M 106 - INTEGRAL CALCULUS
Dr. Tariq A. AlFadhel 1
Solution of the first mid-term exam
First semester 1432-1433 H
Multiple choice questions (One mark for each question)
n
X
(3 + k)2 is equal to :
Question 1. The sum
k=1
1
2n3 + 21n2 + 54n
6
1 3
n + 21n2 + 73n
(b).
6
1
2n3 + 19n2 + 73n
(c).
6
1
(d).
2n3 + 21n2 + 73n
6
(a).
Answer:
n
X
(3 + k 2 ) =
k=1
n
X
(9 + 6k + k 2 ) =
n
X
k=1
k=1
9+6
n
X
k=1
k+
n
X
k2
k=1
2n3 + n2 + 2n2 + n
n(n + 1) n(n + 1)(2n + 1)
+
= 9n + 3n2 + 3n +
= 9n + 6
2
6
6
18n2 + 72n + 2n3 + 3n2 + n
2n3 + 3n2 + n
2
=
= 3n + 12n +
6
6
3
2
2n + 21n + 73n
1
3
2
=
=
2n + 21n + 73n
6
6
The right answer is (d)
Question 2. The value of the integral
Z
sin(1 + 3x) dx is equal to :
1
(a). − cos(1 + 3x) + c
3
(b). 3 cos(1 + 3x) + c
(c).
1
cos(1 + 3x) + c
3
(d). − cos(1 + 3x) + c
Z
Z
1
1
Answer:
sin(1 + 3x) dx =
sin(1 + 3x) 3 dx = − cos(1 + 3x) + c
3
3
The right answer is (a)
1 E-mail
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1
Question 3. The number z that satisfies the conclusion of the Mean Value
Theorem for f (x) = x2 on [−2, 0] is :
r
8
(a). −
3
r
8
(b).
3
−2
(c). √
3
2
(d). √
3
Z
0
x2 dx
Answer : f (z) = −2
0 − (−2)
3 0
1 8
−8
1
4
1 x
2
0−
=
=
=
z =
2 3 −2
2
3
2 3
3
r
2
4
= ±√
z=
3
3
2
2
Note that − √ ∈ [−2, 0] but √ ∈
/ [−2, 0]
3
3
The right answer (c)
Question 4. The average value of f (x) =
(a).
−3
2
(b).
−2
3
(c).
2
3
(d).
3
2
Z
0
√
x + 1 on [−1, 0] is equal to :
√
x + 1 dx Z 0
1
Answer : fav = −1
(x + 1) 2 dx
=
0 − (−1)
−1
i
i 2
3 0
3
3
2h
2
2h
(x + 1) 2
(0 + 1) 2 − (−1 + 1) 2 = (1 − 0) =
=
fav =
3
3
3
3
−1
The right answer is (c)
Question 5. If F (x) =
Z
2x
f ′ (t) dt then F ′ (x) is equal to :
x
(a). f (2x) − f (x)
(b). 2f (2x) − f (x)
2
(c). 2f ′ (x)
(d). 2f ′ (x) − f ′ (x)
d
dx
The right answer is (d)
Answer : F ′ (x) =
Z
2x
x
f ′ (t) dt = f ′ (2x) (2) − f ′ (x) (1) = 2f ′ (2x) − f ′ (x)
Question 6. The value of the integral
Z
5cosh x
dx is equal to :
cschx
(a). 5cosh x + c
(b). (ln 5) 5sinh x + c
(c).
5cosh x
+c
ln 5
5sinh x
+c
ln 5
Z
Z cosh x
5cosh x
5
dx = 5cosh x sinh x dx =
+c
Answer :
cschx
ln 5
The right answer is (c)
(d).
√
Question 7. The derivative of the function f (x) = cosh−1 ( x) is equal to :
(a).
1
√
2 x2 − x
(b). √
1
2x2 − x
(c).
1
√
2x x + 1
(d).
