Unformatted text preview: lim w /2 g. Þ /2 x cos x sin x
x2 0 sin x dx lim 2
w /2
cos x 2. 2 cos w dx This integral is
/2 Þ
0 w lim w h. Þ Ý
1 x cos x sin x dx lim sin /2
w 0
/2
x2 sin w
w 2
1 e x sin xdx This integral is
lim
Þ 1 e x sin xdx w Ý
w lim
wÝ 2. 1 e w cos w
2
cos 1 sin 1
2e
2e 1e
2 w sin w 1 e 1 cos 1
2 1 e 1 sin 1
2 a. Prove that the integral Þ0 1 converges when p 1 and diverges when p
As long as p 1 we have 1 dx
xp 1. Þ0 1 dx lim Þ 1 1 dx lim
1
p
xp
1p
w0
w0
wx
if p 1 and is Ý if p 1. We see also that
1 This limit is 1
1p Þ0 1 1 dx lim log 1
x
w0 351 1
1 log w Ý. p w1 p . b. Prove that the integral Þ1
converges when p 1 and diverges when p
As long as p 1 we have Ý 1 dx
xp 1. Ý Þ1 This limit is 1
p1 1 dx lim Þ w 1 dx lim
1 w1
w Ý 1 xp
wÝ 1
xp
p
if p 1 and is Ý if p 1. We see also that
Ý Þ1 1 dx lim log w
x
wÝ 1 p 1 . p log 1 Ý. 3. Prove that the integral Þ2 Ý 1
x log x
1. dx p is convergent when p 1 and divergent when p
As long as p 1 we have
Ý
w
1
1
lim
Þ 2 x log x p dx w Ý Þ 2 x log x p dx 1 log w 1
1p
when p 1 and is Ý when p 1.
wÝ
lim This limit is p11 log 2
We see also that 1p Ý Þ2 1 dx lim log log w
wÝ
x log x 1 p 1 p log 2 1p log log 2 Ý. 4. Interpret the integral Þ0 2 1
dx
x 1 1/3
as the sum of two improper integrals and evaluate it.
1
2
1
1
dx and Þ
dx and can easily be seen to be 0.
The two integrals are Þ
1/3
1/3
0 1 x1 x1 5. Prove that if f is bounded on an interval a, b and is improper Riemann integrable on a, b then f is Riemann
integrable on a, b and Þa b f Þ a f.
b 0. Using the fact that f To show that f satisfies the second criterion for integrability, suppose that
is integrable on the interval a, b 2 choose a partition
P x 0 , x 1 , , x n
of the interval a, b 2 such that if
E w P, f x
2
2
then m E 2 . Now we extend the partition P to make a partition Q of the interval a, b by adding
the point b. Thus
Q x 0 , x 1 , , x n , b
and since
x x a, b a, b w Q, f x
2
the measure of the latter set is less than .
Now to show that 352 2 EÞ b 2 ,b lim Þa f Þa f Þw f w wb we need only note that if a b w b then Þa f Þa f
b w b and that the latter expression approaches 0 as w supf  b w b. 6. In the discussion of improper integrals that appears above we used the words
... Somewhat less precisely, we also say that the integral Þ b
a f x dx is convergent. ... Why is the statement that the integral Þ f x dx is convergent less precise than the statement that f is
a
improper Riemann integrable on a, b ? Hint: In our study of infinite series we made a careful distinction
Ý
between the symbols a n and n1 a n .
b The symbol Þ b f x dx stand for the limit a Þ a f x dx
w lim wb which is a number. When we say that Þ b
a f x dx is convergent we do not really mean what we are saying. Literally, the statement that Þ f x dx is convergent is the statement that the value of the
a
limit
b Þ a f x dx
w lim wb is convergent. Thus, if the l...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

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