Practice Problems: Exam 1 – MATH 230, Spring 2011
Instructor: Dr. Zachary Kilpatrick
Show all your work. Simplify as much as possible.
1. Evaluate the following integrals:
(a)
Z
2x − 1
dx =
2
x −x−2
Z
A
B
+
dx
x−2 x+1
must solve for A and B, and A(x + 1) + B(x − 2) = 2x − 1 means
A+B =2
and
A − 2B = −1,
thus B = 1 and A = 1
therefore integrate
Z
Z
1
1
dx +
dx = ln |x − 2| + ln |x + 1| + C
x−2
x+1
(b)
Z
4
2
1
dx =
2
x −4
4
Z
2
A
B
+
dx
x+2 x−2
solve for A and B where A(x − 2) + B(x + 2) = 1 means
A+B =0
and
− 2A + 2B = 1,
thus B = 1/4 and A = −1/4
therefore integrate
4
4
Z
Z
1
1 4 1
1 4 1
ln |x + 2|
+ lim+
−
dx +
dx = −
ln |x − 2|
t→2
4 2 x+2
4 2 x−2
4
4
2
t
ln(2/3) ln 2 1
+
− lim+ ln |x − 2| = ∞
=
4
4
4 t→2
integral does not converge
(c)
Z
2
0
x+2
dx =
(x + 1)2
Z
0
2
A
B
+
dx
x + 1 (x + 1)2
solve for A and B where A(x + 1) + B = x + 2 means A = 1 and B = 1. therefore integrate
Z
0
2
1
dx +
x+1
Z
0
2
2
2
1
1
2
= ln 3 +
dx = [ln |x + 1|]0 + −
2
(x + 1)
x+1 0
3
1
(d)
Z
4
dx =
(x + 2)(x2 + 1)
Z
A
Bx + C
+ 2
dx
x+2
x +1
solve for A, B, C where A(x2 + 1) + (Bx + C)(x + 2) = 4 means
A + B = 0 and 2B + C = 0 and A + 2C = 4,
therefore integrate
Z
Z
4
dx
x−2
4
4
−
dx =
ln |x + 2| −
2
5
x+2 5
x +1
5
4
=
ln |x + 2| −
5
(e)
Z
0
2
5
dx =
(x + 3)(x2 + 4)
Z
2
0
A
dx +
x+3
Z
0
2
thus A = 4/5, B = −4/5, C = 8/5
Z
Z
xdx
dx
4
8
+
2
2
5
x +1 5
x +1
2
8
ln |x2 + 1| + tan−1 (x) + C
5
5
Bx + C
dx
x2 + 4
solve for A, B, C where A(x2 + 4) + (Bx + C)(x + 3) = 5 to find A = 5/13, B = −5/13, and
C = 15/13. thus integrate
Z 2
Z 2
Z
5
5
15 2 dx
5
5
15
dx
xdx
−
+
=
ln
5/3
−
ln
2
+
tan−1 (1)
2
2
13 0 x + 3 13 0 x + 4 13 0 x + 4
13
26
26
(f)
Z
6
dx =
4
x −1
Z
Ax + B
C
D
+
+
dx
2
x +1
x+1 x−1
solve for A, B, C, D where (Ax + B)(x2 − 1) + C(x2 + 1)(x − 1) + D(x2 + 1)(x + 1) = 6 gives
the system
A+C +D
B−C +D
−A + C + D
−B − C + D
=
=
=
=
0
0
0
0
which can be solved with elimination and/or substitution to find A = 0, B = −3, C = −3/2,
D = 3/2. thus integrate
Z
Z
Z
3dx
3dx
3dx
3
3
−1
−
−
+
=
−3
tan
x
−
ln
|x
+
1|
+
ln |x − 1| + C
x2 + 1
2(x + 1)
2(x − 1)
