Calculus Assignment: Proving Integral Convergence/Divergence

Verified

Added on  2023/01/23

|10
|1170
|22
Homework Assignment
AI Summary
Document Page
1
Calculus
Student’s Name:
University Affiliation:
tabler-icon-diamond-filled.svg

Paraphrase This Document

Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser
Document Page
2
Part a
If f(x)=|f(x)|
Then a f ( x ) bx 1
Now let us determine whether f(x) converges or not

1

f ( x ) dx=¿ lim
b

1
b
f ( x ) dx ¿
¿ lim
b
¿
¿ lim
b [ f ( b )a+1
a+1 ][ f ( 1 )a +1
a+ 1 ]
¿ lim
b [ f ( 1 ) a +1
a+1 ] +[ f ( b ) a +1
a+1 ]
¿ [ 1
a+1 ]
Therefore, |f(x)| diverge since P<1
Part b
Prove that
1

cosx
x2 dx converges .
Solution
Using integration by parts
u v' dx=uv u' vdx
Document Page
3
Where u= 1
x2 , v'= cosxdx ,u'=2
x3 , v=sinx
2
x3 cosxdx = sinx
x2 +2 sinx
x3 dx
Introducing limits

1

cosx
x2 =lim
b

1
b
cosx
x3 dx
¿ lim
b
¿
¿ lim
b {[ sin ( b )
( b )3 ] [ sin ( 1 )
( 1 )3 ] }+ lim
b
2
1
b
sinx
x3 dx
¿ lim
b
[ 0 ]sin [ 1 ] +lim
b
2
1
b
sinx
x3 dx
From the equation above, lim
b
2
1
b
sinx
x3 dx lim
b
2
1
b
¿ sinx ¿
x3 dx ¿
The absolute value |sin x| when graphed produces a positive graph
with 1 maximum and 0 minimum.
Considering the highest value: 1, it becomes
lim
b
2
1
b
1
x3 dx which converges.
Therefore, the
1

cosx
x2 dx converge .
Part c
Document Page
4
Prove that
1

sinx
x dx converges .
Solution
We will use integration by parts which is stated as:
u v' dx=uv u' vdx
Let u= 1
x , v'=sinxdx , v=cosx
1
x . sinxdx= [ 1
x ( cosx ) ] 1/ x2 (cosx) dx
¿ [ cosx
x ](cosx )/ x2 dx
Introducing the limits

1

sinx
x dx= lim
n

1
n
sinx
x dx
¿ lim
n ([cosx
x2 ) ] b
1
1
n
cosx
x2 dx ¿
¿ lim
n
{
([ cos (n)
(n)2 ) ] +¿
lim
n
( 0 ) +cos ( 1 ) lim
n

1
n
cosx
x2 dx
tabler-icon-diamond-filled.svg

Paraphrase This Document

Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser
Document Page
5
Now taking lim
n

1
n
cosx
x2 dxand testing, we will find that
lim
n

1
n
cosx
x2 dx lim
n

1
n
¿ cosx ¿
x2 dx ¿
And the absolute values of |cos x| ranges between 0 and 1.
Taking the highest value i.e. 1, it will become
lim
n

1
n
1
x2 dx which converges.
Since, lim
n

1
n
cosx
x2 dx lim
n

1
n
¿ cosx ¿
x2 dx ¿ , hence
1

sinx
x dx converges .
Part d
Prove that
1

1cos 2 x
x dx di verges .
Hint: Split the expression inside the integral into two parts.
Solution.

1

1cos 2 x
x dx =
1

1
x dx
1

cos 2 x
x dx
The second part of the equation would be solved using integration by
parts u v' dx=uv u' vdx
u= 1
x , u' = 1
x2 , v'=cos 2 x dx , v =sin 2 x
2
Document Page
6

1

1cos 2 x
x = [ 1
x2 ] {( sin 2 x
2 x ) sin 2 x
2 x2 dx }
Let us introduce limits

1

1cos 2 x
x =lim
b

1
b
1
x dxlim
b

1
b
cos 2 x
x dx
¿ [1
x2 ]{( sin 2 x
2 x ) sin 2 x
2 x2 dx }
¿ lim
b [1
x2 ] b
1{lim
b
( sin 2 x
2 x )
b
1 lim
b
1
b
sin 2 x
2 x2
dx }
¿ lim
b
[ 0 ] 1lim
b
[ 0 ] sin [ 2 ] lim
b

1
b
sin2 x
2 x2 dx
Now takin
lim
b

1
b
sin2 x
2 x2 dxtesting , we will find that lim
b

1
b
sin2 x
2 x2 dx lim
b

1
b
¿ sin 2 x ¿
2 x2 dx ¿
The absolute |sin 2x| ranges between 0 and 2.5
Taking the largest value, we will obtain
lim
b

1
b
5
4 x2 dx
Hence
1

1cos 2 x
x dx diverges .
Document Page
7
Part e
Prove that
1

¿ sinx
x ¿ dx converges . ¿
Hint: Use the fact that |sin x| Sin2 x ϵR
Solutions.
When dealing with absolute values, we would only consider positive
values.
We use integration by parts to obtain or solutions
u v' dx=uv u' vdx
1
x . sinxdx=[ cox
x ¿ ] cosx
x2 dx ¿
We now introduce our limits

1

¿ sinx
x ¿=lim
b

1
b
sinx
x2 dx ¿
Solving the above we obtain
¿ lim
b
¿
¿ lim
b
( [ cos ( b )
b2 ] + cos ( 1 )
1 )
1
b
cosx
x2 dx
tabler-icon-diamond-filled.svg

Paraphrase This Document

Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser
Document Page
8
¿ lim
b
( 0 ) 1 lim
b
cos ( 1 )
1
b
cosx
x2 dx
We now test
1
b
cosx
x2 dx ¿ know the lowest values are 0thehighest values areinfinite .

1
b
cosx
x2 dx>
1
b
¿ cosx
x2 dx
Hence,
1

¿ sinx
x ¿ dx converge ¿
Document Page
9
References
Grosser, D. (2014). Improper Integrals.
Lay, D. (2012). Linear Algebra and Its Applications. University of Mary Land.
Document Page
10
chevron_up_icon
1 out of 10
circle_padding
hide_on_mobile
zoom_out_icon
[object Object]