Page 258 - Calculus Workbook For Dummies

242 Part IV: Integration and Infinite Series

2. Turn into limits.
2 b
= lim # dx + lim # dx
2
2
a " 1 + x x - 1 b " 3 x x - 1
a 2
3. Integrate.
2 b
= lim arc secx + lim arc secx@
@
6
6
a " 1 + a b " 3 2
= lim arc sec2 - arc seca + lim arc secb - arc sec2h
h
^
^
a " 1 + b " 3
π
0
= arc sec2 - + - arc sec2
2
π
=
2
3
A # 1 dx 3=
x
1
c
1. Turn into a limit: lim # 1 dx.
c " 3 x
1
c
=
2. Integrate and finish: lim lnx = lim lnc - ln1 = 3.
^
6
@
h
c " 3 1 c " 3
3
*B # 1 arctanx dx = 3
x
1 3
No work is required for this one, “just” logic. You know from problem 26 that # 1 dx 3= .
x
3 3 3 1
Now, compare # 1 arctanx dx to # 1 dx. But first note that because # 1 dx equals infinity,
x
x
x
1 1 1
3 3 3
so will # 1 x dx, # 1 x dx, or # 1 dx, because the area under 1 x from 1 to any other number
x
,
10 100 , 1 000 000
must be finite.
1 1
From 3 to 3, arctanx $ 1, and therefore arctanx $ 1, and thus x arctanx $ x . Finally,
3
because # 1 dx 3= and because between 3 and 3, 1 arctanx is always equal to or
x x
3 3 3
1
greater than , # 1 arctanx dx must also equal 3 and so, therefore, does # 1 arctanx dx.
x x x
3 1
Aren’t you glad no work was required for this problem?
3
*C # 1 dx is undefined.
x
- 3
Quadrupely improper!
3 - 1 0 1 3
1. Split into four parts: # 1 dx = # 1 dx + # 1 dx + # 1 dx + # 1 dx.
x
x
x
x
x
- 3 - 3 - 1 0 1
- 1 b 1 d
2. Turn into limits: = lim # dx + lim # dx + lim # dx + lim # dx .
x
x
x
x
a " 3- b " 0 - c " 0 + d " 3
a - 1 c 1
3. Integrate:
- 1 b 1 d
= lim ln x B + lim ln x B + lim ln x B + lim ln x B
8
8
8
8
-
a " 3 a b " 0 - - 1 c " 0 + c d " 3 1
= lim ln1 - ln a + lim ln b - ln1 + lim ln1 - ln c j + lim ln d - ln1j
`
`
j
`
`
j
a " 3 b " 0 - c " 0 + d " 3
3 +
4. Finish: = - 3 + - h 3 + 3. Therefore, the limit doesn’t exist, and the definite integral is
^
thus undefined.
1
If you look at the graph of y = x , its perfect symmetry may make you think that
3
# 1 dx would equal zero. But — strange as it seems — it doesn’t work that way.
x
- 3