Solution to Review Problem for Midterm #2
Midterm II: Monday, October 28. in class Topics: 7.1-7.4, 8.1-8.2
and 12.1
No calculator is allowed in the exam. You should know how to solve
these problems without a calculator.
In the following, I will use the follow formula frequently.
R ax
R 1
a
e dx = e ax + C, ax+b
dx = ln |ax+b|
+ C,
Ra
R x
1
2
R ax2 +b dx = 2a · ln |ax + b| + C, sec
R xdx = ln | sec x + tan x| + C,
csc Rxdx = − ln | csc x + cot x| + C, tan xdx = ln | sec x| + C,
and cot xdx = ln | sin x| + C.
1. Evaluate
the following indefinite integrals:
R
x+1
1 (x2 +2x+10)
4 dx
= (x+1)dx.
Solution:
u = x2 +2x+10.
Then
du = (2x+2)dx and du
2
R Let
R
R
x+1
1 du
1
1
1
−4
−3
Thus (x2 +2x+10)4 dx = u4 2 = 2 u du = 2 · (− 3 )u + C = − 6u1 3 +
1
C=−
+ C.
R −3x6(x2 +2x+10)3
2 xe dx Solution: We use integration by parts. Let u = x and
R
R
−3x
dv = e−3x dx. Then du = dx and v = e−3x dx = −e3 . So xe−3x dx =
R
R
−3x
−3x
−3x
−3x
−3x
x( −e3 ) − ( −e3 )dx = − xe 3 + 13 e−3x dx = − xe 3 + 13 · (− e 3 ) + C =
xe−3x
e−3x
−
R 3 − 9 + C.
3 x sin(3x) Solution: Let u = x and dv = sin(3x)dx. Then du = dx
R
R
.
So
x sin(3x)dx = x · (− cos(3x)
)−
and v = sin(3x)dx = − cos(3x)
3
3
R cos(3x)
R
x cos(3x)
x cos(3x)
sin(3x)
1
1
(− 3 )dx = − 3
+ 3 cos(3x)dx = − 3
+ 3 · 3 +C =
x cos(3x)
sin(3x)
− 3
+ 9 + C.
R3 √
√
1
4 0 x x + 1dx
Solution: Let u = x + 1. Then du = 2√x+1
dx,
√
√
2 x + 1du = dx i.e. 2udu = dx. RFrom
√ u = x R+ 1,2 we also have
2
2
u
=
x
+
1.
So
x
=
u
−
1.
Thus
x
1)u · 2udu =
R 4
√ x +5 1dx2 =
√ (u −
2 5
2 3
2
2
3
2u − 2u du = 5 u − 3 u + C = 5 ( x + 1) − 3 ( x + 1) + C.
3
R3 √
√
√
√
√
So 0 x x + 1dx = 52 ( x + 1)5 − 23 ( x + 1)3 = 25 ( 3 + 1)5 − 32 ( 3 + 1)3 −
0
√
√
2
2
2
2
2
2
5
3
5
3
(
0
+
1)
−
(
0
+
1)
=
(2)
−
(2)
−
−
= 64
− 16
− 25 + 23 =
5
3
5
3
5
3
5
3
62
14
R5 −dx 3 =
186−70
15
= 116
.
15
√
√
5 1+√x Solution: Let u = x + 1. Then du = 2√1 x dx., i.e. 2 xdu =
√
dx and 2(u − 1)du = dx where we have used x = u − 1.
R
R
R
Hence 1+dx√x dx = 2(u−1)
du = 2 − u2 du = 2u − 2 ln |u| + C
u
√
√
=
2(
x
+
1)
−
2
ln
|1
+
x| + C.
R
√
1 √
6 √x(1+
dx
Solution:
Let u = x + 1. Then du = 2√1 x dx and
x)
R
R
R
1 √
2du = √1x dx. Hence √x(1+
dx = (1+1√x) · √1x dx = u1 · 2du =
x)
R 2
√
du
=
2
ln
|u|
+
C
=
2
ln
|1
+
x| + C.
u
MATH 1760: page 1 of 10
Solution to Review Problem for Midterm #2
MATH 1760: page 2 of 10
R
√
7 sin x cos xdx Solution:Let u = cos x. Then du = − sin xdx and
R√
R
√
3
3
udu = − 23 u 2 +C = − 23 (cos x) 2 +
sin(x)dx = −du. sin x cos xdx = −
C.