2
2
3 x − 1 − 3 tan−1 x + C
ln =
2
x+1
2
2. Determine whether the integral is convergent or divergent. Evaluate it if it is convergent.
(a)
Z
∞
Z
∞
0
(b)
0
1
dx = lim
t→∞
x+1
Z
t
0
dx
= lim [ln |x + 1|]t0 = lim ln t = ∞
t→∞
x + 1 t→∞
(divergent)
100
dx
(2x + 1)2
use substitution rule with u = 2x + 1 then integral becomes
Z ∞
1
50
−1 t
50
du
=
lim
= 50 (convergent)
−50u
=
50
−
lim
1
t→∞
t→∞ t
u2
1
(c)
Z
0
x
xe dx =
−∞
[xex ]0−∞
−
Z
0
−∞
ex dx = − lim tet − [ex ]0−∞ = − lim tet − 1
t→−∞
t→−∞
where we used integration by parts. now use L’Hospital’s rule to evaluate the limit
lim tet = lim
t
t→−∞ e−t
t→−∞
1
=0
t→−∞ e−t
= − lim
so the limit term vanishes, and the integral evaluates to −1 and it thus convergent.
(d)
Z
∞
−∞
cos xdx = [sin x]∞
−∞ = lim sin t − lim sin t
t→∞
t→−∞
neither limit exists, so the integral is divergent.
(e)
Z
0
∞
−x
e
∞
cos xdx = −e−x cos x 0 −
Z
∞
e−x sin xdx
0
∞
= 1 + e−x sin x 0 −
Z
0
∞
−x
e
cos xdx = 1 −
Z
∞
e−x cos xdx
0
where we’ve integrated by parts twice. the final expression implies
Z ∞
Z ∞
1
−x
2
e cos xdx = 1
⇒
e−x cos xdx =
(convergent)
2
0
0
3
(f)
1
Z
0
1
du =
3u − 1
Z
1/3
0
du
+
3u − 1
Z
1/3 1
ln |3u − 1|
ln |3u − 1|
du
+
=
3u − 1
3
3
0
1/3
1
1/3
we’ve split up the integral, due to the infinite discontinuity at u = 1/3. notice if we try to
evaluate either of the above expressions
lim ln |3u − 1| = −∞
lim ln |3u − 1| = −∞
and
t→1/3−
t→1/3+
so the integral diverges
(g)
Z
1
0
x2 ln x
x ln xdx =
2
1
0
−
Z
1
0
t2 ln t 1
x
dx = − lim+
−
t→0
2
2
4
where we have integrated by parts. now use L’Hospital’s rule for the limit
lim+
t→0
ln t
1/t
t2
t2 ln t
= lim+
=
−
lim
=
lim
= 0.
t→0 2/t2
t→0+ −4/t3
t→0+ 4
2
so the integral equals −1/4 and is thus convergent.
Z 1 −x
Z ∞ −x
Z 1 −x
Z 1 −1
(h) Z ∞ e−x
e
e
e
e
dx =
dx +
dx ≥
dx ≥
dx = e−1 [ln x]10 = ∞
x
x
x
x
x
0
0
1
0
0
so the integral diverges. the first inequality comes from the fact that e−x /x ≥ 0 for all
x ∈ (1, ∞), the second comes from the fact that e−x ≥ e−1 for x ∈ (0, 1). since the original
integral is greater than or equal to one that it infinite, it too must be infinite.
3. Use the Comparison Theorem to determine whether the integral converges or diverges.
(a)
Z
∞
1
1
√
dx ≤
x + ex
Z
1
∞
1
dx =
ex
Z
∞
e−x = e−1
1
so the
converges. we’ve used that
√ integral
x
1/( x + e ) ≤ 1/ex for x ∈ (1, ∞).
(b)
Z
0
1
x+1
dx =
x2
Z
1
0
x
1
+ 2 dx ≥
2
x
x
Z
0
1
√
x + ex ≥ ex for x ∈ (1, ∞) and therefore
x
dx =
x2
so the integral diverges.
4
Z
0
1
dx
= [ln x]10 = ∞
x
4. Sketch the region bounded by the given curves. Draw an approximating rectangle and
label its height and width. Then find the area of the region.
