MAT1300 Lecture 18
Improper Integrals
Pieter Hofstra
November 24, 2009
Overview
Integration by Parts
Improper Integrals
1
Integration by Parts
2
Improper Integrals
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Integration by parts
Recall from last time the formula
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Integration by parts
Recall from last time the formula
Z
uv 0 dx = uv −
Pieter Hofstra
Z
u 0 v dx
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
Evaluate
R
3xe −5x dx.
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
R
Evaluate 3xe −5x dx.
Solution. We let u = 3x, v 0 = e −5x .
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
R
Evaluate 3xe −5x dx.
Solution. We let u = 3x, v 0 = e −5x .Then
u0 = 3
Pieter Hofstra
1
v = − e 5x
5
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
R
Evaluate 3xe −5x dx.
Solution. We let u = 3x, v 0 = e −5x .Then
u0 = 3
1
v = − e 5x
5
Now we can use the integration by parts formula:
Z
Z
0
uv dx = uv − u 0 v dx
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
R
Evaluate 3xe −5x dx.
Solution. We let u = 3x, v 0 = e −5x .Then
u0 = 3
1
v = − e 5x
5
Now we can use the integration by parts formula:
Z
Z
0
uv dx = uv − u 0 v dx
Z
Z
1 −5x
1
−5x
3xe
dx = 3x − e
− (3) · − e −5x dx
5
5
3 −5x
3 −5x
= − xe
+ e
+C
5
25
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
Evaluate
R
3 ln(x 2 )dx.
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
R
Evaluate 3 ln(x 2 )dx.
Solution. Set u = ln(x 2 ), v 0 = 3
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
R
Evaluate 3 ln(x 2 )dx.
Solution. Set u = ln(x 2 ), v 0 = 3 which gives us
u0 =
2x
2
=
2
x
x
Pieter Hofstra
v = 3x
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
R
Evaluate 3 ln(x 2 )dx.
Solution. Set u = ln(x 2 ), v 0 = 3 which gives us
u0 =
2x
2
=
2
x
x
v = 3x
So integration by parts gives us
Z
Z
0
uv dx = uv − u 0 v dx
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
R
Evaluate 3 ln(x 2 )dx.
Solution. Set u = ln(x 2 ), v 0 = 3 which gives us
u0 =
2x
2
=
2
x
x
v = 3x
So integration by parts gives us
Z
Z
0
uv dx = uv − u 0 v dx
Z
Z
2
3 ln(x 2 )dx = 3x ln(x 2 ) −
· 3xdx
x
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
R
Evaluate 3 ln(x 2 )dx.
Solution. Set u = ln(x 2 ), v 0 = 3 which gives us
u0 =
2x
2
=
2
x
x
v = 3x
So integration by parts gives us
Z
Z
0
uv dx = uv − u 0 v dx
Z
Z
2
3 ln(x 2 )dx = 3x ln(x 2 ) −
· 3xdx
x
Z
= 3x ln(x 2 ) − 6dx
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
R
Evaluate 3 ln(x 2 )dx.
Solution. Set u = ln(x 2 ), v 0 = 3 which gives us
u0 =
2x
2
=
2
x
x
v = 3x
So integration by parts gives us
Z
Z
0
uv dx = uv − u 0 v dx
Z
Z
2
3 ln(x 2 )dx = 3x ln(x 2 ) −
· 3xdx
x
Z
= 3x ln(x 2 ) − 6dx
= 3x ln(x 2 ) − 6x + C
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
R
Another way to find 3 ln(x 2 )dx:
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
R
Another way to find 3 ln(x 2 )dx:
Rewrite 3 ln(x 2 ) = 6 ln(x).
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
R
Another way to find 3 ln(x 2 )dx:
Rewrite 3 ln(x 2 ) = 6 ln(x). Now let u = ln(x), v 0 = 6, so that
u 0 = x1 and v = 6x.
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
R
Another way to find 3 ln(x 2 )dx:
Rewrite 3 ln(x 2 ) = 6 ln(x). Now let u = ln(x), v 0 = 6, so that
u 0 = x1 and v = 6x.Then
Z
Z
uv 0 dx = uv − u 0 v dx
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
R
Another way to find 3 ln(x 2 )dx:
Rewrite 3 ln(x 2 ) = 6 ln(x). Now let u = ln(x), v 0 = 6, so that
u 0 = x1 and v = 6x.Then
Z
Z
uv 0 dx = uv − u 0 v dx
Z
Z
1
· 6xdx
6 ln(x)dx = 6x ln(x) −
x
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
R
Another way to find 3 ln(x 2 )dx:
Rewrite 3 ln(x 2 ) = 6 ln(x). Now let u = ln(x), v 0 = 6, so that
u 0 = x1 and v = 6x.Then
Z
Z
uv 0 dx = uv − u 0 v dx
Z
Z
1
· 6xdx
6 ln(x)dx = 6x ln(x) −
x
Z
= 6x ln(x) − 6dx
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
R
Another way to find 3 ln(x 2 )dx:
Rewrite 3 ln(x 2 ) = 6 ln(x). Now let u = ln(x), v 0 = 6, so that
u 0 = x1 and v = 6x.Then
Z
Z
uv 0 dx = uv − u 0 v dx
Z
Z
1
· 6xdx
6 ln(x)dx = 6x ln(x) −
x
Z
= 6x ln(x) − 6dx
= 6x ln(x) − 6x + C