R
R
R
R
8 cos5 xdx Solution: cos5 xdx = (cos2 x)2 cos xdx = (1−sin2 x)2 cos xdx
where we have used cos2 x = 1 − sin2 x.R
R
Let
x. Then
du = cos xdx. Then cos5 xdx = (1−sin2 x)2 cos xdx =
R u =2sin
R
(1 − u )2 du = (1 − 2u2 + u4 )du
= u − 23 u3 + 15 u5 + C = sin x − 23 (sin x)3 + 51 (sin x)5 + C
Rπ
R
R sin x
9 03 sin x sec2 xdx
Solution: Note that sin x sec2 xdx =
dx.
cos2 x
Let
u
=
cos
x.
Then
du
=
−
sin
xdx
and
−du
=
sin
xdx.
Then
R
R
R
sin x sec2 xdx = cos12 x · sin xdx = − u−2 du = u−1 + C = cos1 x + C.
π
Rπ
3
1
= 2 − 1 = 1. We have
Hence 03 sin x sec2 xdx = cos1 x = cos(1 π ) − cos(0)
0
10
11
12
13
3
= 1 and cos( π3 ) = 12 .
Rused cos(0)
x tan−1 (x)dx Solution: Let u = arctan(x) and dv = xdx. Then
R
R
2
du = x21+1 and v = xdx = x2 . Thus x arctan(x)dx = arctan(x) ·
R 2
R 2
R 2
2
2
x2
− x2 · x21+1 dx = x arctan(x)
− 12 x2x+1 dx = x arctan(x)
− 21 (x x+1)−1
dx =
2 +1
2
2
2
R
2 arctan(x)
x2 arctan(x)
x
− 12 (1 − x21+1 )dx =
− 12 (x − arctan(x)) + C.
2
R 2x3
dx Solution: Let u = 1 + x2 . Then du = 2xdx and xdx = du
.
(1+x2 )5
2
R
R
R
2
3
x
x
·xdx = u−1
· du
=
Also we have x2 = u−1 Thus (1+x
2 )5 dx =
(1+x2 )5
u5
2
R
R
1
1
1 −3
1 −4
1 −3
1 −4
1
u−1
−4
−5
du = 2 (u − u )du = 2 · (− 3 u + 4 u ) + C = − 6 u + 8 u +
2
u5
1
C
(1 + x2 )−3 + 18 (1 + x2 )−4 + C
6
R =−
4
sec x tan3 xdx Solution: Let u R= tan(x). Then du
= sec2 (x)dx.
R
3
2
3
Note that Rsec2 (x) = 1 + tan2 (x). So sec4 x tan
R xdx 2= 3sec x · tan
R 3x ·
2
2
3
2
sec xdx = (1 + tan (x)) · tan x · sec xdx == (1 + u ) · u xdu = (u +
4
6
4
6
5
= u4 + u6 + C = tan4 (x) + tan6 (x) + C.
Ru )du
eax sin(bx)dx This is a typical ”integration by parts” example.
We start with u = eax and dv = sin(bx)dx.
R
Then we have du = aeax dx (use chain rule here) and v = sin(bx)dx =
− cos(bx)
(note that (cos(bx))0 = − sin(bx) · b and (− cos(bx)
)0 = sin(bx).
b
b
Using integration by parts, we have
R ax
R
cos(bx)
cos(bx)
e sin(bx)dx = |{z}
eax · (−
) − (−
) · |aeax
|{z}
{zdx}
| {z }
b
b
| {z }
| {z }
u
dv
u
du
v
v
R ax
a
= − cos(bx)
+
e
cos(bx)dx
b
bR
R ax
ax
a
Now we have eax sin(bx)dx = − e cos(bx)
+
b
Rb e cos(bx)dx. We do
not get the answer now. We need to find eax cos(bx)dx using IBP
again.
eax
Solution to Review Problem for Midterm #2
MATH 1760: page 3 of 10
Let u = eax and dv = cos(bx)dx. Then we have du = aeax dx (use
R
(note that (sin(bx))0 =
chain rule here) and v = cos(bx)dx = sin(bx)
b
cos(bx) · b and ( sin(bx)
)0 = cos(bx).
