254 STEP 4. Review the Knowledge You Need to Score High
- ∫ 0
−∞
dx
(4−x)^2
=klim→−∞
∫ 0
k
dx
(4−x)^2
=klim→−∞
− 1
4 −x
∣∣
∣∣
0
k
=klim→−∞
(
− 1
4 −k
+
1
4
)
=
1
4
23.
∫ 1
0
lnxdx=limk→ 0 +
∫ 1
k
lnxdx
=limk→ 0 +
[
xlnx−x
] 1
k
=limk→ 0 +(− 1 −klnk+k)=− 1
24.
∫ 2
− 2
dx
√
4 −x^2
= 2
∫ 2
0
dx
√
4 −x^2
=2limk→ 2
∫k
0
dx
√
4 −x^2
=2limk→ 2
[
sin−^1
(
x
2
)]k
0
=2limk→ 2
[
sin−^1
(
k
2
)
−sin−^1 (0)
]
=2limk→ 2 −
[
sin−^1
(
k
2
)]
= 2
(
π
2
)
=π
25.
∫ 8
− 1
dx
√ (^3) x=
∫ 0
− 1
dx
√ (^3) x+
∫ 8
0
dx
√ (^3) x
=limk→ 0 −
∫k
− 1
dx
√ (^3) x+limk→ 0 +
∫ 8
k
dx
√ (^3) x
=limk→ 0 −
[
3
2
x^2 /^3
]k
− 1
+limk→ 0 −
[
3
2
x^2 /^3
] 8
k
=limk→ 0 −
[
3
2
k^2 /^3 −
3
2
]
+limk→ 0 +
[
12
2
−
3
2
k^2 /^3
]
=
9
2
11.9 Solutions to Cumulative Review Problems
- Asx→−∞, x=−
√
x^2.
xlim→−∞
√
x^2 − 4
3 x− 9
=xlim→−∞
√
x^2 − 4 /−
√
x^2
(3x−9)/x
=xlim→−∞
−
√
(x^2 −4)/x^2
3 −(9/x)
=xlim→−∞
−
√
1 −(4/x)^2
3 − 9 /x
=
−
√
1 − 0
3 − 0
=−
1
3
- y=ln
∣∣
x^2 − 4
∣∣
,
dy
dx
=
1
(x^2 −4)
(2x)
dy
dx
∣
∣∣
∣x= 3 =
2(3)
(3^2 −4)
=
6
5
- (a) (See Figure 11.9-1.)
x
f′
f decr. incr. decr. incr. decr.
–6–5 –1^378
rel.
min.
rel.
max.
rel.
min.
rel.
max.
–+ – +–
Figure 11.9-1