Download Answer Key for Practice Exam #1, Spring 2012

Midterm Exam #1
Math 121
February 22, 2012
1.
[30 Points] Evaluate each of the following limits. Please justify your answers. Be clear if the
limit equals a value, +∞ or −∞, or Does Not Exist.
(a)
0
0
−2 sinh(2x) 0 L’H
−4 cosh(2x)
−4
1 − cosh(2x) 0 L’H
= lim
= lim
= 4
=
1
1
x→0
x→0
x + ln(1 − x)
−1
1 − 1−x
− (1−x)2
!
lim
x→0
(b)
!
lim
ln
1
4 x 1∞
x
lim e −
= lim e
x→∞
x→∞
x
(c)
lim
= e
3
3
1 + (3x)2
=
4
4
p
2
1 − (4x)
0
arctan(3x) 0 L’H
= lim
x→0
arcsin(4x)
x→0
∞·0
!
1
ln e x −
x→∞
4
x
1
x
1
ex −

lim
L’H
( 00 )
=
e
x→∞


4
x
!x !
1
1
ex −
lim ln
= ex→∞

4
e −
x
1
x
1

· e x − x12 +

4
x
4
x2
x 4
lim x ln e −
x→∞
x
=e
1
x
− x12


1
1
1
4
 1 
 1 
lim 
· e x − 2 + 2 (−x2 )
· ex − 4
lim 


1
1
4
4
x→∞
x→∞
x
x
ex −
ex −
x
x
= e−3
=e
=e

2.
(a)
[30 Points]
Z
ln 7
0

Compute the following definite integral. Please simplify your answer.
ln 7 Z
x sinh x dx = x cosh x −
0
ln 7
0
ln 7
ln 7
cosh x dx = x cosh x − sinh x
0
= ln 7 cosh(ln 7) − 0 − sinh(ln 7) + sinh 0 = ln 7 cosh(ln 7) − sinh(ln 7) + 0
!
!
ln 7
ln 7
7 + 71
7 − 17
e + e− ln 7
e − e− ln 7
= ln 7
−
= ln 7
−
2
2
2
2
!
!
48
50
24
25
7
7
−
= ln 7
−
= ln 7
2
2
7
7
u=x
dv = sinh xdx
I.B.P.
du = dx v = cosh x
1
0
2.
(Continued)
Compute the following definite integral. Please simplify your answer.
Z √3
Z √3
x+1
x
1
√
√
√
dx =
dx +
dx
(b)
2
2
4−x
4−x
4 − x2
1
1
1
Z u=1
Z √3
Z u=1
Z √3
1
1
1
1
1
r
√ du +
r dx = −
= − 21
u− 2 du +
2 u=3
u
2
u=3
1
1
4 1 − x4
2
1−
√
Z
3
√3
2
√ 1
1
√
dw = − u + arcsin w
1
1 − w2
w= 12
3
3
2
√ !
√
√
√
√
3
1
π π
π
− arcsin
= −1 + 3 + − = −1 + 3 +
= − 1 + 3 + arcsin
2
2
3
6
6
1 Z
1
= − 2 · 2u +
1
2
Here
w=
√
3
2
u = 4 − x2
du = −2xdx
− 21 du = xdx
and
x=
√1 =⇒ u = 3
x = 3 =⇒ u = 1
x
2
1
dw = dx
2
and
x = 1 =⇒ w =
√
x = 3 =⇒ w =
w =
Also here
1
2
√
3
2
x
1
dx = arcsin
+ C.
2
4 − x2
OR you could use a trigonometric substitution:
OR you can use the fact that
Z
√
2
x = 2 sin θ
θ
√
4 − x2
Trig. Substitute
dx = 2 cos θdθ
Z
√
3
1
=
Z
π
6
π
3
Z π
3 2 sin θ + 1
2 sin θ + 1
p
2 cos θ dθ
2 cos θ dθ =
2
π
π
2 cos θ
4 − 4 sin θ
6
6
π
3
π π
π π
2 sin θ + 1 dθ = −2 cos θ + θ = −2 cos + + 2 cos −
3
3
6
6
π
x+1
√
dx =
4 − x2
π
3
x
= −1 +
Z
6
π
3
+
√
3−
π
6
= −1 +
√
3+
π
6
2
x 2
2
dx
2.
(Continued)
Compute the following definite integral. Please simplify your answer.
Z 1
1 −2x 1