b
Using integration by parts, we have
R ax
sin(bx) R sin(bx)
eax ·
e cos(bx)dx = |{z}
−
· |aeax
{zdx}
|{z}
| {z }
b
b
| {z }
| {z }
u
dv
u
dv
u
du
v
v
R ax
sin(bx)
a
−
e
sin(bx)dx
=
b
b R
R
ax
Now we have eax cos(bx)dx = e sin(bx)
− ab eax sin(bx)dx.
b
R
R ax
ax
a
Now we combine eax sin(bx)dx = − e cos(bx)
e cos(bx)dx and
+
b
b
R ax
R ax
eax sin(bx)
a
e cos(bx)dx =
− b e sin(bx)dx to get
b
Z
Z
eax cos(bx) a eax sin(bx) a
ax
e sin(bx)dx = −
+
−
eax sin(bx)dx
b
b
b
b
Z
ax
ax
e cos(bx) a e sin(bx) a a
(0.0.1)
=−
+ ·
− ·
eax sin(bx)dx
b
b
b
b b
Z
eax cos(bx) aeax sin(bx) a2
eax sin(bx)dx
=−
+
− 2
b
b2
b
Thus
Z
Z
a2
−beax cos(bx) + aeax sin(bx)
ax
e sin(bx)dx + 2
+ c,
eax sin(bx)dx =
b
b2
R
ax
ax sin(bx)
2
+ c,
(1 + ab2 )( eax sin(bx)dx) = −be cos(bx)+ae
2
b
R ax
−beax cos(bx)+aeax sin(bx)
b2 +a2
( e sin(bx)dx) =
+ c and
b2
R b2ax
ax
ax sin(bx)
b2 −beax cos(bx)+aeax sin(bx)
e sin(bx)dx = a2 +b2
+ C = −be cos(bx)+ae
+
b2
a2 +b2
C.
R
14 eax cos(bx)dx Solution:Let u = eax and dv = cos(bx)dx. Then we
R
have du = aeax dx (use chain rule here) and v = cos(bx)dx = sin(bx)
b
0
(note that (sin(bx))0 = cos(bx) · b and ( sin(bx)
)
=
cos(bx).
b
Using integration by parts, we have
R ax
sin(bx) R sin(bx)
ax
e
cos(bx)dx
=
e
·
−
· |aeax
|{z} | {z } |{z}
{zdx}
b
b
| {z }
| {z }
eax
u
du
v
v
R ax
sin(bx)
a
=
− b e sin(bx)dx.
b
R
R ax
ax
a
Now we haveR eax cos(bx)dx = e sin(bx)
−
e sin(bx)dx.
b
b
ax
ax
To integrate e sin(bx)dx, we try u = e and dv = sin(bx)dx.
Then
R
we have du = aeax dx (use chain rule here) and v = sin(bx)dx =
− cos(bx)
(note that (cos(bx))0 = − sin(bx) · b and (− cos(bx)
)0 = sin(bx).
b
b
Using integration by parts, we have
eax
Solution to Review Problem for Midterm #2
MATH 1760: page 4 of 10
R
cos(bx)
cos(bx)
eax sin(bx)dx = |{z}
) − (−
) · |aeax
eax · (−
|{z}
{zdx}
| {z }
b
b
|
{z
}
|
{z
}
u
u
du
dv
v
v
R
ax
= − e cos(bx)
+ ab eax cos(bx)dx
b
R
R
ax
Now we combine eax cos(bx)dx = e sin(bx)
− ab eax sin(bx)dx and
b
R ax
R ax
ax
a
e sin(bx)dx = − e cos(bx)
+
e cos(bx)dx.
b
b
to get
R
Z
eax sin(bx) a eax cos(bx) a
ax
e cos(bx)dx =
−
−
+
e cos(bx)dx
b
b
b
b
Z
eax sin(bx) a
eax cos(bx)
a a
(0.0.2)
=
− · (−
)− ·
eax cos(bx)dx
b
b
b
b b
Z
eax sin(bx) aeax cos(bx) a2
=
+
− 2
eax cos(bx)dx
b
b2
b
Thus
Z
Z
eax sin(bx) aeax cos(bx)
a2
ax
eax cos(bx)dx =
+
+ c,
e cos(bx)dx + 2
b
b
b2
R
ax
ax
2
(1 + ab2 )( eax cos(bx)dx) = e sin(bx)
+ ae bcos(bx)
+ c,
2
b
R ax
eax sin(bx)
aeax cos(bx)
b2 +a2
( e cos(bx)dx) =
+
2
b
b2
+ c and
R b ax
ax sin(bx)
ax cos(bx)
2
e
ae
e cos(bx)dx = a2b+b2
+
+C
b
b2
Z
15
16
17
18
19
ax
beax sin(bx)+aeax cos(bx)
+ C.