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
Which techniques do you need for the following integrals?
R x
1
e x dx
R x2
2
xe dx
R
3
ln(3x)dx
R 1
4
2 dx
R xx
5
dx
x 2 +1
R 1
6
x ln(x) dx
R
7
2xe 3 ln(x) dx
R 3
8
x ln(4x 2 )dx
R ln(x)
9
dx
x2
R x2
10
e dx
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
1
R
x
e x dx
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
1
R
x
e x dx
=⇒ by parts, let u = x, v 0 = e −x
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
1
R
2
R
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
1
R
2
R
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
=⇒ substitution, let u = x 2
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
1
R
2
R
3
R
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
=⇒ substitution, let u = x 2
ln(3x)dx
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
1
R
2
R
3
R
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
ln(3x)dx
=⇒ substitution, let u = x 2
=⇒ by parts, let u = ln(3x), v 0 = 1
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
1
R
2
R
3
R
ln(3x)dx
4
R
1
dx
x2
=⇒ substitution, let u = x 2
=⇒ by parts, let u = ln(3x), v 0 = 1
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
1
R
2
R
3
R
ln(3x)dx
4
R
1
dx
x2
=⇒ substitution, let u = x 2
=⇒ by parts, let u = ln(3x), v 0 = 1
=⇒ power rule
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
1
R
2
R
3
R
ln(3x)dx
4
R
5
R
1
dx
x2
x
dx
x 2 +1
=⇒ substitution, let u = x 2
=⇒ by parts, let u = ln(3x), v 0 = 1
=⇒ power rule
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
1
R
2
R
3
R
ln(3x)dx
4
R
5
R
1
dx
x2
x
dx
x 2 +1
=⇒ substitution, let u = x 2
=⇒ by parts, let u = ln(3x), v 0 = 1
=⇒ power rule
=⇒ substitution, let u = x 2 + 1
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
1
R
2
R
3
R
ln(3x)dx
4
R
5
R
6
R
1
dx =⇒ power rule
x2
x
dx =⇒ substitution,
x 2 +1
1
x ln(x) dx
=⇒ substitution, let u = x 2
=⇒ by parts, let u = ln(3x), v 0 = 1
Pieter Hofstra
let u = x 2 + 1
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
1
R
2
R
3
R
ln(3x)dx
4
R
5
R
6
R
1
dx =⇒ power rule
x2
x
dx =⇒ substitution, let u = x 2 + 1
x 2 +1
1
=⇒ substitution, let u = ln(x)
x ln(x) dx
=⇒ substitution, let u = x 2
=⇒ by parts, let u = ln(3x), v 0 = 1
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
1
R
2
R
3
R
ln(3x)dx
4
R
5
R
6
R
7
R
1
dx =⇒ power rule
x2
x
dx =⇒ substitution, let u = x 2 + 1
x 2 +1
1
=⇒ substitution, let u = ln(x)
x ln(x) dx
2xe 3 ln(x) dx
=⇒ substitution, let u = x 2
=⇒ by parts, let u = ln(3x), v 0 = 1
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
1
R
2
R
3
R
ln(3x)dx
4
R
5
R
6
R
7
R
1
dx =⇒ power rule
x2
x
dx =⇒ substitution, let u = x 2 + 1
x 2 +1
1
=⇒ substitution, let u = ln(x)
x ln(x) dx
R
2xe 3 ln(x) dx =⇒ rewrite as 2x · x 3 dx
=⇒ substitution, let u = x 2
=⇒ by parts, let u = ln(3x), v 0 = 1
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
1
R
2
R
3
R
ln(3x)dx
4
R
5
R
6
R
7
R
1
dx =⇒ power rule
x2
x
dx =⇒ substitution, let u = x 2 + 1
x 2 +1
1
=⇒ substitution, let u = ln(x)
x ln(x) dx
R
2xe 3 ln(x) dx =⇒ rewrite as 2x · x 3 dx
8
R
=⇒ substitution, let u = x 2
=⇒ by parts, let u = ln(3x), v 0 = 1
x 3 ln(4x 2 )dx
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
1
R
2
R
3
R
ln(3x)dx
4
R
5
R
6
R
7
R
1
dx =⇒ power rule
x2
x
dx =⇒ substitution, let u = x 2 + 1
x 2 +1
1
=⇒ substitution, let u = ln(x)
x ln(x) dx
R
2xe 3 ln(x) dx =⇒ rewrite as 2x · x 3 dx
8
R
=⇒ substitution, let u = x 2
=⇒ by parts, let u = ln(3x), v 0 = 1
x 3 ln(4x 2 )dx
=⇒ by parts or rewrite
Pieter Hofstra
R
MAT1300 Lecture 18
2x 3 ln(2x)dx
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
1
R
2
R
3
R
ln(3x)dx
4
R
5
R
6
R
7
R
1
dx =⇒ power rule
x2
x
dx =⇒ substitution, let u = x 2 + 1
x 2 +1
1
=⇒ substitution, let u = ln(x)
x ln(x) dx
R
2xe 3 ln(x) dx =⇒ rewrite as 2x · x 3 dx
8
R
x 3 ln(4x 2 )dx
9
R
ln(x)
dx
x2
=⇒ substitution, let u = x 2
=⇒ by parts, let u = ln(3x), v 0 = 1
=⇒ by parts or rewrite
Pieter Hofstra
R
MAT1300 Lecture 18
2x 3 ln(2x)dx
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
1
R
2
R
3
R
ln(3x)dx
4
R
5
R
6
R
7
R
1
dx =⇒ power rule
x2
x
dx =⇒ substitution, let u = x 2 + 1
x 2 +1
1
