254 STEP 4. Review the Knowledge You Need to Score High
- ∫ 0
−∞dx
(4−x)^2
=klim→−∞∫ 0kdx
(4−x)^2=klim→−∞− 1
4 −x∣∣
∣∣0k=klim→−∞(
− 1
4 −k+
1
4
)
=1
4
23.
∫ 10lnxdx=limk→ 0 +∫ 1klnxdx=limk→ 0 +[
xlnx−x] 1
k
=limk→ 0 +(− 1 −klnk+k)=− 124.
∫ 2− 2dx
√
4 −x^2= 2
∫ 20dx
√
4 −x^2=2limk→ 2∫k0dx
√
4 −x^2=2limk→ 2[
sin−^1(
x
2)]k0=2limk→ 2[
sin−^1(
k
2)
−sin−^1 (0)]=2limk→ 2 −[
sin−^1(
k
2)]
= 2(
π
2)
=π25.
∫ 8− 1dx√ (^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.)
xf′f decr. incr. decr. incr. decr.–6–5 –1^378rel.
min.rel.
max.rel.
min.rel.
max.–+ – +–Figure 11.9-1