a2 +b2
dx
Solution:Let u = ex + 1. Then du = ex dx = (u − 1)dx. So
ex +1
R
R 1
du
1
1
dx = u−1
. exdx+1 = u(u−1)
du. By partial fractions, u(u−1)
= u−1
− u1 .
R dx
= ln |u − 1| − ln |u| + C = ln |ex | − ln |ex + 1| + C = x − ln |ex + 1| + C.
R elnx x+1
dx Solution:LetR u = ln(x)Rand dv = x12 dx. Then du = x1 Rdx and
x2 R
v = x12 dx = − x1 . So lnx2x dx = ln(x) · x12 dx = ln(x) · (− x1 ) − (− x1 ) ·
R 1
ln(x)
1
dx
=
−
+
dx = − ln(x)
− x1 + C.
x
x
x2
x
R 2
x ln xdx Solution:Let u = ln(x) and dv = x2 dx. Then du = x1 dx
R
R
R
R 3
3
3
and v = x2 dx = x3 . So x2 ln xdx = ln(x) · x2 dx = ln(x) · x3 − x3 ·
R x2
3
x3 ln(x)
x3 ln(x)
1
dx
=
−
dx
=
− x9 + C.
3
3
3
Rx ln x
R
dx
Solution:Let u = ln(x) Then du = x1 dx and lnxx dx =
x
R
2
2
udu = u2 + C = (ln(x))
+ C.
2
R√
√
√
√
x sin( x)dx Solution: Let u = x. Then du = 2√1 x dx, 2 xdu =
R√
R
R 2
√
=
R
dx and 2udu = dx.
x sin( x)dx = u sin u(2udu) = 2 u sin(u)du.
By
integration
by
parts
twice, we have
R 2
2
u sin udu = −u cos u + 2u sin u + 2 cos u + C.
MATH 1760: page 5 of 10
Solution to Review Problem for Midterm #2
R√
√
Thus√ x sin(
√ x)dx √
√
√
2
x) + 2 cos( x) + C.
= −( x) cos( x) + 2( x) sin(
R sin(√x)
R sin(√x)
√
√
√
20
dx Solution:
dx = −2 cos( x) + C.
x
x
√
(By substitution, u = x, du = 2√1 x dx.)
R −1 √
√
21 sin √(x x) dx Solution:By substitution, u = x.
R sin−1 (√x)
R
√
dx
=
2
sin−1 udu + C
x
By
by parts,
√ we have
R integration
sin−1 udu =√u sin−1 u + 1 − u2 + C..
R −1
√
√
√
Thus sin √(x x) dx = 2 x sin−1 ( x) + 1 − x + C..
R
22 x sec2 (x)dx Solution:Let u = x and dv = sec2 (x)dx. Then du = dx
and
v = tan x.
R
R
2
x
sec
(x)dx
=
x
tan
x
−
tan xdx = x tan x − ln | sec x| + C.
R
23 arcsin(2x)dx Solution: u = arcsin(2x) and dv = dx. Recall that
d
a
(arcsin(ax)) = √1−a
2 x2 .
dx
R
2
Then du = √1−4x2 dx and v = dx = x. Using integration by parts,
R
R
2
we have arcsin(2x)dx = arcsin(2x)x − x · √1−4x
2 dx = arcsin(2x)x −
R 2x
√
dx. We can use the substitution u = 1 − 9x2 , du = −8xdx
1−4x2
and − du
= xdx to find
√
√
R 2
R 2x 8
R 1
u
1−4x2
√
√ · (− du ) = −
√ du = −
dx
=
+
C
=
−
+ C. Thus
8
2
2
u
4 u
1−4x2
√
R
2
1−4x
R arcsin(3x)dx = x arcsin(2x) + 2 + C.
