Improper Integrals: Over Infinite Intervals and Divergent Integrands, Exercises of Elementary Mathematics

In this lecture, the concept of improper integrals is explored, focusing on integrals over infinite intervals and integrands that become infinite. The lecture explains how to define improper integrals as limits and discusses the convergence and divergence of such integrals. Geometric and algebraic interpretations are provided with examples.

Typology: Exercises

2011/2012

Uploaded on 08/08/2012

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Lecture 39
Improper Integral
In this lecture we will study
Integrals over Infinite Interval
Integrals whose Integrands become Infinite
Integrals over Infinite Interval
• As we saw be fore, for a given conti nuous functi on f,
the definite integral is
It is assumed that the interval [a,b] is finite.
• W hat if we look at [a, +¥) and the corresponding
integral
ò
b
a
dxxf )(
ò
a
dxxf )(
• In this case we define what is called an improper
integral over an infinite interval.
What does it mean to integrate all the
way to + ¥?The answer will be clear if we define
this integral as a limit in the following way.
What this does is to first turn the integral into
more familiar form of over a finite interval, and then
we let the upper limits of the interval approach 0
and see what happen to the answer we had got
earlier.
òò ®
=
l
a
l
a
dxxfdxxf )(lim)(
• If this limi ts exists, t hen we say that t he Improper
Integral Converges, and the value of the limit is
assigned to the integral.
• If the limit does not exist, then we say that t he
Improper Integral Diverges, and no finite value is
assigned.
• Lets get some geometric ideas t o understand
things.
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Lecture 39

Improper Integral

In this lecture we will study

  • Integrals over Infinite Interval
  • Integrals whose Integrands become Infinite

Integrals over Infinite Interval

  • As we saw before, for a given continuous function f , the definite integral is

It is assumed that the interval [a,b] is finite.

  • What if we look at [a, +¥ ) and the corresponding integral

ò

b

a

f ( x ) dx

ò

a

f ( x ) dx

  • In this case we define what is called an improper integral over an infinite interval. What does it mean to integrate all the way to + ¥? The answer will be clear if we define this integral as a limit in the following way.

What this does is to first turn the integral into more familiar form of over a finite interval, and then we let the upper limits of the interval approach 0 and see what happen to the ans wer we had got earlier.

ò (^) ®+¥ ò

l

a

l a

f ( x ) dx lim f ( x ) dx

  • If this limits exists, then we say that the Improper Integral Converges , and the value of the limit is assigned to the integral.
  • If the limit does not exist, then we say that the Improper Integral Diverge s , and no finite value is assigned.
  • Lets get some geometric ideas to understand things.

Improper Integral

What's happening here in these examples?

  • In the first case with f(x) = 1/ x^2, we get fin ite answer

over the same interval?

  • In the second case we get a divergent limit and so

we were unable to calculate the area under graph of

1/x over [1,+ ¥)

Look at the graphs of the two functions

  • We can see geometrically that the graph of 1/ x^2 is

approaching y = 0 much faster than that of 1/ x.

  • Algebraically also, if you divide 1 by the square of a

number, the result is much smaller than if you divide by

the number itself.

for example ½ > ¼ and 1/8 > 1/

  • SO the idea is that as x goes to + ¥, 1/x^2 goes

to 0 MUCH faster than 1/x, so much so that when

we attempt to find the area under the graph over

the infinite interval [ 1, + ¥) the first is convergent,

and the other is divergent.

Lets think about Volume of second

case.

  • Lets rotate the graphs of 1/x over [1. + ¥) around the x-axis.

We get solid of revolution that look like funnels with no lo wer point as shown in figure below.

  • We would like to find the volume o f this solid.
  • The cross section is a disk with radius f(x).
  • For f(x) = 1/ x, we get for volu me
    • So we can find out how much paint can be held in this solid, but we cannot paint the inside of the solid!!!