An example from the last lecture revisited
Some homeowner insurance policies include automatic inflation coverage based on the construction cost index (CCI) as published by the National
Department of Commerce. Each year the property insurance coverage is increased by an amount based on the change in CCI. Let f ( t ) be the CCI at time t years since January 1, 2010 and let f (0) = 100. Suppose the construction cost index is rising at a rate proportional to the CCI and the index was e 4 times its starting value on January 1, 2012. Construct and solve the differential equation satisfied by f ( t ). Then determine when the
CCI will reach e
7 times its starting value.
A.
never
B.
in 7 years
C.
in 3.5 years
D.
in 5 years
Math 105 (Section 204)
Improper integrals
2011W T2 1 / 6

Improper integrals
All the definite integrals we have encountered so far are of the form
Z b f ( x ) dx , a where a < b are finite real numbers (in other words, the domain of integration [ a , b ] is an interval of finite length) and f is a continuous, finite-valued function on [ a , b ] (i.e., the function does not blow up inside or near the boundary of the interval).
Sometimes definite integrals can be computed even if either or both of these assumptions fail to hold. Such integrals are called improper .
Math 105 (Section 204)
Improper integrals
2011W T2 2 / 6

Improper integrals: interval of integration infinite
These are integrals of the form
Z
∞ f ( x ) dx , or a
Z b
−∞ f ( x ) dx or
Z
∞
−∞ f ( x ) dx , where a and b are finite and f is continuous on the interval of integration.
Math 105 (Section 204)
Improper integrals
2011W T2 3 / 6
Examples:
1.
2.
Z
∞ arctan x dx
1 x 2 + 1
Z
∞ cos x dx
0

Improper integrals: interval of integration infinite
These are integrals of the form
Z
∞ f ( x ) dx , or a
Z b
−∞ f ( x ) dx or
Z
∞
−∞ f ( x ) dx , where a and b are finite and f is continuous on the interval of integration.
Examples:
1.
Z
∞ arctan x dx
1 x 2 + 1
2.
Z
∞ cos x dx
0
Math 105 (Section 204)
Improper integrals
2011W T2 3 / 6

Infinite interval of integration (ctd)
In each case, the integral is evaluated using a limiting procedure:
Improper integrals as limits of proper integrals
Z
∞
Z
N f ( x ) dx = lim
N →∞ a
Z b
−∞ f ( x ) dx = lim
M →∞ a
Z b
− M f ( x ) dx , f ( x ) dx ,
Z
∞
−∞ f ( x ) dx = lim
M →∞
Z c
− M f ( x ) dx + lim
N →∞
Z
N c f ( x ) dx .
provided the limit on the right exists. If it does, the improper integral is said to converge ; if not, the improper integral diverges . In the third case, both limits need to exist for the integral to converge.
Math 105 (Section 204)
Improper integrals
2011W T2 4 / 6

Improper integral: integrand is unbounded
These are integrals of the form
Z b f ( x ) dx a where a , b are finite but there is one point p ∈ [ a , b ] where f is unbounded.
Definition via limits
1 .
If p = a , define
2 .
If p = b , define
3 .
If a < p < b , define
Z b a
Z b a
Z b f
Z b f ( x ) dx = lim h → 0+ a + h
Z b − h f ( x ) dx .
( x ) dx = lim h → 0+
Z p a f ( x ) dx .
Z b f ( x ) dx = f ( x ) dx + f ( x ) dx .
a a p provided the limits or improper integrals on the right exist.
Math 105 (Section 204)
Improper integrals
2011W T2 5 / 6

Example
Find the value of the three integrals
Z
2
− 2 x 2
2 x
− 1 dx
Z
3
− 2 x 2
2 x
− 1 dx
Z
3
2 x 2
2 x
− 1 dx .
A.
0, ln(8 / 3), ln(8 / 3)
B.
0, ∞ , ln(8 / 3)
C.
∞ , ln(8 / 3), ln(8 / 3)
D.
∞ , ∞ , ln(8 / 3)
Math 105 (Section 204)
Improper integrals
2011W T2 6 / 6