
3x2
=S" - 24th du
6
-2.
44
4
-1241
1/2164
0
64
0
(64-x)164
= 4[(64-(64)³) - (64-03)]
= 4 ((64-(64))-64)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a0802a2-9dd1-43c0-9b3f-3c3accbadbd6%2F96da1dd0-66b2-4682-b952-cad78c44b67b%2Fga3f75q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Determine if the following improper integrals converge or diverge. If an integral converges, find the
value it converges to.
(a)√(x-2) (x²+1) dx
(Hint: The initial resulting limit here will be an indeterminate form so be careful)
4
(b) f
6x2
0
√64-23dx
Q4)
-5
tim
x²-9
X2-9
x²+x-3
N
6-700√3
dx
хро
x²-9
dy
Xh-2x+4x-3
x²+x-2x²-3
3x²-4x+1
S
6x2
dx
57-7791
24 6х2
(64-x³½2 dx
4-64-X3
du=-3x²dx
-3x²du = dx
CH
3x²-41-+1
LH.
2
6x-4
증
lim
2
dx
6x-4
2
dx
+-700 6x-4
f
=
3
14
lim po
U=6x-4
du- bdx
du-dy
+ 4 = • du
14
làm 186
и
Tim
=>]
23
€300
du
34
bt-4
lim
t700
In 16x-41
61-4
- Tim In (6 (6+-4)) - 3 In (6 (14)-4))
t->00
= im 1/3 In (36-24) - 1/3 (84-24)
= lim / In (366-24) - In (60)
=
=
12
Vo
64
6x2
64
ه داد
X
D
64
4
I du
3x2
=S" - 24th du
6
-2.
44
4
-1241
1/2164
0
64
0
(64-x)164
= 4[(64-(64)³) - (64-03)]
= 4 ((64-(64))-64)
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