1
2x x2 − 1
√
1
√ 2
( x) − 1
1
1
= p
=
= √ √
2 x x−1
2 x(x − 1)
The right answer is (a)
Answer : f ′ (x) = q
1
√
2 x
1
√
2 x2 − x
Question 8. The value of the integral
(a).
1
+c
cos x
(b).
1
+c
sin x
Z
3
(sin x)(sec x)2 dx is equal to :
(c).
1
+c
sec x
1
(sec x)3 + c
3
Z
Z
2
Answer : (sin x)(sec x) dx =
(d).
sin x
dx = −
(cos x)2
(cos x)−1
1
+c=
+c
−1
cos x
The right answer is (a)
Z
(cos x)−2 (− sin x) dx
=−
Question 9. If
(a). ecos
−1
−1
(d). esin
−1
−1
ecos x
√
dx = f (x) + c , then f (x) is equal to :
1 − x2
−1
x
(b). e− cos
(c). −ecos
Z
x
x
x
ecos x
√
Answer :
dx = −
1 − x2
The right answer is (c).
Z
−1
Z
ecos
−1
x
Question 10. The value of the integral
−1
−1
√
dx = −ecos x + c
2
1−x
Z
e2x
dx is equal to :
e4x − 1
1
sin−1 e2x + c
2
1
(b). sinh−1 e2x + c
2
1
(c). cosh−1 e2x + c
2
(d). cosh−1 e2x + c
Z
Z
e2x
1
2e2x
1
p
Answer :
dx
=
dx = cosh−1 (e2x ) + c
4x
e −1
2
2
(e2x )2 − (1)2
The right answer is (c)
(a).
4
Full questions
Question 11. Approximate the integral
Z
1
e4x dx using Simpson’s rule for
0
n = 4. [3 marks]
Answer :
f (x) = e4x , [a, b] = [0, 1] and n = 4 .
∆x = 1−0
4 = 0.25
x
=
0
, x1 = 0.25 , x2 = 0.5 , x3 = 0.75 and x4 = 1
Z0 1
1−0
[f (0) + 4f (0.25) + 2f (0.5) + 4f (0.75) + f (1)]
e4x dx ≈
3(4)
0
1
=
[1 + 4(2.7183) + 2(7.3891) + 4(20.086) + 54.598]
12
1
1
[1 + 10.873 + 14.778 + 80.344 + 54.598] =
[161.59] = 13.466
=
12
Z 112
e4x dx ≈ 13.466
0
2x+1
Question 12. If y = (cosh x)
, then find y ′ . [2 marks]
2x+1
2x+1
Answer : y = (cosh x)
⇒ ln y = ln (cosh x)
= (2x + 1) ln(cosh x)
Differentiate
both
sides
sinh x
y′
= 2 ln(cosh x) + (2x + 1)
= 2 ln(cosh x) + (2x + 1) tanh x
y
cosh x
′
y = y [2 ln(cosh x) + (2x + 1) tanh x]
2x+1
y ′ = (cosh x)
[2 ln(cosh x) + (2x + 1) tanh x]
Question 13. Evaluate the integral
Z
√
x−2
dx [3 marks]
8 − 2x2
Answer :
Z
Z
Z
x−2
x
2
√
√
√
dx =
dx −
dx
2
2
8−
8 − 2x2
Z 2x
Z 8 − 2x
1
1
= (8 − 2x2 )− 2 x dx − 2 q √ √ 2 dx
2
8 −
2x
√
Z
Z
1
1
2
2
2 −2
q √ =−
(8 − 2x ) (−4x) dx − √
√ 2 dx
2
4
2
2 2 −
2x
!
