Calc1131W11T2 - Improper integrals

Sunday, 20 May 2018

9:06 PM

Untitled picture.png Machine generated alternative text:
One kind of improper integral occurs when we integrate a function over an infinite region; going to infinity. If the function decays fast enough then the total area under its graph can still be finite.
Consider the function y
4
3
2
1
1
3
Area
2
100's
1000's
10000's
0.4999999999
1
3
99006
1
3
4
5
6
dc
By experimenting with the slider, the integer value n for which
• is closest to 0.45 is n
• is closest to 0.487 is n
6
3
DO 1
dc appears to approach
As n increases, the value of the integral
3
We can verify this algebraically by first computing a finite integral, namely
1
Now as we take the limit of this expression we get
oo 1
dc
7
0.5
1
dc
lim
-1/(2RA2) + 1/2
1
dc
05
1

Untitled picture.png Machine generated alternative text:
1
This is a special case of the p-integral
1
which converges to a finite value precisely when

Untitled picture.png Machine generated alternative text:
1
This is a special case of the p-integral
1
which converges to a finite value precisely when

Untitled picture.png Machine generated alternative text:
Evaluate the following improper integrals; or write undefined if they don't converge:
i)
xe¯3Xdx =
-1(7)
1
2
Note: use arctan (7) to write tan
3
, use pi to write T, and use exp (3) to write e

Untitled picture.png Machine generated alternative text:
1
Consider the function y
1
4
3
2
10's
100's
1000's
10000's
X 10100—4
Area —
2
1
3
99999 x 10100
4
1
—dr 241.7714
5
6
7
By experimenting with the sliders, we see that an integer value n for which
• is close to 4 is n
• is close to 10 is n
• is close to 173 is n
Note:
54
19999
99999*10A70
dc is within 1% of the required value.
try to give an answer so
• you can enter 12345 x 10678
in Maple notation as 12345*10A (678)
DO 1
dc appears to be undefined
As n increases, the value of the integral

Untitled picture.png Machine generated alternative text:
1
Consider the function y
1
4
3
2
10's
100's
1000's
10000's
X 10100—4
Area —
2
1
3
99999 x 10100
4
1
—dr 241.7714
5
6
7
By experimenting with the sliders, we see that an integer value n for which
• is close to 4 is n
• is close to 10 is n
• is close to 173 is n
Note:
54
19999
99999*10A70
dc is within 1% of the required value.
try to give an answer so
• you can enter 12345 x 10678
in Maple notation as 12345*10A (678)
DO 1
dc appears to be undefined
As n increases, the value of the integral

Untitled picture.png Machine generated alternative text:
We can verify this algebraically by first computing a finite integral, namely
RI
Now as we take the limit of this expression we get
Ill(R)
RI
lim
R*D0 1 X
undefined
dc which converges to a finite value precisely when
This is a special case of the p-integral

Untitled picture.png Machine generated alternative text:
Evaluate the area of the region under the graph y
Note: remember to use pi (capital p) for T, and exp (2) for e
The area is

Ink Drawings

Untitled picture.png Machine generated alternative text:
1
Let's consider the curve y
from x
called Gabriel's Horn.
04
Gabnel With horn, from
'The Wonders of Creation and
the Oddities of Existence', 14th century.
The volume contained in S is given by the integral of the cross-sectional area. Namely
Volume
1 to DO. If we take this curve and rotate it about the x-axis we get an infinite tubelike surface s i sometimes
T y dc
1
dc
Our understanding of the convergence of p-integrals leads us to conclude that the integral for the volume converges precisely when 2P > 1
p>1/2
O
in other words when
The surface area of S is given by the formula
Surface Area
Since the integrand can be bounded by
The integral for the surfarce area converges precisely when p > 1
So; in the case for p = 1 , we note that the volume is finite
2
1
O
27ry
< 2K
dy
dc
1
dc
p2
1
< 27T
but the surface area is infinite
O
p2
dc
This is perhaps surprising'

Untitled picture.png Machine generated alternative text:
1
Let's consider the curve y
from x
called Gabriel's Horn.
04
Gabnel With horn, from
'The Wonders of Creation and
the Oddities of Existence', 14th century.
The volume contained in S is given by the integral of the cross-sectional area. Namely
Volume
1 to DO. If we take this curve and rotate it about the x-axis we get an infinite tubelike surface s i sometimes
T y dc
1
dc
Our understanding of the convergence of p-integrals leads us to conclude that the integral for the volume converges precisely when 2P > 1
p>1/2
O
in other words when
The surface area of S is given by the formula
Surface Area
Since the integrand can be bounded by
The integral for the surfarce area converges precisely when p > 1
So; in the case for p = 1 , we note that the volume is finite
2
1
O
27ry
< 2K
dy
dc
1
dc
p2
1
< 27T
but the surface area is infinite
O
p2
dc
This is perhaps surprising'

Untitled picture.png Machine generated alternative text:
The Comparison Test is also useful to show than an improper integral such as
O S f(x) g(x) on the interval [a, DO), then the comparison text states that if
g(x) dc diverges; again by comparing it against another divergent integral which is easier to calculate. If
f(x) dx diverges; then so does
g@) dc
To show that
diverges
Since x > 2
2
O we will compare it to
2
1
x3/5
which we know
diverges
1
dc
(x3 — 4)1/5
O since it is a p-integral with p
x3 — 4 < implies that
(x3 -4)1/5 <
and hence
1
x3/5
Integrating both sides of this inequality over [2, 00) completes the argument.

Created with Microsoft OneNote 2016.