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Unformatted text preview: imit is 2 then we should be saying that the number 2 is convergent. Of
course such an interpretation is not intended. Instead, we mean that the limit exists and is finite. In
b
that sense, saying that Þ f x dx is convergent is not as precise as saying that f is improper
a
Riemann integrable on a, b . Some Exercises on the Comparison Test for Integrals
1. Determine the convergence or divergence of the following integrals:
a. Þ1 Ý x2 x
dx
x1 Since
lim
xÝ
and since Þ Ý
1 1
x 3/2 x2 x
x1 1
1
x 3/2 dx is convergent, the integral Þ Ý
1 x2 x
dx is convergent.
x1 Þ0 1 b. 1 dx
x x2
Since
1
x x 2
1
x 0
x
1 lim and since Þ
c. Þ1 Ý 1
0 1
x dx is divergent, the integral Þ sin 2 x dx
x2 353 0 1 1 dx is divergent.
x x2 Since
sin 2 x
x2 0
for all x 1 and since the integral Þ 1
x2 dx is convergent, the integral Þ Ý1
1 x2 convergent.
d. Þ0 Ý sin 2 x dx
x2 x We break thisintegral into the two parts Þ Ý
1 sin 2 x dx is
x2 sin 2 x dxand Þ Ý sin 2 x dx.
1
x2 x
x2 x 1
0 Since
sin 2 x
x2 x
lim 1
x 0
x and since the integral Þ 1 1
x 0 1 dx is convergent, the integral Þ 1
0 sin 2 x dx is convergent.
x2 x Since
sin 2 x
x2 x 0
whenever x 1 and since the integral Þ Ý convergent. Therefore the given integral Þ
e. Þ0 /2 1
x 5/2 1 Ý
0 1
x 5/2 dx is convergent, the integral Þ Ý
1 sin 2 x dx is
x2 x sin 2 x dx is convergent.
x2 x tan x dx From the fact that
lim x /2 cos x 1
x
2 we see that
tan x
1
x lim
2 x /2 and since the integral Þ /2 1
2 0 x dx is convergent, the integral Þ /2
0 tan x dx is convergent. Alternatively, we could have evaluated this integral directly. It is an elementary integral that can
be evaluated by making the substitution u tan x and then breaking the integral down with
partial fractions. Þ1 2 f. 1 dx
log x
Since
1
log x
lim 1
x 1
x1 and since theintegral Þ
g. /2 Þ0 2
1 1
x1 1 dx is divergent, the integral Þ 2
1 1 dx is divergent.
log x log sin x dx We observe that log sin x 0for each x 0,
2 and so the comparision test can’t be applied to this integral as it stands. On the other hand, we can apply the comparison test to the integral
/2
log sin x dx. Since Þ0 354 lim log sin x x0 and since the integral Þ
therefore the integral Þ
h. Þ2 Ý 1
log x log x 1 0
/2 0 1
x 1
x 0 dx is convergent, the integral Þ /2 log sin x dx is convergent and 0 log sin xdx is convergent. dx Hint: This problem is the integral analogue of the problems that occurred earlier on the
comparison test for series.
For x sufficiently large we have
1
1
exp log x log log x
log x log x
Since the integral Þ
i. Þ3 Ý 1
log x 2 1
dx
log log x log x
For x sufficiently large we have
1
1
exp log x log log log x
log log x log x log x dx is convergent. Ý1
2
x2 dx is convergent, the integral Þ 1
exp 2 log x
Ý 1
log log x 2 log x 1.
x2
dx is convergent. Ý Þ 30 1
dx
log log log x log x
For x sufficiently large we have
1
1
exp log x log log log log x
log log log x log x
Since the integral Þ k. dx is convergent...
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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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