M216
TEST #4
Name:
SHOW YOUR WORK: WITHOUT DETAILED EXPLANATIONS , NO FULL CREDIT
For each problem, qualify your answer. See table below for points.
ANSWER (A) certain (B) fairly sure (C) not sure
EXTRA POINTS for each problem:
Correct
+1
+ 12
− 12
Wrong
−1
− 12
+ 12
Z

? Pick ONE of these answers:
2 − 4 + 8
¯
¯
(a) (4 points) This is just the derivative of a natural logarithm, so the answer is ln ¯2 − 4 + 8¯ + 
1. (10 points) How would you solve the integral
(b) (10 points) By completing the square in the denominator and substituting with  tan ; (in this case
 − 2 = 2 tan )
(c) (8 points) By factoring the denominator 2 − 4 + 8 = ( − ) ( − ) and then using partial fractions;
(d) (4 points)Add and subtract 2 on the top of the fraction and then split into two integrals;
(e) (0 points) This is equal to
− 1 − 14 ln || + 18  + ;
R

2
−
R

4
+
R

8 
which are all easy integrals, so the answer is
¤ How sure are you about your answer to the problem above?
(A) I am certain I am right!
(B) Fairly sure
2. (10 points) Solve using partial fractions (not tables: SHOW WORK):
OR
Z
(C) Not sure at all
Z
3
2
 =
−



2
=
+
+
 ( − 1) ( + 1)

−1 +1
2 =  ( − 1) ( + 1) +  ( + 1) +  ( − 1)

= Z0 :  = −2;  = 1 :  = 1;  = −1 :  = 1
Z
Z
−2
1
1
=
 +
 +
 = −2 ln || + ln | − 1| + ln | + 1| + 

−1
+1
⎤
⎡
Z
Z
 = sec 
cos 
sec  tan 

⎦
⎣
=2
=  = sec  tan  = 2

2
2
 ( − 1)
sin 
sec  tan 
2 − 1 = tan2 
¯
¯√
¯ 2 − 1 ¯
¯
¯
= 2 ln |sin | +  = 2 ln ¯
¯+
¯
¯

2
 = 2
3
 −
Z
¤ How sure are you about your answer to the problem above?
(A) I am certain I am right!
(B) Fairly sure
(C) Not sure at all
3. (10 points) When we want to solve the integral
Z
32 +7−12
5 −23 −22 −3−2 
using the method of partial fractions, we
¡
¢
first factor the denominator 5 − 23 − 22 − 3 − 2 = ( − 2) 2 + 1 ( + 1)2  and then split it into simpler
fractions. Which of the following partial fractions decompositions is correct? Pick ONE of these answers (no
need to evaluate the integral):
(a) (0 points)
32 +7−12
(−2)(2 +1)(+1)2
=

−2
+

2 +1
(b) (5 points)
32 +7−12
(−2)(2 +1)(+1)2
=

−2
+
+
2 +1
+

+1
(c) (3 points)
32 +7−12
(−2)(2 +1)(+1)2
=

−2
+
+
2 +1
+

(+1)2
(d) (10 points)
(e) (5 points)
32 +7−12
(−2)(2 +1)(+1)2
32 +7−12
(−2)(2 +1)(+1)2
=
=

−2

−2
+
+
+
+
2 +1

2 +1
+

(+1)2
+
+

+1

+1
+
+
(+1)2
+

(+1)2
+
(+1)2
¤ How sure are you about your answer to the problem above?
(A) I am certain I am right!
(B) Fairly sure
(C) Not sure at all
4. (10 points) Evaluate :
¸ Z µ
∙ 3 −2
¶
Z 3
Z
Z
2
2
 − 2
2
2 −4 =  + 2 −4
=

+

=

=

+

2
2
2
 −4
 −4
 −4
(Long division)
¸
∙
¯
¯
1
1
 = 2 − 4
= 2 + ln || +  = 2 + ln ¯2 − 4¯ + 
=
 = 2
2
2
1 2
=
 + ln | − 2| + ln | + 2| + 
2
¤ How sure are you about your answer to the problem above?
(A) I am certain I am right!
(B) Fairly sure
(C) Not sure at all
5. Consider the integral
Z
2
 ()  . In order to ease your calculations, here is the table of some selected values
−2
for this function (you don’t have to use ALL of these numbers - just pick the ones you need).

 ()

 ()

 ()

 ()

 ()

 ()
-3.00
0.0
-2.00 35.0
-1.00
0.0
0.00
0.0
1.00
8.0
2.00 -15.0
-2.75 29.4
-1.75 25.3
-0.75 -3.6
0.25
3.5
1.25
5.2
2.25 -21.6
-2.50 41.3
-1.50 15.2
-0.50 -4.4
0.50
6.6
1.50
0.0
2.50 -24.1
-2.25 41.5
-1.25
6.4
-0.25 -2.9
0.75
8.3
1.75 -7.1
2.75 -18.5
y

3.00
3.25
3.50
3.75
 ()
0.0
37.8
102.4
202.9
40
20
-3
-2
Z
-1
2
3
x
 ()
2
−2
( − 3) (Z− 15) () ( − (−1)) ( − (−3))  = 17 6
2
 ()  using the Midpoint rule and  = 4 intervals.
(a) (3 points) Estimate the integral
 ()
 (−15) = 152
 (−05) = −44
 (05) = 66
 (15) = 0

−15
−05
05
15
1
-20
−2
 = 2−(−2)
( (−15) +  (−05) +  (05) +  (15))
4
= 44 ((152) + (−44) + (66) + (0)) = 17 4
¤ How sure are you about your answer to the problem above?
(A) I am certain I am right!
(B) Fairly sure
(C) Not sure at all
Z 2
(b) (3 points) Estimate the integral
 ()  using the Trapezoid rule and  = 4 intervals.

