34 STEP 2. Determine Your Test Readiness
40.
∫∞
0
e−xdx=klim→∞
∫k
0
e−xdx=klim→∞[−e−x]k 0
=klim→∞
[
−e−k+e^0
]
=klim→∞
[
−e−k+ 1
]
= 1
Chapter 12
- Total distance=
∫ 4
0
v(t)dt+
∣∣
∣
∣
∫ 6
4
v(t)dt
∣∣
∣
∣
=
1
2
(4)(20)+
∣∣
∣∣^1
2
(2)(−10)
∣∣
∣∣
= 40 + 10 =50 feet
42.
∫ 5
− 1
f(x)dx=
∫ 1
− 1
f(x)dx+
∫ 5
1
f(x)dx
=−
1
2
(2)(1)+
1
2
(2+4)(1)
=− 1 + 3 = 2
- To find points of intersection, set
y=x^2 −x= 0
⇒x(x−1)= 0 ⇒x=0orx=1.
See Figure DS-12.
01
y
x
y=x^2 – x
Figure DS-12
Area=
∣
∣∣
∣
∫ 1
0
(
x^2 −x
)
dx
∣
∣∣
∣=
∣∣
∣∣
∣
x^3
3
−
x^2
2
] 1
0
∣∣
∣∣
∣
=
∣∣
∣∣
(
1
3
−
1
2
)
− 0
∣∣
∣∣=
∣∣
∣∣−^1
6
∣∣
∣∣
=
1
6
44.
∫k
−k
f(x)dx= 0 ⇒ f(x) is an odd function,
i.e., f(x)=−f(−x). Thus the graph in choice
(D) is the only odd function.
- Area=
∫k
1
√
xdx=
∫k
1
x^1 /^2 dx
=
[
x^3 /^2
3 / 2
]k
1
=
[
2
3
x^3 /^2
]k
1
=
2
3
k^3 /^2 −
2
3
(1)^3 /^2
=
2
3
k^3 /^2 −
2
3
=
2
3
(
k^3 /^2 − 1
)
.
Since A=8, set
2
3
(
k^3 /^2 − 1
)
= 8 ⇒k^3 /^2 − 1
= 12 ⇒k^3 /^2 =13 ork= 132 /^3.
- See Figure DS-13.
Figure DS-13
Using the [Intersection] function of the
calculator, you obtain the intersection
points atx= 0 .785398, 3.92699, and
7.06858.
Area=
∫ 3. 92699
0. 785398
(sinx−cosx)dx
+
∫ 7. 06858
3. 92699
(cosx−sinx)dx
= 2. 82843 + 2. 82843 ≈ 5. 65685 ≈ 5. 657
You can also find the area by:
Area=
∫ 7. 06858
. 785398
∣∣
sinx−cosx
∣∣
dx
≈ 5. 65685 ≈ 5. 657.