Version 066 L FINAL EXAM Henry (54974) This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 Determine if
x0
1
As f and g are differentiable functions such that
x0
Rehman (aar638) HW04 sachse (56620) This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points Thus we need to nd
2 1
1
f (x)dx = F (2) F (1) .
Now
2 2
The gr
Version 038 L FINAL EXAM Neitzke (56585) This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points 1. I = 52
1
1
respectively. Use these to nd the value of t
jones (bwj276) HW02 neitzke (55460)
1
4
This print-out should have 13 questions.
Multiple-choice questions may continue on
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before answering.
3
2
2
1
001
10.0 points
If g (x) is continuous on [a,b] and m is a
constant
pham (pp8458) HW07 neitzke (55460)
This print-out should have 23 questions.
Multiple-choice questions may continue on
the next column or page nd all choices
before answering.
001
(ii) f has a vertical asymptote at one or
more of x = a, x = b or x = c for
elghonimi (sie94) HW03 hong (52950)
7
in which case
2. I = 13
I =
3. I = 12 correct
1
cos u du = sin u + C
3
1
3
with C an arbitrary constant. Consequently,
4. I = 14
1
I = sin(8 t3 ) + C .
3
5. I = 15
Explanation:
To reduce I to an integral of the form
elghonimi (sie94) HW03 hong (52950)
1. 13 hours later
2. I = 0
2. 12 hours later correct
3. I =
1
4
4. I =
3
4
5. I =
1
4
3. 11 hours later
4. 14 hours later
5. car never returns to Austin
Explanation:
Since the car leaves Austin at time t = 0,
the positi
elghonimi (sie94) HW02 hong (52950)
determine the largest interval(s) on which F
is decreasing.
2
6
2. F (x) = 5x 2x
3. F (x) = 5x2 + 3x
1. [6, 3]
4. F (x) = 7x2 + 3x
2.
3,
5. F (x) = 7x2 3x correct
3.
6,
6. F (x) = 7x2 2x
4.
1
,
4
1
,3
4
Explanation:
O
elghonimi (sie94) HW03 hong (52950)
In this case,
6
when f (0) = 1 and f (1) = 3.
2
I = 12
1
1
1
du = 12
2
(u + 3)
u+3
2
1
.
1. I = 3
2. I = 2
Thus
I = 12
1 1
5 4
=
3
.
5
3. I = 0
4. I = 1 correct
013
10.0 points
5. I = 1
Determine the integral
2
I =
(1
elghonimi (sie94) HW03 hong (52950)
5. I = 2
x2
2
3
4
+ C correct
1
2
u1/2 du =
1 3/2
u +C
3
1
(2x2 + 1)3/2 + C ,
3
=
2x
dx,
2
du =
To evaluate the integral set u = 2x2 + 1. For
then du = 4x dx, in which case
f (x) =
Explanation:
x2
4. Then
Set u =
2
5
w
elghonimi (sie94) HW03 hong (52950)
2
it follows that
Explanation:
If
dy
3
= x1/2 5x1/2 ,
dx
2
then
2x 1,
x |x 1| =
1,
y = 1x
10x
1/2
+C
with C an arbitrary constant. The value of
C is determined by the condition y(1) = 12,
for then
y(1) = 9 + C = 12 ,
i
elghonimi (sie94) HW03 hong (52950)
3
What will be the position of the car after 2
hours of driving?
Consequently, I is the
increase in Miras weight from age 2 to 8 .
1. 1.0 miles north of Austin
keywords: integral, rate growth, FTC, weight,
net change,
0
elghonimi (sie94) HW03 hong (52950)
This print-out should have 22 questions.
Multiple-choice questions may continue on
the next column or page nd all choices
before answering.
001
1
1. I = sin x + C
2. I = 2 cos x + C
10.0 points
Determine the integral
5
elghonimi (sie94) HW03 hong (52950)
6. I = 2( e2 1 )
6. I =
Explanation:
Since
e2x
2
2
+ln x
2
= eln x e2x = xe2x ,
3
5+1
4
Explanation:
Set u = 4ex 3. Then du = 4ex dx, while
x = 0
=
u = 1
x = ln 2
the integral can be rewritten as
=
u = 5.
In this case,
elghonimi (sie94) HW05 hong (52950)
This print-out should have 22 questions.
Multiple-choice questions may continue on
the next column or page nd all choices
before answering.
1
Find the volume, V , of the solid obtained
by rotating the region bounded by
elghonimi (sie94) HW05 hong (52950)
Consequently,
2
the x-axis is given by
9
V = .