√
1
2
1 (8 − 2x2 ) 2
2x
√
+c
− √ sin−1
=−
1
4
2
2 2
2
x
1p
2
=−
8 − 2x2 − √ sin−1
+c
2
2
2
Z
1
q
Question 14. Evaluate the integral
dx [2 marks]
2
x 4 + (ln x)
Answer :
Z
Z
1
1
ln x
−1
x
q
q
+c
dx =
dx = sinh
2
2
2
x 4 + (ln x)
(2)2 + (ln x)
5
M 106 - INTEGRAL CALCULUS
Dr. Tariq A. AlFadhel 2
Solution of the second mid-term exam
First semester 1432-1433 H
Multiple choice questions (One mark for each question)
ln x
Question 1. lim
is equal to
x→0+ ln(sin x)
(a) ∞
(b) 0
(c) 1
(d) −∞
ln x
−∞
−∞
x→0 ln(sin x)
Apply L’Hôpital’s rule 1
ln x
1 sin x
x
lim
= lim cos x =
= (1)(1) = 1
cos x x
x→0+ ln(sin x)
x→0+
sin x
The right answer is (c)
Answer: lim+
Question 2. If
(a)
1
A
B
=
+
then the value of A is equal to
(x − 4)(x + 2)
x−4 x+2
1
6
(b) 4
(c)
1
4
(d) − 16
A
B
1
=
+
⇒ 1 = A(x + 2) + B(x − 4)
(x − 4)(x + 2)
x−4 x+2
Put x = 4 then 1 = A(4 + 2) + B(4 − 4) ⇒ 1 = 6A ⇒ A = 61
The right answer is (a)
Answer:
Question 3. To evaluate the integral
(a) x = 4 sec θ
Z p
(b) x = 2 cos θ
(c) x = 2 sec θ
(d) x = 2 tan θ
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4x2 − 16 dx , we use the substitution
Answer :
Z p
4x2 − 16 dx =
Z p
(2x)2 − (4)2 dx
So we use the substitution 2x = 4 sec θ ⇒ x = 2 sec θ
The right answer (c)
Question 4. The value of the integral
Z
π
2
cos5 x sin x dx is equal to
0
(a) 0
(b)
1
3
(c) 3
(d)
1
6
Z π2
Z π2
Answer :
cos5 x sin x dx = −
(cos x)5 (− sin x) dx
0 π
0
1
(cos x)6 2
1
=− 0−
=−
=
6
6
6
0
The right answer is (d)
x
2
Question 5. The substitution u = tan
into
(a)
Z
2 du
(b)
Z
du
(c)
Z
1
du
1+u
(d)
Z
1
du
1 + u2
transforms the integral
Z
Answer : Using the half angle substitution u = tan x2
2
1 − u2
and dx =
du
cos x =
2
1+u
1 + u2
Z
Z
1
2
1
du
dx =
1−u2
1 + cos x
1
+
u2
1 + 1+u2
Z
Z
Z
2
2
1 + u2
1 + u2
du =
du = du
=
(1 + u2 ) + (1 − u2 ) 1 + u2
2
1 + u2
The right answer is (b)
Question 6. If
Z
3
2
(sec x) tan x dx =
Z
(a) u = tan x
3
(b) u = (sec x) 2
7
√
u du then
1
dx
1 + cos x
(c) u =
√
sec x
(d) u = sec x
Z
Z
1
3
Answer :
(sec x) 2 tan x dx = (sec x) 2 sec x tan x dx
Put u = sec x ⇒ du = Zsec x tan x dx