(a) x = y 2/4, y = x2 /4
4
3
Z
∆x
y
0
2
4
4
3/2
16
x3
4x
x2
2 x − dx =
=
−
4
3
12 0
3
√
√
2 x − x 2 /4
1
0
0
1
2
3
x
4
(b) x = −1, y = 1/(x + 2), y = x1/3 /20
y
1
1
x+2
0.5
∆x
− x 1/3/20
Z
8
Z
1/2
−1
x1/3
1
−
dx =
x+2
20
3x4/3
ln |x + 2| −
80
9
= ln 10 −
16
0
0
2
x
4
6
8
(c) y = |x + 1|, y = 2, y = x + 2
2.5
2
y
1.5
∆y
1
1
0.5
∆y
0
−0.5
0
2y
−1
0
x
1
5
2ydy +
2
1/2
7
dy = y 2 0 + [y]21/2 =
4
1/2
Z
8
−1
5. Find the volume of the solid obtained by rotating the region bounded by the given curves
about the specified line. Sketch the region, the solid, and a typical disk or washer.
(a) y = 1 − x3 , x = 0, y = 0; about the x-axis
1.5
Z
1
0
1
3 2
dx = π
2 2
Z
π 1−x
y
0.5
r = 1 − x3
0
−0.5
−0.5
0
0.5
1
x
Z
1
1 − 2x3 + x6 dx
0
1
9π
x4 x7
=
+
= π x−
2
7 0
14
1.5
(b) y = x3/2 , x = y 2; about the y-axis
1
Z
0.8 r = y 2
I
1
π y
0
−π y
y
0.6
2/3 2
0.4
dy =
1
0
π y 4/3 − y 4 dy
3y 7/3 y 5
−
= π
7
5
1
0
=
8π
35
r O = y 2/3
0.2
0
0
0.5
1
x
(c) y = x2 /2, y = x; about x = 3
2
Z
0
y
1.5
rI = 3 −
1
√
1
2
π(3 − y) − π(3 −
2y
rO = 3 − y
0.5
0
0
1
x
2
3
6
p
√
1/2
2
6 2y − 8y + y dy
2y) dy = π
0
1
√ 3/2
y3
2
= π 4 2y − 4y +
3 0
√
11
= 4 2−
3
2
Z
1
6. Use the method of cylindrical shells to find the volume generated by rotating the region
bounded by the given curves about the specified axis. Sketch the region and a typical shell.