24 R tan x ln(cos x)dx RSolution:Let u = cos
Then du = − sin xdx.
R x.
ln u
sin x
tan x ln(cos x)dx = cos x ln(cos x)dx = − u du = − ln | ln u| + C (By
making
a substitution w = ln u.) = − ln | ln cos x| + C.
R
2
25 ln(x + 1)dx
Solution:Let u = ln(x2 + 1) and dv
R = dx. Then
du = x22x+1 and v = x. By integration by parts, ln(x2 + 1)dx =
R 2
2
x ln(x2 + 1) − x2x
= 2(x2 + 1) − 2
2 +1 . By long division, we have 2x
2
2x2
= 2(xx2+1)−2
= 2 − x22+1 .
x2 +1R
+1 R
2
2
So x2x
R2 +1 dx 2= (2 − x2 +1 )dx 2=
2x − 2 arctan(x) + C.
Thus
R 3 √ ln(x + 1)dx = x ln(x + 1) − 2x +2 22 arctan(x) + C.
26 x 1 + x2 dx Solution:Let u = 1R + x√. Then du = R2xdx and xdx =
√
du
. Note that x2 = u − 1. Then x2 1 + x2 xdx = (u − 1) u du
=
2R
2
3
1
5
3
5
3
5
1
1
2
2
1
1
1
1
2
(u 2 − u 2 ) = 2 ( 5 u 2 − 3 u 2 ) + C = 5 u 2 − 3 u 2 + C = 5 (1 + x ) 2 − 3 (1 +
2
2 32
x
+ C.
R )x2 +10x+12
2 +10x+12 x2 +10x+12
x2 +10x+12
A
B
27 x3 +8x2 +12x dx Solution: xx3 +8x
2 +16x x(x2 +8x+16) = x(x+2)(x+6) = x + x+2 +
C
. Multiply x(x + 2)(x + 6) to both sides, we have x2 + 10x + 12 =
x+6
A(x + 2)(x + 6) + Bx(x + 6) + Cx(x + 2).
Solution to Review Problem for Midterm #2
28
29
30
31
32
33
MATH 1760: page 6 of 10
Plug in x = 0, we have 12 = 12A and A = 1. Plug in x = −2, we
have 4 − 20 + 12 = −8B, 8B = −4 and B = − 21 Plug in x = −6, we
have 36 − 60 + 12 = 24C, 24C = −12 and C = − 12
2 +10x+12
1
1
1
So xx3 +8x
2 +16x = x − 2(x+2) − 2(x+6) .
R 2 +10x+12
1
1
Thus xx3 +8x
2 +16x dx = ln |x| − 2 ln |x + 2| − 2 ln |x + 6| + C.
R e4t
dt Solution: Let u = e2t − 1. Then du = 2e2t dt and e2t =
(e2t −1)3
R
R
R
R
R
4t
2t
u + 1. (e2te−1)3 dt = (e2te−1)3 e2t dt = u+1
du = u12 du + u13 du = − u1 −
u3
1
+ C = − e2t1−1 − 21 (e2t 1−1)2 + C
2u2
R x2
2
x2
x2
A
B
dx Solution: x4x−1 = (x2 −1)(x
2 +1) = (x−1)(x+1)(x2 +1) = x−1 + x+1 +
x4 −1
Cx+D
.
x2 +1
One should get A = 14 , B = − 14 , C = 0 and D = 12 .
x2
1
1
= 41 · x−1
− 14 · x+1
+ 12 · x21+1 .
x4 −1
R x2
Thus x4 −1 dx = 41 ln |x − 1| − 41 ln |x + 1| + 21 tan−1 x + C.
R x2
Solution: Let u = x + 2. Then du = dx and x = u − 2.
10 dx
R (x+2)
R
R u2 −4u+4
R −8
2
(u−2)2
x
dx
=
du
=
du
=
u − 4u−9 + 4u−10 du
10
10
10
(x+2)
u
u
1
1
4
= − 17 u−7 + 84 u−8 − 49 u−9 + C = − 7(x+2)
7 + 2(x+2)8 − 9(x+2)9 + C .
R 2x−6
dx Solution: There is a typo in the original problem.
x2 +4x+13
R 2x−6
This problem should be x2 +4x+13
dx.