Z
Z
√
1
3
(sec x) 2 tan x dx = (sec x) 2 sec x tan x dx =
u du
The right answer is (d)
Question 7. The improper integral
Z
∞
0
1
dx
x2 + 4
(a) converges to 0
(b) diverges
(c) converges to
π
4
(d) converges to π2
Z ∞
Z t
Z t
1
1
1
Answer :
dx
=
lim
dx
=
lim
dx
2
2
2
t→∞ 0 x + 4
t→∞ 0 (x) + (2)2
x +4
0
x t
1
1
1
t
1π 1
π
= lim
− tan−1 (0) =
tan−1
= lim
tan−1
− (0) =
t→∞ 2
2 0 t→∞ 2
2
2
22 2
4
The right answer is (c)
Question 8. The value of the integral
Z
√
1
dx is equal to
4x − x2
x−2
+c
(a) sinh
2
x−2
(b) sin−1
+c
2
1
x−2
(c) sin−1
+c
2
2
x+2
(d) sin−1
+c
2
Z
Z
1
1
√
p
Answer :
dx
dx =
2 − 4x + 4) + 4
4x − x2
−(x
Z
x−2
1
p
+c
dx = sin−1
2
2
2
(2) − (x − 2)
The right answer is (b)
−1
Question 9. The area of the region bounded by the graphs of equations y = x2
and y = −x is equal to
(a)
5
6
8
(b) 2
(c)
1
6
(d)
1
3
Answer :
1
y = x2
1
-1
y = -x
-1
Points of intersection between y = x2 and y = −x
x2 = −x ⇒ x2 + x = 0 ⇒ x(x + 1) = 0 ⇒ x = −1 , x = 0
0
2
Z 0
x3
x
−x − x2 dx = − −
Area =
2
3 −1
−1 1 1
1 1
1
Area = 0 − − +
= − =
2 3
2 3
6
The right answer is (c).
Question 10. The value of the integral
(a)
Z
tan3 x sec x dx is equal to
1
sec3 x + sec x + c
3
1
(b) − sec3 x − sec x + c
3
1
(c) − sec3 x + sec x + c
3
1
sec3 x − sec x + c
3
Z
Z
Answer :
tan3 x sec x dx = tan2 x sec x tan x dx
Z
= (sec2 x − 1) sec x tan x dx
(d)
Put
Z u = sec x then du
Z = sec x tan x dx 3
1
u
3
− u + c = sec3 x − sec x + c
tan x sec x dx = (u2 − 1) du =
3
3
The right answer is (d)
9
Full questions
Question 11. Evaluate
Z
2x tan−1 x dx . [2 marks]
Answer : Using integration by parts
u = tan−1 x
dv = 2x dx
1
dx v = x2
du =
1 + x2
Z
Z
Z
(1 + x2 ) − 1
x2
2
−1
−1
2
−1
dx
=
x
tan
x−
dx
2x tan x dx = x tan x−
2
1 + x2
1+x
Z 1
1−
= x2 tan−1 x −
dx = x2 tan−1 x − x + tan−1 x + c
1 + x2
Question 12. Sketch the region R bounded by the graphs of y = x2 − 2 ;
y = −x2 and find its area . [2 marks]
Answer :
y = x2 - 2
1
-1
-1
y = -x 2
-2
Note that y = x2 − 2 is a parabola with vertex (0, −2) and opens upward ,
y = −x2 is a parabola with vertex (0, 0) and opens downward .