=⇒ substitution, let u = ln(x)
x ln(x) dx
R
2xe 3 ln(x) dx =⇒ rewrite as 2x · x 3 dx
8
R
x 3 ln(4x 2 )dx
9
R
ln(x)
dx
x2
=⇒ substitution, let u = x 2
=⇒ by parts, let u = ln(3x), v 0 = 1
=⇒ by parts or rewrite
R
2x 3 ln(2x)dx
=⇒ by parts, let u = ln(x), v =
Pieter Hofstra
MAT1300 Lecture 18
1
x2
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
1
R
2
R
3
R
ln(3x)dx
4
R
5
R
6
R
7
R
1
dx =⇒ power rule
x2
x
dx =⇒ substitution, let u = x 2 + 1
x 2 +1
1
=⇒ substitution, let u = ln(x)
x ln(x) dx
R
2xe 3 ln(x) dx =⇒ rewrite as 2x · x 3 dx
8
R
x 3 ln(4x 2 )dx
9
R
10
R
ln(x)
dx
x2
2
x
e dx
=⇒ substitution, let u = x 2
=⇒ by parts, let u = ln(3x), v 0 = 1
=⇒ by parts or rewrite
R
2x 3 ln(2x)dx
=⇒ by parts, let u = ln(x), v =
Pieter Hofstra
MAT1300 Lecture 18
1
x2
Overview
Integration by Parts
Improper Integrals
Mixed Integral Quiz
x
e x dx
2
xe x dx
=⇒ by parts, let u = x, v 0 = e −x
1
R
2
R
3
R
ln(3x)dx
4
R
5
R
6
R
7
R
1
dx =⇒ power rule
x2
x
dx =⇒ substitution, let u = x 2 + 1
x 2 +1
1
=⇒ substitution, let u = ln(x)
x ln(x) dx
R
2xe 3 ln(x) dx =⇒ rewrite as 2x · x 3 dx
8
R
x 3 ln(4x 2 )dx
9
R
10
R
ln(x)
dx
x2
2
x
e dx
=⇒ substitution, let u = x 2
=⇒ by parts, let u = ln(3x), v 0 = 1
=⇒ by parts or rewrite
R
2x 3 ln(2x)dx
=⇒ by parts, let u = ln(x), v =
=⇒ can’t do this one!
Pieter Hofstra
MAT1300 Lecture 18
1
x2
Overview
Integration by Parts
Improper Integrals
Improper Integrals
Consider y = e −x , and the area under it and above [1, 7].
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
We know how to calculate this are already - it’s just
Pieter Hofstra
MAT1300 Lecture 18
R7
1
e −x dx.
Overview
Integration by Parts
Improper Integrals
R7
We know how to calculate this are already - it’s just 1 e −x dx.
R∞
But suppose what we want is 1 e −x dx, the area under the
graph between x = 1 and x = ∞. But what does this mean?
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
R7
We know how to calculate this are already - it’s just 1 e −x dx.
R∞
But suppose what we want is 1 e −x dx, the area under the
graph between x = 1 and x = ∞. But what does this mean?
R 100
R 1000000 −x
We can calculate 1 e −x dx, 1
e R dx and so on.
∞
Each provides a better approximation to 1 e −x dx.
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
R7
We know how to calculate this are already - it’s just 1 e −x dx.
R∞
But suppose what we want is 1 e −x dx, the area under the
graph between x = 1 and x = ∞. But what does this mean?
R 100
R 1000000 −x
We can calculate 1 e −x dx, 1
e R dx and so on.
∞
Each provides a better approximation to 1 e −x dx.
R ∞ −x
So our intuition should be that 1 e dx is the limit of this.
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
R7
We know how to calculate this are already - it’s just 1 e −x dx.
R∞
But suppose what we want is 1 e −x dx, the area under the
graph between x = 1 and x = ∞. But what does this mean?
R 100
R 1000000 −x
We can calculate 1 e −x dx, 1
e R dx and so on.
∞
Each provides a better approximation to 1 e −x dx.
R ∞ −x
So our intuition should be that 1 e dx is the limit of this.
Definition
Suppose f (x) is continuous on [a, ∞) then we define
Z
∞
Z
f (x)dx = lim
b→∞ a
a
b
f (x)dx
and call this the improper integral of f from a to ∞.
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
R7
We know how to calculate this are already - it’s just 1 e −x dx.
R∞
But suppose what we want is 1 e −x dx, the area under the
graph between x = 1 and x = ∞. But what does this mean?
R 100
R 1000000 −x
We can calculate 1 e −x dx, 1
e R dx and so on.
∞
Each provides a better approximation to 1 e −x dx.
R ∞ −x
So our intuition should be that 1 e dx is the limit of this.
Definition
Suppose f (x) is continuous on [a, ∞) then we define
Z
∞
Z
f (x)dx = lim
b→∞ a
a
b
f (x)dx
and call this the improper integral of f from a to ∞.
So an improper integral is an ordinary integral followed by an
ordinary limit.
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
Evaluate
R∞
1
e −x dx.
Solution.
Z
∞
e
−x
Z
dx
=
1
Pieter Hofstra
lim
b→∞ 1
b
e −x dx
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
Evaluate
R∞
1
e −x dx.
Solution.
Z
∞
e
1
−x
Z
dx
b
e −x dx