Completing the square, we get x2 + 4x + 13 = x2 + 4x + 4 + 9 =
(x + 2)2 + 32 . Let x + 2 = 3u. Then dx = 3du and x = 3u −
R 2x−6
R 2x−6
R 2·(3u−2)−6
R 6u−10
2.
dx
=
dx
=
3du
=
3du =
2
2
3
2
2
(x+2) R
+3
3 u+3
9(u2 +1)
R 6u−10x +4x+13 R 2u
1
10
10
2
du = u2 +1 du − 3 u2 +1 du = ln |u + 1| − 3 arctan(u) + C =
3(u2 +1)
x+2 2
ln |( 3 ) + 1| − 10
arctan( x+2
) + C. In the last step, we have used
3
3
x+2
u= 3 .
R x3 −1
dx
Solution: From the long division, we have x3 − 1 =
x3 +x
R 3 −1
R 3
R
(x3 + x) − x − 1. xx3 +x
dx = x +x−x−1
= (1 − xx+1
3 +x )dx
x3 +x
x+1
x+1
A
Bx+C
= x(x2 +1) = x + x2 +1 .
x3 +x
Multiply x(x2 + 1) to both sides,
we have x + 1 = A(x2 + 1) + (Bx + C)x.
Plug in x = 0, we have A = 1.
By comparing the coefficient, we have B = −1 and C = 1.
(−1x+1)
1
= x1 − x2x+1 + x21+1 .
Thus xx+1
3 +x = x +
x2 +1
R x+1
So x3 +x dx = ln |x| − 12 ln |x2 + 1| + arctan x + C.
R 3 −1
Therefore xx3 +x
dx = x − ln |x| + 12 ln |x2 + 1| − arctan x + C.
R x+1
x+1
A
B
C
dx Solution: xx+1
3 −x2 = x2 (x−1) = x + x2 + x−1 .
x3 −x2
Solution to Review Problem for Midterm #2
MATH 1760: page 7 of 10
Multiply x2 (x − 1) to both sides, we have
x + 1 = Ax(x − 1) + B(x − 1) + Cx2 .
Plug in x = 0, we have B = −1.
Plug in x = 1, we have C = 2.
Thus x + 1 = Ax(x − 1) − 1(x − 1) + 2x2 = Ax2 − Ax − x + 1 + 2x2 =
(A + 2)x2 − (A + 1)x + 1.
By comparing the coefficient, we have A = −2.
2
1
2
So xx+1
3 −x2 = − x − x2 + x−1 .
R x+1
Thus x3 −x2 dx = −2 ln |x| + x1 + 2 ln |x + 1| + C
2. Determine whether each integral is convergent or divergent. If the
integral is convergent, compute its value.
R∞
Rb
Rb 1
R∞
(a) 1 11 dx Solution: 1 11 dx = limb→∞ 1 11 dx = limb→∞ 1 x− 3 dx =
x3
x3
x3
b
R
2
2
∞
limb→∞ 23 x 3 == limb→∞ 23 b 3 − 32 = ∞. So 1 11 dx diverges. Here we
x3
1
p
have
0.
b→∞ b = ∞ if
R ∞ lim
R bp >
Rb 5
R ∞ 1 used the fact that
1
1
(b) 1 5 dx Solution: 1 5 dx = limb→∞ 1 5 dx = limb→∞ 1 x− 4 dx =
x4
x4
x4
b
R∞ 1
− 41 − 14
limb→∞ −4x = limb→∞ −4b +4 = 4. So 1 5 dx converges. Here
x4
1
have used the fact that limRb→∞ bq = 0 if q < 0.
Rwe
2
∞ x2
(c) 0 x3 +1 dx Solution: Note that x3x+1 dx = 31 ln |x3 + 1| + C by substiRb 2
R∞ 2
tuting u = x3 +1 and du
= x2 dx. Thus 1 x3x+1 dx = limb→∞ 1 x3x+1 dx =
3
b
R∞ 2
3
limb→∞ ln |x + 1| = limb→∞ ln |b3 + 1| − ln 1 = ∞. So 0 x3x+1 dx di1
3
verges.
