Math 220
November 1
I. A ball was thrown straight up in the air on Mars with a speed of 20 ft/s. If the acceleration
due to gravity on Mars is 10 ft/s2 , when will the ball hit the ground?
II. What constant acceleration is required to increase the speed of a car from 35 mi/h
to 55 mi/h in 5 seconds?
III. What is the constant acceleration needed to change the speed of a car from v0 mi/h
to v1 in t seconds?
IV. The acceleration of a car in m/s2 is given by:
(
0.6t if 0 ≤ t ≤ 10
a(t) =
0
if t > 10.
If the car is stopped at t = 0, how long will it take to travel 1 km?
V. A car is traveling at 60 mi/h when the brakes are fully applied, producing a constant
deceleration of 15 ft/s2 . What is the distance traveled before the car comes to a stop?
VI. The speed of a runner increased steadily during 3 seconds of a race. Her speeds are
given below. Find lower and upper estimates for the distance that she traveled during these
three seconds.
t(s)
0 0.5 1.0 1.5 2.0 2.5 3.0
v(ft/s) 20 18 14 10 8
4
0
VII. Use the form of the definition of the inegral given in Theorem 4 to evaluate the integral.
1.
3
Z
5dx
0
2.
Z
2
(3 − 2x − 3x2 )dx
0
3.
Z
b
5dx
a
4.
Z
b
xdx
a
1
5.
b
Z
x2 dx
a
6.
b
Z
x3 dx
a
7.
b
Z
(cx3 + dx2 + ex + f )dx
a
VIII. Evaluate the following integrals
1.
π/2
Z
sin(x) + cos(2x)dx
−π/2
2.
Z
√
0
(2 +
4 − x2 ) +
p
4 − (x + 2)2 dx
−2
3.
Z
2
|x + 1|dx
−1
4.
Z
2
|x + 1| − 2|x − 1|dx
−1
5.
Z
2
|x2 + 1|dx
0
2
1
Solutions
I. A ball was thrown straight up in the air from the ground on Mars with a speed of 20 ft/s.
If the acceleration due to gravity on Mars is 10 ft/s2 , when will the ball hit the ground?
Answer:
We assume that up is positive.
a(t) = −10
v(t) = −10t + v0
From the problem we have v0 = 20
s(t) = −5t2 + 20t + s0
Since the ball starts on the ground s0 = 0
s(t) = −5t2 + 20t = 5t(4 − t)
It will hit the ground when s(t) = 0 when t = 4.
The ball will hit the ground 4 seconds after it is thrown.
II. What constant acceleration is required to increase the speed of a car from 35 mi/h
to 55 mi/h in 5 seconds?
Answer:
5
hours
5 seconds = 3600
We assume acceleration is constant a(t) = a
v(t) = at + v0
We’re given v0 = 35.
v(t) = at + 35
v(
5
5
)=a
+ 35
3600
3600
5
55 = a
+ 35
3600
5
20 = a
3600
14400 = a
The car must accelerate at 14400 mi/h2 or 4 mi/h/s.
III. What is the constant acceleration needed to change the speed of a car from v0 mi/h
to v1 in t seconds?
Answer:
v(t) = at + v0
3
v1 = at + v0
v1 − v0 = at
v1 − v0
=t
a
It will take
v1 −v0
a
hours and therefore
(v1 −v0 )3600
a
seconds.
IV. The acceleration of a car in m/s2 is given by:
(
0.6t if 0 ≤ t ≤ 10
a(t) =
0
if t > 10.
If the car is stopped at t = 0, how long will it take to travel 1 km?
Answer:
(
0.3t2 + C1
v(t) =
C2
if 0 ≤ t ≤ 10
if t > 10.
Since the car is stopped at t = 0, we have v(0) = 0 which gives us C1 = 0. Thus
(
0.3t2 if 0 ≤ t ≤ 10
v(t) =
C2
if t > 10.
Velocity is continuous therefore.
lim 0.3t2 = lim+ C2
t→10−
t→10
30 = c2
Thus
(
0.3t2
v(t) =
30
if 0 ≤ t ≤ 10
if t > 10.
Now we find position.
(
0.1t3 + C1
s(t) =
30t + C2
4
if 0 ≤ t ≤ 10
if t > 10.
We assume s(0) = 0 which gives us C1 = 0 Position is continuous which gives us.
lim− 0.1t3 = lim+ 30t + C2
t→10
t→10
100 = 300 + C2
−200 = C2
(
0.1t3
if 0 ≤ t ≤ 10
s(t) =
30t − 200 if t > 10.
Now we solve for t in s(t) = 1000.