Points of intersection between y = x2 − 2 and y = −x2
x2 − 2 =Z−x2 ⇒ 2x2 = 2 ⇒ x2 = 1Z⇒ x = −1 , x = 1
1
1
[2 − 2x2 ] dx
[−x2 − (x2 − 2)] dx =
1
−1
−1
2
4
2
8
2x3
− −2 +
=4− =
=
2−
Area = 2x −
3 −1
3
3
3
3
Z
1
√
dx [3 marks]
Question 13. Evaluate
2
x 4 − x2
Answer : Using trigonometric substitution
x
Put x = 2 sin θ ⇒ sin θ =
2
π
π
dx = 2 cos θ dθ , where
Z
Z −2 < θ < 2
Z
2 cos θ dθ
2 cos θ dθ
1
p
√
p
=
dx =
2
2
2
2
2
(2 sinZθ) 4 − (2 sin θ)
4 sin θ 4 − 4 sin2 θ
Z x 4−x
2 cos θ dθ
2 cos θ dθ
q
=
2
4
sin
θ (2 cos θ)
2
2
4 sin θ 4(1 − sin θ)
Z
Z
1
1
1
1
dθ =
csc2 θ dθ = − cot θ + c
=
2
4
4
4
sin θ
Area =
10
2
x
Θ
4 - x2
√
1 4 − x2
1
1
+c
dx = − cot θ + c = −
4
4
x
x2 4 − x2
Z
x−1
dx [3 marks]
Question 14. Evaluate
x2 + x
Answer : Using Partial fractions
x−1
x−1
A
B
=
= +
x2 + x
x(x + 1)
x
x+1
x − 1 = A(x + 1) + Bx
Put x = 0 then 0 − 1 = A(0 + 1) + B(0) ⇒ A = −1
Put x = −1 then −1 − 1 = A(−1 + 1) + B(−1) ⇒ B = 2
x−1
1
2
=− +
2
x Zx +
1
Zx + x
1
2
x−1
dx
− +
dx =
x2Z+ x
x x+1
Z
1
1
dx + 2
dx = − ln |x| + 2 ln |x + 1| + c
=−
x
x+1
Z
√
11
M 106 - INTEGRAL CALCULUS
Dr. Tariq A. AlFadhel 3
Solution of the Final Exam
First semester 1432-1433 H
Multiple choice questions (one mark for each question)
2
1. The sum
n
X
(k − 1) is equal to
k=1
(a)
n2 (n − 1)
2
(b)
n(n − 1)
2
The answer :
(k − 1) =
k=1
=
k=1
k −
n
X
k=1
1=
(d)
n2 (n2 − 1)
2
n2 (n2 + 1)
− n2
2
n (n + 1 − 2)
n2 (n2 − 1)
n (n + 1) − 2n
=
=
2
2
2
2
2
2
2
n
X
n2 (n2 + 1)
2
2
2
2
n
X
(c)
2
The right answer is (d) .
1
2. The average value of the function f (x) = (x + 1) 3 on [−2, 0] is equal to
(a) 3
(b) 0
(c) −1 (d) −3
Z 0
1
0
(x + 1) 3 dx
4
1 3
−2
3
=
(x + 1)
The answer : fav =
0 − (−2)
2 4
−2
4
4
1 3
1 3
3
3
fav =
(0 + 1) 3 − (−2 + 1) 3 =
(1) − (1) = 0
2 4
4
2 4
4
The right answer is (b) .
3. The integral
Z
4x dx is equal to
22x
4x
4x+1
2x
+ c (b)
+ c (c)
+ c (d)
+c
2 ln 2
ln 2
ln 4
ln 4
x
Z
22
4x
22x
The answer :
4x dx =
+c=
+c=
+c
2
ln 4
ln 2
2 ln 2
(a)
The right answer is (a) .
4. If F (x) =
3 E-mail
Z
x2
1
p
3
t4 + 1 dt , then F ′ (x) is equal to
: [email protected]
12
(a)
p
3
(b) x2
x8 + 1
The answer : F ′ (x) =
F ′ (x) = 2x
p
3
p
3
x8 + 1
d
dx
Z
x2
1
(c) 2x
p
3
p
3
x8 + 1
t4 + 1 dt =
q
3
(d) 2x
4
p
3
x4 + 1
(x2 ) + 1 (2x)
x8 + 1
The right answer is (c)
Z
5. The integral
2sin x cos x dx is equal to
(a) 2sin x + c
The answer :
(b)
Z
2sin x
+c
ln 2
(c) (ln 2)2sin x + c
(d)
−2sin x
+c
ln 2
2sin x
+c
ln 2
2sin x cos x dx =
The right answer is (b)
6. If f (x) = xln x then f ′ (e) is equal to
(a) 2
(b) 2e
(c) 0
(d) e
The answer : f (x) = xln x ⇒ ln |f (x)| = ln xln x = (ln |x|)2
Diffrentiating both sides :
f ′ (x)
1
2 ln |x|
= 2 ln |x| =
f (x)
x
x
f ′ (x) = f (x)
f ′ (e) =
2 ln |x| xln |x|
2 ln |x|
=
x
x
2 ln(e) eln e
2e
=
=2
e
e
The right answer is (a)
7. lim
x→0
sin x − x
x3
(a) ∞
(b) − 61
is equal to
(c)
1
6
(d) 0
sin x − x
0
The answer : lim
3
x→0
x
0
Apply L’Hôpital’s rule
cos x − 1
sin x − x
= lim
lim
3
x→0
x→0
x
3x2
Apply L’Hôpital’s rule
13
0
0
lim
x→0
cos x − 1
− sin x
1 sin x
1
1
= lim
= lim −
= − (1) = −
x→0
x→0
3x2
6x
6 x
6
6
The rightanswer is (b) .
