Page 257 - Calculus Workbook For Dummies

241
Chapter 12: Who Needs Freud? Using the Integral to Solve Your Problems

1 /x
x
3. I give you permission to pull the limit to the outside: lny = lim ln 1 + h k.
a
^
x " 0
1
x
4. Use the log of a power rule: lny = limc x ln 1 + hm.
^
x " 0
x
ln 1 + h
^
5. You’ve got a 3 $ 0 case so rewrite: lny = lim .
x " 0 x
0
6. You get — I’m down with it.
0 1
1 + x
7. Replace with derivatives: lny = lim = 1.
x " 0 1
8. Your original limit equals y, so you’ve got to solve for y.
lny = 1
y = e
1
w Evaluate # dx = - 18 .75.
5 x
- 32 1 0 1
1. Undefined at x = 0, so break in two: # dx = # dx + # dx .
5 x 5 x 5 x
- 32 - 32 0
a 1
2. Turn into limits: lim # dx + lim # dx .
=
a " 0 - 5 x b " 0 + 5 x
- 32 b
a 1
5 / 4 5 5 / 4 5 5 5
=
3. Integrate: lim; x E + lim; x E = 0 - $ 16 + - 0 = - 18 .75.
a " 0 - 4 b " 0 + 4 4 4
- 32 b
6
x # x lnx dx 18 ln6 - 9
=
0
1. The integral is improper because it’s undefined at x = 0, so turn it into a limit:
6
= lim # x lnx dx
c " 0 +
c
2. Integrate by parts. Hint: lnx is L from LIATE. You should obtain:
6
1 2 1 2
= lim; x lnx - x E
c " 0 + 2 4
c
1 1 2 1 2
= limc $ 36 ln6 - - c lnc + c m
9
c " 0 + 2 2 4
1 2
9
= 18 ln6 - - lim c _ lnci
2 c " 0 +
- 3
3. Time to practice L’Hôpital’s Rule. This is a 0 $ ^ - 3h limit, so turn it into a 3 one:
1 lnc
= 18 ln6 - - lim
9
2 c " 0 + 1
c 2
4. Replace numerator and denominator with derivatives and finish:
1
1 c 1 c 2
9
= 18 ln6 - - lim = 18 ln6 - - lim - m = 18 ln6 - 9
9
c
2 c " 0 + 2 2 c " 0 + 2
- 3
c
3
y # dx = π
2
1 x x - 1 2
This is a doubly improper integral because it goes up and right to infinity. You’ve got to split it
up and tackle each infinite impropriety separately.
1. It doesn’t matter where you split it up; how about 2, a nice, easy-to-deal-with number.
2 3
= # dx + # dx
2
2
x x - 1 x x - 1
1 2