30t − 200 = 1000
30t = 1200
t = 40
So it will take 40 seconds for the car to travel 1 km.
V. A car is traveling at 60 mi/h when the brakes are fully applied, producing a constant
deceleration of 15 ft/s2 . What is the distance traveled before the car comes to a stop?
mi 5280f t 1h
60 · 5280f t
5280f t
264f t
88f t
mi
= 60
=
=
=
=
h
h 1mi 3600s
3600s
60s
3s
s
The velocity and position functions for the car are:
60
v(t) = −15t + 88
s(t) = −15t2 /2 + 88t
So we need to solve
v(t) = −15t + 88 = 0 =⇒ t =
88
15
It will take t = 88
seconds for the car to stop. The car will travel s( 88
)=
15
15
88
88·44
3872
(−44 + 88) = 15 = 15
15
The car will travel 3872
feet.
15
−15 882
2 152
+ 88 88
=
15
VI. The speed of a runner increased steadily during 3 seconds of a race. Her speeds are
given below. Find lower and upper estimates for the distance that she traveled during these
three seconds.
5
t(s)
0 0.5 1.0 1.5 2.0 2.5 3.0
v(ft/s) 20 18 14 10 8
4
0
The lower estimate:
1
1
1
1
1
1
LE = v(0.5) + v(1) + v(1.5) + v(2) + v(2.5) + v(3)
2
2
2
2
2
2
1
= (v(0.5) + v(1) + v(1.5) + v(2) + v(2.5) + v(3))
2
1
= (18 + 14 + 10 + 8 + 4 + 0)
2
1
= (54)
2
= 27
The upper estimate:
1
1
1
1
1
1
U E = v(0) + v(0.5) + v(1) + v(1.5) + v(2) + v(2.5)
2
2
2
2
2
2
1
= (v(0) + v(0.5) + v(1) + v(1.5) + v(2) + v(2.5))
2
1
= (20 + 18 + 14 + 10 + 8 + 4)
2
1
= (74)
2
= 37
The runner traveled at least 27 feet but less than 37 feet.
VII. Use the form of the definition of the inegral given in Theorem 4 to evaluate the integral.
1.
Z
3
5dx
0
6
Answer:
Z
3
5dx = lim
n→∞
0
= lim
n→∞
= lim
n→∞
= lim
n→∞
n
X
f (xi )∆x
i=1
n
X
i=1
n
X
3 3
f (0 + i )
n n
5
i=1
3
n
n
X
15
i=1
n
n
15 X
= lim
1
n→∞ n
i=1
15
n
n
= lim 15
= lim
n→∞
n→∞
= 15
2.
Z
2
(3 − 2x − 3x2 )dx
0
7
Answer:
Z
2
2
(3 − 2x − 3x )dx = lim
0
n→∞
= lim
n→∞
= lim
n→∞
n
X
f (xi )∆x
i=1
n
X
i=1
n
X
2 2
f (0 + i )
n n
2 2
2
(3 − 2(i ) − 3(i )2 )
n
n n
i=1
n
X
6
8
24
= lim
( − i 2 − i2 3 )
n→∞
n
n
n
i=1
n
n
n
X
X
6 X 8
24
−
= lim
i 2−
i2 3
n→∞
n i=1 n
n
i=1
i=1
n
n
n
6X
24 X 2
8 X
i− 3
i
1− 2
n→∞ n
n
n
i=1
i=1
i=1
= lim
8 n(n + 1) 24 n(n + 1)(2n + 1)
6
n− 2
− 3
n→∞ n
n
2
n
6
=6−4−8
= −6
= lim
3.
Z
b
5dx
a
8
Answer:
Z
b
5dx = lim
a
n→∞
= lim
n→∞
= lim
n→∞
n
X
f (xi )∆x
i=1
n
X
i=1
n
X
i=1
f (a + i
5
b−a b−a
)
n
n
b−a
n
n
b−aX
= lim 5
1
n→∞
n i=1
b−a
n
n→∞
n
= lim 5(b − a)
= lim 5
n→∞
= 5(b − a)
4.
Z
b
xdx
a
9
Answer:
Z
b
xdx = lim
a
n→∞
= lim
n→∞
= lim
n→∞
= lim
n→∞
n
X
f (xi )∆x
i=1
n
X
i=1
n
X
f (a + i
(a + i
i=1
b−a b−a
)
n
n
b−a b−a
)
n
n
n
X
a(b − a)
n
i=1
+i
(b − a)2
n
n
n
a(b − a) X
(b − a)2 X
= lim
1+
i
n→∞
n
n
i=1
i=1
(b − a)2 n(n + 1)
a(b − a)
n+
n→∞
n
n
2
(b − a)2
= a(b − a) +
2
2
2ab a2
b
+
= ab − a2 + −
2
2
2
b 2 a2
−
=
2
2
= lim
5.