b 
= lim −e −x 
b→∞
=
lim
b→∞ 1
1
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
Evaluate
R∞
1
e −x dx.
Solution.
Z
∞
e
1
−x
Z
dx
b
e −x dx

b 
= lim −e −x 
b→∞
1
i
h
−b
= lim −e + e −1
=
lim
b→∞ 1
b→∞
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
Evaluate
R∞
1
e −x dx.
Solution.
Z
∞
e
1
−x
Z
dx
b
e −x dx

b 
= lim −e −x 
b→∞
1
i
h
−b
= lim −e + e −1
=
lim
b→∞ 1
b→∞
=
Pieter Hofstra
1
e
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
Evaluate
R∞
1
e −x dx.
Solution.
Z
∞
e
1
−x
Z
dx
b
e −x dx

b 
= lim −e −x 
b→∞
1
i
h
−b
= lim −e + e −1
=
lim
b→∞ 1
b→∞
=
1
e
So the area under the curve y = e −x and above [1, ∞) is e1 .
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
Calculate
R∞
2
√1 dx.
x
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
R∞
Calculate 2 √1x dx.
Solution. The area in question looks like this:
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
Z
2
∞
1
√ dx
x
Z
=
Pieter Hofstra
lim
b→∞ 2
b
1
√ dx
x
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
Z
2
∞
1
√ dx
x
Z
b
1
√ dx
b→∞ 2
x

b 
1/2 x

= lim 
b→∞ 1/2 =
lim
2
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
Z
2
∞
1
√ dx
x
Z
b
1
√ dx
b→∞ 2
x

b 
1/2 x

= lim 
b→∞ 1/2 2
h √
√ i
= lim 2 b − 2 2
=
lim
b→∞
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
Z
2
∞
1
√ dx
x
Z
b
1
√ dx
b→∞ 2
x

b 
1/2 x

= lim 
b→∞ 1/2 2
h √
√ i
= lim 2 b − 2 2
=
lim
b→∞
= ∞
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
Z
2
∞
1
√ dx
x
Z
b
1
√ dx
b→∞ 2
x

b 
1/2 x

= lim 
b→∞ 1/2 2
h √
√ i
= lim 2 b − 2 2
=
lim
b→∞
= ∞
Thus the area under the curve is infinite. In cases where the limit
doesn’t exist (i.e. is infinite) we say the improper integral diverges.
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
Calculate
R0
−∞ e
x dx.
Solution. Here we approach negative infinity, but we deal with it
in the same way
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
We have
Z
0
x
e dx
Z
=
−∞
Pieter Hofstra
lim
b→−∞ b
0
e x dx
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
We have
Z
0
x
e dx
−∞
Z
0
e x dx
 