8. The integral
Z
1
√
(x + 1) x2 + 2x
dx is equal to
(a) ln |x2 + 2x| + c
(b) sin−1 (x + 1) + c
−1
(c) sech (x + 1) + c (d) sec−1 (x + 1) + c
Z
Z
dx
dx
p
√
=
The answer :
2
2
(x + 1) x + 2x
(x + 1) (x + 2x + 1) − 1
Z
dx
p
=
= sec−1 (x + 1) + c
(x + 1) (x + 1)2 − (1)2
The right answer is (d) .
9. The improper integral
Z
√
2x
dx
16 − x4
(a) converges to π2 (b) converges to π4
(c) converges to π (d) diverges
Z 2
Z
2x
2x
√
√
dx = lim
dx
The answer :
t→2 0
16 − x4
16 − x4
2 t
Z 2
x
2x
p
dx = lim sin−1
= lim
t→2
t→2 0
4
(4)2 − (x2 )2
0
2
π
π
t
− sin−1 (0) = sin−1 (1) − sin−1 (0) = − 0 =
= lim sin−1
t→2
4
2
2
The right answer is (a)
10. To evaluate the integral
Z
1
dx , we use the substitution
x x6 − 1
√
(a) u = x6
(b) u = x2 (c) u = x3 (d) u = x6 − 1
Z
Z
1
1
p
√
The answer :
dx
dx =
6
x x −1
x (x3 )2 − 1
So , we use the substitution u = x3
The right answer is (c)
2
1
Note : u = x3 ⇒ x = u 3 ⇒ dx = 13 u− 3 du
Z
1
1
√
dx =
3
x x6 − 1
Z
2
u− 3
1
du =
1√
3
u 3 u2 − 1
14
Z
du
1
√
= sec−1 u + c
3
u u2 − 1
11. The partial fraction decomposition of
A
B
(a) 2 + + 2
x
c −1
B
C
A
+
+
(c)
x
x−1 x+1
The answer :
2x3
takes the form
x(x2 − 1)
A Bx + C
+ 2
x
x −1
A
B
(d) 2 +
+
x−1 x+1
(b) 2 +
2x3
2x2
(2x2 − 2) + 2
2
=
=
=2+
x(x2 − 1)
x2 − 1
x2 − 1
(x − 1)(x + 1)
A
B
2x3
=2+
+
2
x(x − 1)
x−1 x+1
The right answer is (d)
12. If
Z
1
x2
dx =
1
6 x3 − 1
(a) x = u2
u8
du then
u2 − 1
Z
(b) x = u3
(c) x = u6
(d) x = u8
Z
1
x2
dx involves
The answer : Since the integrand in the integral
1
6 x3 − 1
1
1
x 2 and x 3 then we use the substitution x = u6 (note that l.c.m(2, 3) = 6)
.
The right answer is (c) .
13. The area of the region bounded by the graphs of y = 2x , y = x , 0 ≤ x ≤ 1
is equal to
(a)
1
2
(b) 2
(c)
1
4
1
3
(d)
The answer :
y =2x
1
y =x
1
15
Note that y = 2x and y = x are two straight lines passing through the
origin .
