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Material Type: Notes; Class: CALCULUS II; Subject: Mathematics; University: University of Kentucky; Term: Spring 2004;
Typology: Study notes
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Can an region have infinite perimeter and finite area? If so, we can color it in, but we cannot draw its outline.
Example. Find the area bounded by x = 1, y = 0 and y = e−x.
Solution. This is a new problem–until now, we have found the area of regions which were of finite length in every direction. We attempt to solve this problem, by first finding the area bounded by x = 1, y = 0, y = e−x^ and x = N. This is
∫ (^) N
1
e−x^ dx =
e
− e−N^.
If we take the limit as N goes to infinity, we expect to obtain a reasonable value for the infinite area.
area = lim N →∞
e
− e−N^ =
e
Definition. If a is a real number and f is a function which is Riemann integrable on each interval [a, N ] for N > a, then we define the improper integral as
∫ (^) ∞
a
f (t) dt = lim N →∞
∫ (^) N
a
f (t) dt.
If this limit exists and is finite, we say the improper integral is convergent. If the limit is infinite or does not exist, we say the improper integral is divergent.
Exercise. Write out the corresponding definition for improper integrals at −∞.
Exercise. Re-read the definition of the Riemann integral
∫ (^) b a f^ (t)^ dt^ and explain what difficulties would arise if we let b = ∞ in this definition.
Example. How should we treat the integral ∫ (^) ∞
−∞
x dx?
Solution. If we break the integral up as follows:
∫ (^) ∞
−∞
x dx =
∫ (^0)
−∞
x dx +
∫ (^) ∞
0
x dx,
then each of the integrals on the right is divergent.
Example. Evaluate the following improper integrals, or state if the integral diverges.
∫ (^0)
−∞
e^2 x^ dx
∫ (^0)
−∞
e−^2 x^ dx
∫ (^) ∞
1
1 + x^2
dx.
Example. Determine for which p, the integral
∫ (^) ∞
1
xp^
dx
converges.
Solution. If p 6 = 1, then
∫ (^) ∞
1
xp^
dx = lim N →∞
∫ (^) N
1
x−p^ dx = lim N →∞
x−p+ −p + 1
∣∣ ∣∣ ∣
x=N
x=
= lim N →∞
N −p+ −p + 1
−p + 1
This limit will exist and be finite if the exponent of N is negative. That is the integral converges if p > 1. The limit will be infinite and the integral diverges if p < 1. The case p = 1 is treated separately because it involves the logarithm. If p = 1, then the integral is divergent.