R ∞ 1 We have used limb→∞ b + 1 = ∞ and limx→∞ ln x = ∞.1
(d) e x(ln x)3 dx Solution: We have used u = ln x and du = x dx to
R
R 1
integrate x(ln1x)3 dx =
du = − 12 u−2 + C = − 12 (ln1x)2 + C and
u3
R∞ 1
dx = limb→∞ − 21 (ln1x)2 |be = limb→∞ − 21 (ln1b)2 + 21 (ln1e)2 = 21 . We
e x(ln x)3
used ln e = 1.
Rhave
R∞
Rb
4
∞
3
(e) −∞ x dx Solution: Since 0 x3 dx = limb→∞ 0 x3 dx = limb→∞ b4 = ∞,
R∞ 3
so
x dx diverges.
R ∞ −∞
R
3
2 −x3
(f) −∞ x e dx Solution: We first integrate x2 e−x dx. Let u = −x3 .
R
R
3
. Hence x2 e−x dx = eu ·(− du
)=
Then du = −3x2 dx and x2 dx = − du
3
3
3
R
R
u
−x
3
3
∞
b
− e3 + C = − e 3 + C. Thus 0 x2 e−x dx = limb→∞ 0 x2 e−x dx =
b
R0
−b3
−x3 3
limb→∞ − e 3 = limb→∞ − e 3 + 13 = 13 . Now we look at −∞ x2 e−x dx.
0
0
R 0 2 −x3
R0
−x3 3
We have −∞ x e dx = limb→−∞ b x2 e−x dx = limb→−∞ − e 3 =
−b
limb→−∞ − 31 + e 3
limx→∞ ex = ∞.
3
b
3
= ∞. Here have have used limb→−∞ −b = ∞ and
R
R∞
3
3
Thus x2 e−x dx diverges. Note that −∞ x2 e−x dx
Solution to Review Problem for Midterm #2
MATH 1760: page 8 of 10
R∞
R0
3
3
converges if both −∞ x2 e−x dx and 0 x2 e−x dx converge. Note that
R ∞ 2 −x3
R0
R∞
3
3
x e dx diverges if either −∞ x2 e−x dx or 0 x2 e−x dx diverges.
−∞
dy
dy
3. Solve the differential equation (a) dx
= y 2 − 4y + 3 with y(0) = 3 (b) dx
=
2
y − 4y + 3 with y(0) = 2
Solution: (a) Note that y 2 − 4y + 3 = (y − 1)(y − 3) So y = 1 and y = 3 are
dy
equilibrium of the differential equation dx
= y 2 − 4y + 3 So if y(0) = 3
then y(x) = 3 for all x.
dy
(b)Note
= y2 −
dx
R 4y + 3 = 1(y − 1)(y A− 3). BSeparating the variable, we
R that
1
have (y−1)(y−3) dy = dx. (y−1)(y−3) = y−1 + y−3 . Multiplying (y − 1)(y − 3)
to both sides, we have 1 = A(y − 3) + B(y − 1). Plugging y = 1 and
1
1
1
y = 3, we have A = − 21 and B = 21 . Thus (y−1)(y−3)
= − 21 · y−1
+ 12 · y−3
and
R
R
y−3
1
1
1
1
1
1
1
1
dy = (− 2 · y−1 + 2 · y−3 )dy = − 2 ln |y−1|+ 2 ln |y−3| = 2 ln | y−1 |+c.
(y−1)(y−3)
R
R
y−3
1
dy = dx can be integrated to get 12 ln | y−1
| = x + c,
Thus (y−1)(y−3)
y−3
y−3
ln | y−1 | = 2x + c1 (here c1 = 2c) y−1 = e2x+c1 = ec1 · e2x = Ce2x where C = ec1 .
From y−3
= Ce2x , we have y − 3 = Ce2x (y − 1), y − 3 = Ce2x y − Ce2x ,
y−1
2x
3−Ce
y − Ce2x y = 3 − Ce2x , y(1 − Ce2x ) = 3 − Ce2x and y = 1−Ce
2x . Using the
0
initial condition y(0) = 2, we have 3−Ce
= 2, 3−C
= 2, 3 − C = 2 − 2C,
1−Ce0
1−C
3+e2x
−C + 2C = 2 − 3 and C = −1. Thus y = 1+e2x .
Remark: Note that if you try the initial condition y(0) = 3, we have
3−C
= 3, 3 − C = 3 − 3C, 2C = 0 and C = 0. Thus y = 3
1−C
4. Suppose that dy
= −y 2 (y − 3)(y − 5)
dt
(a) Determine the equilibria of this differential equation. Solution:
Solving y 2 (y − 3)(y − 5) = 0, we have we have y = 0, y = 3 or y = 5.