4
1
(x1/4 )2 (x4 )2 dx
V =
0
1
003
1
=
10.0 points
0
The shaded region shown in
=
x 2 x8 dx
1
2 3 1 9
x2 x
3
9
.
0
Consequently,
V =
004
2 1
3 9
=
5
.
9
10.0 points
Fi
elghonimi (sie94) HW04 hong (52950)
7
y
3. Area = 81
4. Area = 54
5. Area = 72 correct
/2
cos :
sin :
of y = cos x and y = sin x on [0, ] show,
0,
cos sin
Explanation:
The graph of f is a parabola opening to the
left, while the graph of g is a parabola
elghonimi (sie94) HW04 hong (52950)
while
Consequently, the shaded region has
Area =
009
6
37
sq. units .
12
6
3
10.0 points
3 2 1 3
x x
2
3
(3x x2 ) dx =
6
3
=
45
.
2
Consequently,
Find the area between the graph of f and
the x-axis on the interval [0,
elghonimi (sie94) HW04 hong (52950)
is bounded by the graphs of
105
5. A =
sq.units
2
f (x) = x3 + x2 x ,
Explanation:
The graph of f is a parabola opening upwards and having y-intercept at y = 9. In
addition, since
1
the graph has x-intercepts at x = and
elghonimi (sie94) HW04 hong (52950)
is the area of the shaded region
4
x = 3. Thus the required area is similar to
the shaded region in the
6
y
.
4
x
2
2
2
4
6
2
keywords: AreaBetween, AreaBetweenExam,
006
In terms of denite integrals, therefore, the
requ
elghonimi (sie94) HW04 hong (52950)
As the parabola opens downwards, the area of
the region enclosed by the parabola and the
straight line is thus given by
3
6
3.
4
2
3
(6 x x2 ) dx
2
1
1
6x x2 x3
2
3
=
2
2
.
3
2
4
6
2
4
6
2
Consequently,
2
125
sq. units.
elghonimi (sie94) HW04 hong (52950)
This print-out should have 11 questions.
Multiple-choice questions may continue on
the next column or page nd all choices
before answering.
1
The shaded region in
y
x
001
10.0 points
Find the area of the region enclosed
elghonimi (sie94) HW04 hong (52950)
Consequently, the
2
is given by
area =
4
4
sq.units .
3
Area =
0
(g(x) f (x) dx
4
003
=
10.0 points
0
Find the area bounded by the graphs of f
and g when
(12x 3x2 ) dx.
Now
4
f (x) = x2 4x,
1. area =
g(x) = 8x 2x2 .
0
(
elghonimi (sie94) HW06 hong (52950)
5
On the other hand
Consequently,
I =
009
3
.
4
/4
3 dx =
3x
0
10.0 points
/4
0
=
3
.
4
Consequently,
Find the value of
/4
I=
1
(3 8) .
4
I =
4
3 tan x dx .
0
1. I =
3
(3 2)
8
2. I =
1
(3 4)
4
010
10.0 points
Evaluate
elghonimi (sie94) HW02 hong (52950)
2.
Explanation:
Since
1
0,
3
3. (3, ) correct
1
,
3
4.
0,
5.
7
(3, )
1
,3
3
1
= sec2 ,
2
cos
d
tan = sec2 ,
d
we see that
/6
I =
0
(3 sec2 2 cos ) d
6. (0, 3)
Explanation:
The graph of F will be concave down when
F (x)
elghonimi (sie94) HW02 hong (52950)
6. F (x) = 6x 1 +
4
By the Fundamental Theorem of Calculus
and the Chain Rule,
x2
Explanation:
By the Fundamental Theorem of Calculus
and the Chain Rule,
g(x)
d
dx
a
When
g(x)
d
dx
f (t) dt
= f (g(x)g (x) .
f (t) dt
= f
elghonimi (sie94) HW01 hong (52950)
This print-out should have 20 questions.
Multiple-choice questions may continue on
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before answering.
001
002
1
10.0 points
Determine
lim
x1
x 0 x2 (x +
10.0 points
Determine if
6)
elghonimi (sie94) HW06 hong (52950)
1. I =
2
9
3. I =
2. I =
2
4
4. I =
2
3. I =
8
2
3
4
4
5. I =
4. I =
3
3
4
3
3
6. I =
2
72
017
2
5. I =
18
10.0 points
Determine the integral
015
3 2x
dx .
x2 1
I =
10.0 points
Determine the integral
1
dx .
2 x2 1
x
I =