Z
b
a
Answer:
10
x2 dx
Z
b
2
x dx = lim
a
n→∞
= lim
n→∞
= lim
n→∞
n
X
i=1
n
X
f (xi )∆x
f (a + i
i=1
n
X
i=1
n
X
(a + i
b−a b−a
)
n
n
b − a 2b − a
)
n
n
2
a(b − a)
2 (b − a) b − a
(a + i2
= lim
+i
)
n→∞
n
n2
n
i=1
= lim
n→∞
n
X
i=1
2
(a2
3
(b − a)
a(b − a)2
2 (b − a)
+ i2
+
i
)
n
n2
n3
a(b − a)2 n(n + 1) (b − a)3 n(n + 1)(2n + 1)
b−a
n+2
+
)
n→∞
n
n2
2
n3
6
(b − a)3
= a2 (b − a) + a(b − a)2 +
3
b 3 a3
−
=
3
3
= lim (a2
6.
Z
b
a
Answer:
11
x3 dx
Z
b
3
x dx = lim
n→∞
a
= lim
n→∞
= lim
n→∞
n
X
i=1
n
X
f (xi )∆x
f (a + i
i=1
n
X
i=1
n
X
(a + i
b−a b−a
)
n
n
b − a 3b − a
)
n
n
2
3
a2 (b − a)
2 a(b − a)
3 (b − a) b − a
(a + i3
= lim
+ 3i
+i
)
n→∞
n
n2
n3
n
i=1
= lim
n→∞
n
X
i=1
3
(a3
3
4
b−a
a2 (b − a)2
2 a(b − a)
3 (b − a)
+ i3
+
3i
+
i
)
n
n2
n3
n4
a2 (b − a)2 n(n + 1)
a(b − a)3 n(n + 1)(2n + 1) (b − a)4 (n(n + 1)
b−a
n+3
+
3
+
n→∞
n
n2
2
n3
6
n4
22
4
3
(b − a)
= (a3 (b − a) + a2 (b − a)2 + a(b − a)3 +
)
2
4
b 4 a4
=
−
4
4
= lim (a3
7.
b
Z
(cx3 + dx2 + ex + f )dx
a
Answer:
Z
b
3
Z
2
b
3
(cx + dx + ex + f )dx = c
x dx + d
a
a
4
= c(
Z
b
2
x dx + e
a
Z
b
Z
xdx + f
a
b
dx
a
b
a4
b 3 a3
b 2 a2
− ) + d( − ) + e( − ) + f (b − a)
4
4
3
3
2
2
VIII. Evaluate the following integrals
1.
Z
π/2
sin(x) + cos(2x)dx
−π/2
12
Answer:
Z
π/2
Z
π/2
sin(x) + cos(2x)dx =
Z
π/2
sin(x)dx +
−π/2
−π/2
cos(2x)dx
−π/2
=0+0
=0
2.
Z
√
0
(2 +
4 − x2 ) +
p
4 − (x + 2)2 dx
−2
Answer:
Z
0
(2 +
√
Z 0
Z
p
√
(2 + 4 − x2 )dx +
4 − x2 ) + 4 − (x + 2)2 dx =
−2
−2
−2
1
1
= 2 ∗ 2 + π22 + π22
4
4
=4+π+π
= 4 + 2π
3.
Z
2
|x + 1|dx
−1
Answer:
Z
2
Z
2
|x + 1|dx =
−1
x + 1dx
−1
1
= 3∗3
2
9
=
2
4.
Z
2
|x + 1| − 2|x − 1|dx
−1
13
0
p
4 − (x + 2)2 dx
Answer:
Z
2
Z
2
Z
|x − 1|dx
−1
Z 1
−1
Z 2
−1
2
|x + 1|dx − 2
|x + 1| − 2|x − 1|dx =
(x − 1)dx − 2
(x + 1)dx + 2
=
−1
Z
−1
9
1
1
+2 2∗2−2 1∗1
2
2
2
9
= +4−1
2
9
= +3
2
15
=
2
5.
Z
2
|x2 + 1|dx
0
Answer:
Z
2
2
Z
2
x2 + 1dx
Z
Z0 2
2
x dx +
=
|x + 1|dx =
0
0
3
2
1dx
0
2
03
−
+ 1(2 − 0)
3
3
8
= +2
3
14
=
3
=
14
(x − 1)dx
1
=
2