(a) y = 2 + x − x2 , y = 2 − x; about the y-axis
2.5
Z
2
1.5
2
2
0
(2πx) 2x − x
y
1
h = 2x − x 2
0.5
Z
2
2x2 − x3 dx
0
2
3
8π
x4
2x
=
−
= 2π
3
4 0
3
dx = 2π
0
−0.5
0
0.5
1
1.5
x
2
(b) x = 4y − y 3, x = 0 about the x-axis
2
2
Z
1
(2πy) 4y − y
0
1.5
y
Z
h = 4y − y 3
1
3
Z
2
4y 2 − y 4 dy
0
2
3
128π
y4
4y
=
−
= 2π
3
5 0
15
dx = 2π
0.5
0
0
1
x
2
3
(c) y = x, y = 1, x = 2; about x = 3
y
2
h = 2−x
r = 3−x
2
1
1.5
2
x
2.5
1
6 − 5x + x2 dx
2
1
5x2 x3
= 2π 6x −
+
2
3 2
7π
=
6
(2π(3 − x)) (2 − x) dx = 2π
1.5
1
Z
3
7
7. Find the arc length of the specified curve.
(a) x = 51 y 5/2 +
√1 ,
y
1≤y≤4
dx
1
1
= y 3/2 − y −3/2 ⇒ 1 +
dy
2
2
dx
dy
2
2
1 1
1 3/2
1
y + y −3/2
= y 3 + + y −3 =
4
2 4
4
so the arc length of this curve will be
4
Z
Z 4r
67
1 2 5/2
1 4 3/2
1 3/2
2
−3/2
−1/2
−3/2
=
y +y
dy =
(y + y
) dy =
y − 2y
4
2 1
2 5
10
1
1
(b) y = ln 2x,
dy
1
=
dx
x
√
√
3≤x≤2 2
2
x2 + 1
dy
1
⇒ 1+
=1+ 2 =
dx
x
x2
so the arc length of this curve will be
Z 2√2 r 2
Z tan−1 2√2 r
Z tan−1 2√2
x +1
tan2 θ + 1
sec3 θ
2
dx
=
sec
θdθ
=
dθ
√
√
√
x2
tan2 θ
tan θ
tan−1 3
tan−1 3
3
Z tan−1 2√2
Z tan−1 2√2
sec θ
2
1 + tan θ dθ =
csc θ + tan θ sec θdθ
=
√
√
tan θ
tan−1 3
tan−1 3
√
−1 2 2
1 3
√
= 1 + ln
= [− ln | csc θ + cot θ| + sec θ]tan
−1
tan
3
2 2
8. Suppose 3 J of work are needed to stretch a spring from its natural length of 20cm to 30cm.
(a) What is the spring constant k?
2 0.3
Z 0.3
x
= 0.025k = 3
kxdx = k
2 0.2
0.2
⇒
k = 120
N
m
(b) How much work is needed to stretch the spring from 30cm to 40cm?
2 0.4
Z 0.4
x
120(0.035) = 5.6 J
120xdx = 120
2 0.3
0.3
(c) How far beyond its natural length will a force of 20N keep the spring stretched?
20 = 120x
⇒
x = 1/6 m ≈ 16.6 cm
8
9. An upright conical tank (tip pointing up) with height 10ft and radius 5ft is full of water.
Given that water weighs 62.5lb/ft3 , find the work required to pump the water out of the
tank.
W = 62.5
Z
10
π
0
62.5π
=
4
Z
10
10 − y
2
2
(10 − y)dy
1000 − 300y + 30y 2 − y 3 dy
0
10
62.5π
y4
2
3
=
1000y − 150y + 10y −
= 39062.5π
4
4 0
At a particular level, y, the cone has cross-sectional area A = π((10 − y)/2)2 and it is 10 − yft
underwater.
10. Treasure chest with half-cylindrical top and rectangular prism base (pictured) sits at the
bottom of a bay. Its top is 20m beneath the bay’s surface. Considering the density of water
is 1000 kg/m3 , calculate the total hydrostatic pressure on one vertical side (shown) of the
chest.
Z 3
Z 4
p
2
W = 9800
8(27 − y)dy +
(24 − y)2 16 − y dy
0
0
" 3
Z 4p
y2
= 9800 8 27y −
+ 48
16 − y 2dy
4m
2 0
0
Z 4 p
−2
y 16 − y 2dy
0
"
#
Z π/2
Z 0
3m
√
udu
= 9800 612 + 48
16 cos2 θdθ +
8m
0
16
16738400
128
=
+ 1881600π
= 9800 612 + 192π −
3
3
Notice we must compute two separate integrals for the different shapes, rectangular and
semicircular. The rectangle has constant
p width 8, whereas the semicircle has width that
changes as the level moves up given by 16 − y 2 .
9
11. Find the distance from P (4, −3, 1) to: (a) Q(−1, 1, 5); (b) xy-plane; (c) yz-plane; (d)
xz-plane; (e) x-axis; (f) y-axis; (g) z-axis.
