Take a Diagnostic Exam 35
- Width of a rectangle=
6 − 0
3
=2.
Midpoints arex=1, 3, and 5 and f(1)=2,
f(3)=10 andf(5)=26.
∫ 6
0
f(x)dx≈2(2+ 10 +26)≈2(38)= 76
- The intersection of the circlesr=2 cosθand
r=2 sinθcan be found by adding the area
swept out byr=2 sinθfor 0≤θ≤
π
4
and
the area swept byr=2 cosθfor
π
4
≤θ≤
π
2
.
A=
1
2
∫π/ 4
0
4 sin^2 θdθ+
1
2
∫π/ 2
π/ 4
4 cos^2 θdθ
= 2
∫π/ 4
0
sin^2 θdθ+ 2
∫π/ 2
π/ 4
cos^2 θdθ
=
∫π/ 4
0
(1−cos 2θ)dθ
+
∫π/ 2
π/ 4
(1+cos 2θ)dθ
=
(
θ−
1
2
sin 2θ
)∣∣
∣
∣
π/ 4
0
+
(
θ+
1
2
sin 2θ
)∣∣
∣∣
π/ 2
π/ 4
=
[
π
4
−
1
2
− 0
]
+
[(
π
2
+ 0
)
−
(
π
4
+
1
2
)]
=
π
2
− 1
- Differentiatex= 3 t−t^3 ⇒
dx
dt
= 3 − 3 t^2 and
y= 3 t^2 ⇒
dy
dt
= 6 t. The length of the curve
fromt=0tot=2is
L=
∫ 2
0
√
( 3 − 3 t^2 )^2 +(6t)^2 dt
=
∫ 2
0
√
9 − 18 t^2 + 9 t^4 + 36 t^2 dt
=
∫ 2
0
√
9 + 18 t^2 + 9 t^4 dt
=
∫ 2
0
3
√
( 1 +t^2 )^2 dt= 3
∫ 2
0
(
1 +t^2
)
dt
= 3
(
t+
1
3
t^3
)∣∣
∣∣
2
0
= 3 t+t^3
∣∣ 2
0
=(6+8)−(0)= 14.
Chapter 13
50.
dy
dx
=2 sinx⇒dy=2 sinxdx
∫
dy=
∫
2 sinxdx⇒y=−2 cosx+C
Atx=π,y= 2 ⇒ 2 =−2 cosπ+C
⇒ 2 =(−2)(−1)+C
⇒ 2 = 2 +C= 0.
Thus,y=−2 cosx.
- Amount of water leaked
=
∫ 5
0
10 ln (t+1)dt.
Using your calculator, you obtain
10(6 ln 6−5), which is approximately
57.506 gallons.
52.
dy
dx
=ky⇒y=y 0 ekt
Half-life= 5730 ⇒y=
1
2
y 0
whent= 5730.
Thus,
1
2
y 0 =y 0 ek(5730)⇒
1
2
=e^5730 k.
ln
(
1
2
)
=ln
(
e^5730 k
)
⇒ln
(
1
2
)
= 5730 k
ln 1−ln 2= 5730 k⇒−ln 2= 5730 k
k=
−ln 2
5730
