Download Solutions to Test One sample questions.

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Page 1 of 6:
..
1. Find the derivatives
(a,
$[ x4 .e3x 1 =
(d)
$(cosh(arctan x))
-
-
3
4% Y
33
-+ 2. 3 e
=
&(arc+&
x2+ I
n)_
I
3 X
CALCULUS TWO TEST ONE
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Page 2 of 6
4
2. Find
Y =
by logarithmic differentiation.
e3X.(x3+ I ) ~
"Jm-
AlaJ=
-
+ 7.L
3s
lq 1 = A
- R (x7r v -'Is
/ x 3 + ~-I -$&h2+'+)
2X
-
-
3. Find the indefinite integrals.
X rr~%
C
(a)
1 x 3 InXdx
/
br
ysr'C;*1~.
(Useintegrationbyparts.)
dd
= x ~ $/X & -- AK &
\
=
IXY_R*)(
t'
.
o r
GO TO PAGE 3.
-
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CALCULUS TWO TESTONE
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-
Page 3 of 6.
= 2+x3
olu -. 3xZ&
L(
r ~ ( ~ ) c ~ i - i . u c ~d - * B - I
++C
+C
z
,+g p 2 - 4 ) W 4 -
x3
K .- X
gfi.-2
m 2 & =1 -
L(
Z
&
-
L&3(
7
x
dx
using trigonometric substitution.
2
x = YQx2- Y -- y h 2 .
3 2
,,(=
&O
&=
oaBa
fh.0
- 1-6
, J,-,
- 16
/
IA
\
GO TO PAGE 4.
1
16
,1j-t,
-"&
- /6
-
5 11
- \
7
t
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CALCULUS TWO TEST ONE
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4. Compute
GO TO PAGE 5.
x2
SOL
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Page 4 of 6.
CALCULUS TWO TEST ONE
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a
Page 5 o i b
6. Solve the initial value problem.
j
c
S
!
y,
y =2fi
+I-
Vat~.UIS
CO~(X)'=
dx
where
J2=pmi 4
e -Qulal
e
y,,
;5
so
y C k X
0
J~L~IJ-J.
a
&4t*uf
tk
6 =
=4
( = -L/C"
+X
C,~
1
-& l m r I
Z
if X
aw+;4;ffen*f;ate.
Cc i
/
tt 3 . F .
1s
' f c,& ~
12, = &o".,d(
ecp-'-.
p L + i1 ~
wkRc
:+:&Je dt'f('*:*
- G C
C W-"- -~
t-erP
&
is o U 4
k K W k P .
C =a
7. Let f-be a one to one differentiable function.
f(1)-= 3, f(3) = 5, f(5) =9, f(1) = 4, F(3) = 6, f(5) = 8
Find, (f-' )
'(5) using the above information and a theorem.
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CALCULUS TWO TEST ONE SAMPLE QUESTIONS. ACTUAL TEST MAY VARY.
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8. Compute
¶ 01
2x + 3
2
x + 2x + 2
¶ 01
dx =
2x + 2
2
x + 2x + 2
dx +
¶ 01
1
(x + 1 ) 2 + 1
dx
= [ln x 2 + 2x + 2 ] 10 + [tan −1 (x + 1 )] 10
= ln 1 2 + 2(1 ) + 2 − ln 0 2 + 2(0 ) + 2 + tan −1 (2 ) − tan −1 (1 )
= ln(5 ) − ln(2 ) + tan −1 (2 ) −
¶
9. Compute
4
0
dx
(25 − x 2 ) 5/2
x = 5 sin .
Let
Then
4
= ln(2.5 ) + tan −1 (2 ) −
using trigonometric substitution.
dx = 5 cos d, 25 − x 2 = 25 cos 2 ,
¶
1 =
1
x
.
The
integral
then
equals
5
0
4
4
sin −1 5 . Note that tan sin −1 5
¶ 0
5 cos d
3125 cos 5 = sin −1
1
¶ 0
=
1
625
=
1 1
625 3
GO TO PAGE 2.
1
4
= ¶0
1
d
625 cos 4 sec 2 sec 2 d =
tan 3 + tan 1
0
=
=
1
625
and
5 cos d
where
(25 cos 2 ) 5/2
= 43 . The integral equals
1
625
¶ 0
1
sec 4 d
¶ 0 (tan 2 1
1 1 4 3
625 3 3
+
4
3
+ 1 ) d(tan )
− 0=
172
50625