(a) |P~Q| =
(b)
(c)
(d)
(e)
(f)
(g)
p
(−1 − 4)2 + (1 − (−3))2 + (5 − 1)2 =
√
57
p
(4 − 4)2 + (−3 − (−3))2 + (0 − 1)2 = 1
p
(0 − 4)2 + (−3 + 3)2 + (1 − 1)2 = 4
p
(4 − 4)2 + (0 − (−3))2 + (1 − 1)2 = 3
p
√
(4 − 4)2 + (0 − (−3))2 + (0 − 1)2 = 10
p
(0 − 4)2 + (−3 − (−3))2 + (0 − 1)2 =
p
√
17
(0 − 4)2 + (0 − (−3))2 + (1 − 1)2 = 5
12. Show that x2 + y 2 + z 2 = 10x − 4y + 2z represents a sphere, and find its center and radius.
(x2 − 10x) + (y 2 + 4y) + (z 2 − 2z) = 0
(x2 − 10x + 25) + (y 2 + 4y + 4) + (z 2 − 2z + 1) = 30
(x − 5)2 + (y + 2)2 + (z − 1)2 = 30
√
so the center of the sphere is C(5, −2, 1) and its radius is 30
~ when:
13. Compute a = AB
(a) A(1, −3), B(−1, 1); a = h−1 − 1, 1 − (−3)i = h−2, 4i
(b) A(0, −2, 3), B(4, 0, −1);
a = h4 − 0, 0 − (−2), −1 − 3i = h4, 2, −4i
(c) A(4, −3, −3), B(2, 1, 7);
a = h2 − 4, 1 − (−3), 7 − 4i = h−2, 4, 3i
10
14. Given a = 5i − 2j + k and b = −2i + 2j − 3k, compute:
(a) a + b = 3i − 2k
(b) a − b = 7i − 4j + 4k
(c) 3a − 2b = 19i − 10j + 9k
(d) |b| =
p
(−2)2 + 22 + (−3)2 =
(e) |b − a| = |a − b| =
p
√
17
72 + (−4)2 + 42 =
√
81 = 9
15. For the following pairs of vectors, calculate a · b; determine whether they are parallel,
orthogonal, or neither; and find the scalar and vector projections of b onto a.
(a) a = h−2, 1i, b = h1, 2i,
compa b =
a · b = (−2)(1) + 1(2) = 0, so vectors are orthogonal
0
a·b
=√
= 0,
|a|
−22 + 1
proja b =
0
a·b
a=
h−2, 1i = h0, 0i
2
2
|a|
−2 + 1
(b) a = h−3, 4, 1i, bh1, 3, 2i
a · b(−3)(1)
+ 4(3) + 1(2)
√
√ = 11 6= 0,√so vectors are not
√ orthogonal
2
2
2
2
2
2
|a| = −3 + 4 +
√1 = 26, |b| = 1 + 3 + 2 = 14,
so a · b = 11 6= ± 364 = |a||b|, so vectors are not parallel either
compa b =
a·b
11
=√ ,
|a|
26
proja b =
11
−33 22 11
a·b
a = h−3, 4, 1i = h
, , i
2
|a|
26
26 13 26
(c) a = i + 2j − 4k, b = −2i − 4j + 8k
a · b(1)(−2)
+ 2(−4) + (−4)(8)
p
p 6= 0, so vectors are not
√ = −42
√ orthogonal
|a| = 12 + 22 + (−4)2 = 21, |b| (−2)2 + (−4)2 + 82 = 2 21,
so a · b = −42 = −|a|||b|, so vectors are antiparallel
√
−42
a·b
= √ = −2 21 = −|b|,
|a|
21
−42
a·b
a=
h1, 2, −4i = h−2, −4, 8i = b
proja b =
|a|2
21
compa b =
11
16. Calculate the volume of the parallelepiped with the adjacent edges P Q, P R, and P S.
(a) P (−3, 1, 0), Q(−1, 2, 1), R(−3, 3, 2), S(−3, 0, 1)
a = P~Q = 2i + j + k, b = P~R = 2j + 2k, c = P~S = −j + k
2 1 1 0 2 2 = |2(2 + 2) − 1(0 − 0) + 1(0 − 0)| = 8
0 −1 1 (b) P (1, −2, 1), Q(0, −1, 1), R(−1, 0, 2), S(−2, −2, 3)
a = P~Q = −i + j, b = P~R = −2i + 2j + k, c = P~S = −3i + 2k
−1 1 0 −2 2 1 = | − 1(4 − 0) − 1(−4 + 3) + 0(−2 + 3)| = 3
−3 0 2 17. Determine whether the two specified lines are parallel, skew, or intersecting. If they
intersect, find the point of intersection.