So the equilibria of this differential equation are y = 0, y = 3 or
y = 5.
(b) Graph dy
as a function of y, and use your graph to discuss the
dt
stability of the equilibria. Solution: We plug in the value y = −1 <
0, 0 < y = 1 < 3 and 3 < y = 4 < 5 and 5 < y = 6 to −y 2 (y − 3)(y − 5).
y
-1
1
4
2
2
2
2
−y (y − 3)(y − 5) −(−1) (−1 − 3)(−1 − 5) −1 (1 − 3)(1 − 5) −4 (4 − 3)(4 − 5)
sign
+
y
6
2
2
−y (y − 3)(y − 5) −6 (6 − 3)(6 − 5)
sign
From the graph below, we know that y = 0 is unstable, y = 3 is
unstable, y = 5 is stable.
MATH 1760: page 9 of 10
Solution to Review Problem for Midterm #2
F IGURE 1. Graph for problem 4
(c) What can you say about the solution limt→∞ y(t) if y(0) = 1 or
y(0) = 4? Solution: If y(0) = 1, we know that y is decreasing to 0
and limt→∞ y(t) = 0. If y(0) = 4, we know that y is increasing to 5
and limt→∞ y(t) = 5.
dy
5. Suppose that dx
= g(y) and the graph of dy
as a function of y is given
dt
by the figure above
F IGURE 2. Graph for problem 5
(a) Determine the equilibria of this differential equation. Solution:
The equilibrium of this differential equation are y = −2, y = 1,
y = 5 and y = 8
(b) Use the graph to discuss the stability of the equilibria. Solution:
From the graph on next page, we know that y = −2 is unstable,
y = 1 is stable, y = 5 is unstable and y = 8 is unstable.
Solution to Review Problem for Midterm #2
6.
7.
8.
9.
10.
11.
12.
MATH 1760: page 10 of 10
(c) What can you say about limt→∞ y(t) if y(0) = 3 or y(0) = 6 ?
Solution: If y(0) = 3, we know that y is decreasing to 1 and
limt→∞ y(t) = 1. If y(0) = 6, we know that y is increasing to 8
and limt→∞ y(t) = 8.
A standard deck contains 52 different cards. In how many ways can
you select 7 cards from the deck? The order of the card is not impor52!
52!
tant. So the answer is C(52, 7) = 7!(52−7)!
= 7!45!
= 52·51·50·49·48·47·46
7·6·5·4·3·2·1
Suppose you want to plant a flower bed with 3 different plants. You
can choose from among 5 plants How many different choices do you
have? The order of the flower is not important. So the answer is
5!
5!
C(5, 3) = 3!(5−3)!
= 3!2!
= 5·4·3
= 10.
3·2·1
A committee of 2 people must be chosen from a group of 4. The
committee consists of a president, a vice president. How many committees can be selected? Solution:There are 4 ways to choose the
president and 3 ways to choose the vice president. So the answer is
4 · 3 = 12. (The order is important because the role of the committee
member is different. The answer is P (4, 2) = 4!
= 4 · 3 = 12.)
2!
An amino acid is encoded by triplet nucleotides (three nucleotides).
How many different amino acids are possible if there are 4 different
nucleotides that can be chosen for a triple? Solution: There are 4
possible choices for each nucleotides. The answer is 4 · 4 · 4 = 64.
You have just enough time to play 3 different songs out of 5 from your
favorite CD. In how many ways can you program your CD player to
play the 3 songs? Solution: The order of the song is important. So
5!
the answer is P (5, 3) = (5−3)!
= 5!
= 5 · 4 · 3 = 60.
2!
Suppose that you want to investigate the effects of leaf damage on
the performance of drought-stressed plants. You plan to use 5 levels
of leaf damage and 3 different watering protocol, you plan to to have
4 replicates. What is the total number of replicates? Solution: The
answer is 5 · 3 · 4 = 60.
Ten children are divided up into three groups, of 2, 3 and 3 children,
respectively. In how many ways can this be done if the order within
each group is not important? Solution: The order of the children is
10!
not important. So the answer is 2!3!5!
.