−2
−1
0
1
2
 ()
 (−2) = 35
 (−1) = 0
 (0) = 0
 (1) = 8
 (2) = −15
−2
 = 2−(−2)
( (−2) + 2 (−1) + 2 (0) + 2 (1) +  (2))
2·4
= 12 ((35) + 2 (0) + 2 (0) + 2 (8) + (−15)) = 18
¤ How sure are you about your answer to the problem above?
(A) I am certain I am right!
(B) Fairly sure
(C) Not sure at all
Z 2
(c) (4 points) Estimate the integral
 ()  using the Simpson’s rule and  = 8 intervals.

−200
−150
−100
−050
 ()
350
152
00
−44

000
050
100
150
200
 ()
00
66
80
00
−150
−2
=
2−(−2)
3·8
( (−2) + 4 (−15) + 2 (−1) +
+4 (−05) + 2 (0) + 4 (05) + 2 (1) + 4 (15) +  (2))
= 16 ((1) (35) + 4 (152) + 2 (0) + 4 (−44) +
+2 (0) + 4 (66) + 2 (8) + 4 (0) + (−15)) = 17 6
(or) =
1
3
(2 +  ) =
1
3
(2 (174) + 18) = 17 6
¤ How sure are you about your answer to the problem above?
(A) I am certain I am right!
(B) Fairly sure
(C) Not sure at all
6. (10 points) Evaluate using any method (if you use the table of integrals - state the formula):
Z
2
√
 =
42 − 9
=
∙
¸ Z
∙
¸
1 2
1
 = 2 :  = 12 
  44
4
√
=

=
 = 1 
=3
2 − 32 2
¶
µ p 2
¯
¯
´
³
2
p
p
1 
1 p
3
9
¯
¯
2 − 32 +
ln ¯ + 2 − 32 ¯ +  =  42 − 9 +
ln 2 + 42 − 9 + 
8 2
2
8
16
¤ How sure are you about your answer to the problem above?
(A) I am certain I am right!
(B) Fairly sure
(C) Not sure at all
7. (10 points) Evaluate using any method (if you use the table of integrals - state the formula):
Z
¸ Z
∙
¸
22
 = sin 
  56
√
=
 =
 = cos 
=2:=1
2+
¢√
¢√
4 ¡
2 ¡ 2
= 2
3 − 8 + 8 · 4 2 +  +  =
3 sin2  − 8 sin  + 32 2 + sin  + 
15
15
2 sin2  cos 
√
 =
2 + sin 
∙
¤ How sure are you about your answer to the problem above?
(A) I am certain I am right!
(B) Fairly sure
(C) Not sure at all
8. (10 points) Evaluate using the table of integrals:
Z1
0
1
 =
2 − 4
=
¯¶=1 µ
¯¶
¯
¯
¯
¯
¸ µ
1 ¯¯ 0 − 2 ¯¯
1 ¯¯ 1 − 2 ¯¯
1 ¯¯  − 2 ¯¯
  20
−
=
=
ln ¯
ln
ln
=2
4
 + 2 ¯ =0
4 ¯1 + 2¯
4 ¯0 + 2¯
¯¶
µ¯
¯ 1¯
1
1
1 1
1
¯
ln ¯− ¯¯ − ln |−1| = ln − 0 = − ln 3 ≈ −0275
4
3
4
4 3
4
∙
¤ How sure are you about your answer to the problem above?
(A) I am certain I am right!
(B) Fairly sure
(C) Not sure at all
9. (10 points) Evaluate using partial fractions method:
Z1
0
⎤


¶
Z 1µ 1
= −2
+ +2
1
4
4
⎦
=
−

1 =  ( + 2) +  ( − 2)
−2 +2
0
 = 2 :  = 14 , and  = −2 :  = − 14
µ
¶1 µ
¶ µ
¶
1
1
1
1
1
1
=
ln | − 2| − ln | + 2| =
ln 1 − ln 3 −
ln 2 − ln 2
4
4
4
4
4
4
0
1
= − ln 3 ≈ −0275
4
⎡
1
 = ⎣
2 − 4
1
2 −4
¤ How sure are you about your answer to the problem above?
(A) I am certain I am right!
(B) Fairly sure
(C) Not sure at all
10. (10 points) Evaluate using any method (if you use the table of integrals - state the formula):
¸ Z ?
1
 = ¡3 + 2 +  +¢1
=
 = (ln ||)??
2
+
2
+
1


=
3

0
?
¯¢1
¡ ¯
= ln ¯3 + 2 +  + 1¯ 0 = ln 4
 using partial fractions
¡
¢
32 + 2 + 1

 + 
3 + 2 +  + 1 = 2 + 1 ( + 1) so 3
=
+ 2
 + 2 +  + 1
+1
 +1
¡
¢
32 + 2 + 1 =  2 + 1 + ( + ) ( + 1)
 = −1 : 2 = 2 so  = 1
 = 0 : 1 = 1 + , so  = 0
 = 1 : 6 = 2 + 2, so  = 2
¸
∙
Z 1
Z 1
32 + 2 + 1
1
2
note:R if  = 2 +R1

=

+

=
1
2
 = 2, so 22
3 + 2 +  + 1
+1  =
  = ln 
0 +1
0  +1
¯ 2
¯¢1
¡
= ln | + 1| + ln ¯ + 1¯ 0 = 2 ln 2
¤ How sure are you about your answer to the problem above?
(A) I am certain I am right!
(B) Fairly sure
(C) Not sure at all
Z
Z
0
1
1
32 + 2 + 1
 =
3
 + 2 +  + 1
∙