0
x 

=
lim e b→−∞
=
lim
b→−∞ b
b
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
We have
Z
0
x
e dx
−∞
Z
0
e x dx
 
0
x 

=
lim e b→−∞
b
i
h
=
lim 1 − e b
=
lim
b→−∞ b
b→−∞
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example (Continued)
We have
Z
0
x
e dx
−∞
Z
0
e x dx
 
0
x 

=
lim e b→−∞
b
i
h
=
lim 1 − e b
=
lim
b→−∞ b
b→−∞
= 1
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Application: present value
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Application: present value
Recall our formula for present value:
Z t1
c(t)e −rt dt
0
where t1 is the number of years, c(t) is the income function and r
is the inflation rate (or interest rate).
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Application: present value
Recall our formula for present value:
Z t1
c(t)e −rt dt
0
where t1 is the number of years, c(t) is the income function and r
is the inflation rate (or interest rate).
Improper integrals are used to compute the present value when we
want the process to stop after t1 years but to continue forever.
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
An alumnus wishes to make a gift to the university for a
scholarship fund. The fund will give out $18,000 per year every
year. Assuming an interest rate of 10% (compounded
continuously), what is the present value of the endowment?
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
An alumnus wishes to make a gift to the university for a
scholarship fund. The fund will give out $18,000 per year every
year. Assuming an interest rate of 10% (compounded
continuously), what is the present value of the endowment?
Solution. Since the endowment should be paying out money
forever, we get that the present value is
Z
∞
c(t)e
0
−0.10t
Z
dt =
lim
b→∞ 0
Pieter Hofstra
b
18000e −0.10t dt
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
An alumnus wishes to make a gift to the university for a
scholarship fund. The fund will give out $18,000 per year every
year. Assuming an interest rate of 10% (compounded
continuously), what is the present value of the endowment?
Solution. Since the endowment should be paying out money
forever, we get that the present value is
Z
∞
c(t)e
0
−0.10t
Z
b
18000e −0.10t dt

b 
18000 −0.10t 

= lim
e
b→∞ −0.10
dt =
lim
b→∞ 0
0
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
An alumnus wishes to make a gift to the university for a
scholarship fund. The fund will give out $18,000 per year every
year. Assuming an interest rate of 10% (compounded
continuously), what is the present value of the endowment?
Solution. Since the endowment should be paying out money
forever, we get that the present value is
Z
∞
c(t)e
0
−0.10t
Z
b
18000e −0.10t dt

b 
18000 −0.10t 

= lim
e
b→∞ −0.10
0
h
i
= lim 180000 − 180000e −0.10b
dt =
lim
b→∞ 0
b→∞
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
An alumnus wishes to make a gift to the university for a
scholarship fund. The fund will give out $18,000 per year every
year. Assuming an interest rate of 10% (compounded
continuously), what is the present value of the endowment?
Solution. Since the endowment should be paying out money
forever, we get that the present value is
Z
∞
c(t)e
0
−0.10t
Z
b
18000e −0.10t dt

b 
18000 −0.10t 

= lim
e
b→∞ −0.10
0
h
i
= lim 180000 − 180000e −0.10b
dt =
lim
b→∞ 0
b→∞
= $180,000
Pieter Hofstra
MAT1300 Lecture 18
Overview
Integration by Parts
Improper Integrals
Example
An alumnus wishes to make a gift to the university for a
scholarship fund. The fund will give out $18,000 per year every
year. Assuming an interest rate of 10% (compounded
continuously), what is the present value of the endowment?
Solution. Since the endowment should be paying out money
forever, we get that the present value is
Z
∞
c(t)e
0
−0.10t
Z
b
18000e −0.10t dt

b 
18000 −0.10t 

= lim
e
b→∞ −0.10
0
h
i
= lim 180000 − 180000e −0.10b
dt =
lim
b→∞ 0
b→∞
= $180,000
Pieter Hofstra
MAT1300 Lecture 18