1
x3 1
2x
−x
xe dx − e +2
x + x + 2x dx =
dx =
(c)
e
e
3 0
2
0
0
0
0
!
Z
1
1
1 −2x 1
x3 1
1 −2x x3
−x
−x −x
−x
=
e
dx
−
+
+
2
−xe
e
=
+
2
−xe
−
e
− e 3 0
2
3
2
0
0
0
1
1
1
1
1
1
+ 2 −e−1 − e−1 − e−2 − 0 − 0 − 2 − e0 = − 4e−1 − e−2 + 2 +
3
2
2
3
2
2
1
1
x+ x
e
Z
=
2
Z
1
2
17 4
1
− − 2
6
e 2e
u=x
dv = e−x dx
I.B.P. above (on middle term)
du = dx v = −e−x
3.
[40 Points]
(a)
Compute the following indefinite integral.
Z
Z
1
x2
1
sin2 θ
x2
x2
√
p
arcsin x −
arcsin x −
x arcsin x dx =
dx =
· cos θ dθ
2
2
2
2
1 − x2
1 − sin2 θ
Z
Z
Z
1
1
x2
1
sin2 θ
sin2 θ
x2
√
arcsin x−
arcsin x−
·cos θ dθ =
arcsin x−
sin2 θdθ
·cos θ dθ =
2
2
2
2
cos
θ
2
2
cos θ
Z
Z
1 − cos(2θ)
1
x2
1
1 − cos(2θ)dθ
arcsin x −
dθ =
arcsin x −
2
2
2
4
1
x2
1
1
1
arcsin x −
arcsin x − θ + sin(2θ) + C
θ − sin(2θ) + C =
4
2
2
4
8
Z
=
x2
2
=
x2
2
=
x2
2
=
x2
1
1
1
1 p
x2
arcsin x − θ + 2 sin θ cos θ + C =
arcsin x − arcsin x + x 1 − x2 + C
2
4
8
2
4
4
u = arcsin x
du = √
dv = xdx
x2
1
dx v =
2
1 − x2
1
x = sin θ
θ
√
1 − x2
Trig. Substitute
dx = cos θdθ
(b)
Z
1
(x2
+ 4)
5
2
dx =
Z
x
1
(4 tan2 θ
+ 4)
5
2
· 2 sec2 θ dθ =
3
Z
1
5
(4 sec2 θ) 2
· 2 sec2 θ dθ
=
=
=
=
=
1
1
· 2 sec2 θ dθ
5
2
5
(2
sec
θ)
( 4 sec θ)
Z
Z
Z
1
1
1
sec2 θ
1
dθ = 4
dθ =
cos3 θ dθ
24
sec5 θ
2
sec3 θ
16
Z
Z
Z
1
1
1
2
2
cos θ cos θ dθ =
(1 − sin θ) cos θ dθ =
(1 − w2 ) dw
16
16
16
3 !
1
w3
1
sin3 θ
1
x
1
x
√
√
w−
+C =
sin θ −
+C =
+C
−
16
3
16
3
16
x2 + 4 3
x2 + 4
!
1
x3
x
√
+C
−
3
16
x2 + 4 3(x2 + 4) 2
Z
√
· 2 sec2 θ dθ =
Z
√
x = 2 tan θ
x2 + 4
x
θ
Trig. Substitute
dx = 2 sec2 θdθ
2
Standard w substitution for odd trig. integral
Z
cos3 θ dθ technique:
w = sin θ
dw = cos θ dθ
Z 2
x2
x +1−1
2
(c)
ln(x + 1) dx = x ln(x + 1) − 2
dx = x ln(x + 1) − 2
dx
2
x +1
x2 + 1
Z
Z 2
Z
Z
1
1
x +1
2
2
1 dx −
dx −
dx = x ln(x + 1) − 2
dx
= x ln(x + 1) − 2
x2 + 1
x2 + 1
x2 + 1
Z
2
2
Z
= x ln(x2 + 1) − 2 (x − arctan x) + C = x ln(x2 + 1) − 2x + 2 arctan x + C
u = ln(x2 + 1)
I.B.P.
du =
dv = dx
2x
dx v = x
+1
x2
OR you could do a trigonometric substitution on
=
Z
tan2 θ
sec2 x dθ =
sec2 x
Z
2
tan θ dθ =
Z
Z
x2
dx =
x2 + 1
Z
tan2 θ
sec2 x dθ
tan2 θ + 1
sec2 θ − 1 dθ = tan θ − θ + C = x − arctan x + C
4
√
x2 + 1
x = tan θ
x
θ
Trig. Substitute
dx = sec2 θdθ
1
***************************************************************************
OPTIONAL BONUS