(a) The line through (−2, 4, 1) and (−5, 7, −1) and the line through (−1, 6, −1) and (5, 0, 1).
for the first line, v = h−3, 3, −2i and P (−2, 4, 1), so the line is given as:
L1 : x = −2 − 3t, y = 4 + 3t, z = 1 − 2t
for the second line, u = h6, −6, 2i and Q(−1, 6, −1), so the line is:
L2 : x = −1 + 6s, y = 6 − 6s, z = −1 + 2s
−3
3
−2
=
6=
,
6
−6
2
so lines are not parallel
−2 − 3t = −1 + 6s,
4 + 3t = 6 − 6s,
to see if they intersect, set up the three equations
1 − 2t = −1 + 2s,
and try to solve by substitution, by converting the third equation to t = 1 − s, substituting
this into the second equation gives 7 + s = 6 − 6s so s = −1/7 thus t = 8/7, but upon
substituting this into the first equation, we find −38/7 6= −13/7, so there is no pair t, s such
that the parametric equations have the same x, y, z coordinate
(b) L1 : x = 2 − 3t, y = 1 + t, z = −1 + 2t
L2 : x = −s, y = 4 − 2s, z = −2 + 3s
−3
1
2
6=
6= ,
−1
−2
3
so lines are not parallel
to see if they intersect, set up the three equations
2 − 3t = −s,
1 + t = 4 − 2s,
−1 + 2t = −2 + 3s,
2nd equation implies t = 3 − 2s, plug into 3rd to find 5 − 4s = −2 + 3s so s = 1 and t = 1,
plug into 1st equation to find −1 = −1, so the lines do intersect.
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18. Find the equations of the planes and determine if they are parallel, perpendicular, or
neither. If neither, determine the angle between them.
(a) The plane that passes through the points P (1, 1, 1), Q(2, −1, 0), R(0, 0, 2) and the plane
that passes through the point S(2, 2, 2) that contains the line x = 1 − t, y = 2 + t, z = 3t.
P~Q = h1, −2, −1i, P~R = h−1, −1, 1i
i
j
k
n = P~Q × P~R = 1 −2 −1 = (−2 − 1)i − (1 − 1)j + (−1 − 2)k = −3i − 3k
−1 −1 1 taking P as our point, the equation for the first plane is −3(x − 1) − 3(z − 1) = 0, which can
be rewritten as x + z = 2.
by reading the direction numbers of the line in the second plane, we find it contains the vector
v = h−1, 1, 3i. at t = 0, the line contains the point T (1, 2, 0), so the plane also contains the
vector T~S = h1, 0, 2i.
i j k m = v × T~S = −1 1 3 = 2i + 5j − k
1 0 2 taking S as our point, the equation for the second plane is 2(x − 2) + 5(y − 2) − (z − 2) = 0,
which can be rewritten as 2x + 5y − z = 12.
to find how planes are oriented with one another, simply compute the angle between using
their normal vectors
cos θ =
n·m
1
1(2) + 1(−1)
p
= √
=√
|n||m|
2 15
12 + 12 22 + 52 + (−1)2
⇒
1
θ = cos−1 √ ≈ 1.4413 ≈ 82.58◦
2 15
the planes are neither parallel or perpendicular
(b) 2x + y − z = 2, x − 2y + z = 4
to find how planes are oriented with one another, simply compute the angle between using
their normal vectors
−1
2(1) + 1(−2) + (−1)(1)
n·m
p
=
=p
|n||m|
4
22 + 12 + (−1)2 12 + (−2)2 + 12
−1
θ = cos−1
≈ 1.8235 ≈ 104.48◦
4
cos θ =
the planes are neither parallel or perpendicular
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