Do not attempt these unless you are completely done with the rest of the exam.
***************************************************************************
OPTIONAL BONUS #1
1.
Z
e
√
√
1+ x
dx = 4
Z
2
Compute the following indefinite integral.
w(w − 1)e dw = 4
w
Z
3
(w − w)e dw = 4
w
Z
3 w
w e dw −
Z
w
we dw
w3 ew − 3w2 ew + 6wew − 6ew − (wew − ew ) + C
= 4 w3 ew − 3w2 ew + 6wew − 6ew − wew + ew + C = 4 w3 ew − 3w2 ew + 5wew − 5ew + C
p
p
√ √ p
√ 3 √1+√x
√ 2 √1+√x
√ √1+√x
+5
− 5e 1+ x + C
=4
1+ x e
1+ x e
1+ x e
−3
= 4 ((∗) − (∗∗)) = 4
= 4e
√
√
√
1+ x
p
p
p
√ 3
√ 2
√ 1+ x −3
1+ x +5
1+ x −5 +C
p
i
p
√
√
√ 1 + x − 3 (1 + x) + 5
1+ x −5 +C
p
i
√ √ hp
√
√ p
√ √
√ = 4e 1+ x
1+ x+ x
1+ x −3−3 x+5
1+ x −5 +C
= 4e
= 4e
Here
√ h
1+ x (1
√
+
√
√ h p
1+ x 6 1
x)
+
√
x+
√ p
√ √ i
x
1+ x −8−3 x +C
p
√
√
√
w = 1 + x
=⇒ w2 = 1 + x =⇒ w2 − 1 = x
1
√
√1 √
dx
dw =
2 x
2
p
√ √
4
1+ x
xdw = dx
1+ x
4w(w2 − 1)dw = dx
*****************************************************************************
Z
Z
Z
3 w
3 w
2 w
3 w
2 w
w
(*)Aside:
w e dw = w e − 3 w e dw = w e − 3 w e − 2 we dw
= w3 ew − 3w2 ew + 6
Z
Z
wew dw = w3 ew − 3w2 ew + 6 wew − ew dw
5
= w3 ew − 3w2 ew + 6
Z
wew dw = w3 ew − 3w2 ew + 6wew − 6ew + C
u = w3
u = w2
dv = ew dw
First I.B.P.
dv = ew dw
Second I.B.P.
du = 3w2 dw v = ew
u=w
du = 2wdw v = ew
dv = ew dw
Third I.B.P.
du = dw v = ew
*****************************************************************************
Z
Z
(**)Aside:
wew dw = wew − ew dw = wew − ew + C
u=w
dv = ew dw
I.B.P.
du = dw v = ew
*****************************************************************************
OPTIONAL BONUS #2
Z
Z
Z
IBP
sec x dx = sec x sec x dx = sec x tan x − sec x tan2 x dx
Z
Z
2
= sec x tan x − sec x(sec x − 1) dx = sec x tan x − sec3 x − sec x dx
Z
Z
Z
(∗∗∗)
= sec x tan x − sec3 x dx + sec x dx = sec x tan x + ln | sec x + tan x| − sec3 x dx
2.
3
Compute the following indefinite integral.
2
Regroup:
Z
Z
sec3 x dx = sec x tan x + ln | sec x + tan x| − sec3 x dx
which yields
Z
2 sec3 x dx = sec x tan x + ln | sec x + tan x| + C
and then
Z
1
sec3 x dx = (sec x tan x + ln | sec x + tan x|) + C
2
u = sec x
dv = sec2 xdx
I.B.P.
du = sec x tan x v = tan x
(***) Aside
6
Z
sec x dx =
Z
sec x
sec x + tan x
sec x + tan x
= ln | sec x + tan x| + C
Here
dx =
Z
sec2 x + sec x tan x
dx =
sec x + tan x
w = sec x + tan x
dw = sec x tan x + sec2 xdx
7
Z
1
dw
w