2 1
Z 1
x
1
1
(2x − x) dx =
Area =
= −0=
2
2
2
0
0
The right answer is (a)
14. The arc length of the graph y = 4x frm A(0, 0) to B(1, 4) is equal to
√
√
√
√
(a) 17 (b) 5 (c) 4 17 (d) 4 5
The answer : f (x) = 4x ⇒ f ′ (x) = 4
Z 1p
Z 1√
√
√
√
L=
1 + (4)2 dx =
17 dx = 17[x]10 = 17(1 − 0) = 17
0
0
The right answer is (a)
Note : y = 4x is a straight line , so the arc length here
p is equal to the distance
between
the
two
points
(0,
0)
and
(1,
4)
which
is
(1 − 0)2 + (4 − 0)2 =
√
17
15. The slope of the tangent line at the point corresponding to t = π4 on the
parametric curve given by the equations x = sin t , y = cos t , 0 ≤ t ≤ 2π
is
(a) −1
(b) 1
(c) 0
π
4
dy
t
− sin t
= − tan t
cos t
equals − tan π4 = −1
dy
=
The answer : m =
dx
The slope at t =
(d) 13
dx
dt
=
The right answer is (a)
16. If a graph has poalar equation r = 2 csc θ , then its equation in xy-system
is
(a) x = 2
(b) y = 2
(c) x =
The answer : r = 2 csc θ ⇒ r =
1
2
(d) y = 12
2
⇒ r sin θ = 2 ⇒ y = 2
sin θ
The right answer is (b)
17. The length of the curve C : x = cos 2t , y = sin 2t ; 0 ≤ t ≤
(a) π
(b)
The answer :
π
2
(c) 2π
(d)
π
4
dy
dx
= −2 sin 2t ,
= 2 cos 2t
dt
dt
16
π
2
is equal to
L=
L=
Z
Z
π
2
0
π
2
0
p
(−2 sin 2t)2
+
(2 cos 2t)2
dt =
Z
π
2
0
π
√
π
4 dt = 2[t]02 = 2
−0 =π
2
p
4 sin2 2t + 4 cos2 2t dt
The right answer is (a)
Note : : 0 ≤ t ≤ π2 ⇒ 0 ≤ 2t ≤ π, So the arc length of the given
parametric curve is half of the circumference of the unit circle whic equals
2π(1)
=π
2
18. To evaluate the integral
Z
tan5 x sec5 x dx we use the substitution :
(a) u = tan2 x (b) u = tan x (c) u = sec x (d) u = sin x
Z
Z
5
5
The answer :
tan x sec x dx = tan4 x sec4 x sec x tan x dx
=
Z
tan2 x
2
sec4 x sec x tan x dx =
Z
2
sec2 x − 1 sec4 x sec x tan x dx
So we use the substitutio u = sec x ⇒ du = sec x tan x dx
The right answer is (c)
19. If a point has (r, θ)-coordinates (r, θ) = (1, π6 ) then its (x, y)-coordinates
is
(a) (
√
3 1
2 , 2)
(b) ( 12 ,
√
3
2 )
(c) (
√
√
2
2
2 , 2 )
The answer : (r, θ) = (1, π6 ) ⇒ r = 1 , θ =
x = r cos θ = (1) cos π6 =
y = r sin θ = (1) sin π6 =
(d) (1, 0)
π
6
√
3
2
1
2
So its (x, y)-coordinates is (
√
3 1
2 , 2)
The right answer is (a)
20. The slope of the tangent line to the curve r = cos θ at θ =
(a)
π
2
(b) 0
(c)
π
4
π
4
is
(d) 1
The answer : x = r(θ) cos θ = cos2 θ ⇒
y = r(θ) sin θ = cos θ sin θ ⇒
dy
dθ
cos 2θ
dy
m=
= dx =
dx
−
sin 2θ
dθ
dy
dθ
dx
dθ
= −2 sin θ cos θ = −2 sin 2θ
= cos2 θ − sin2 θ = cos 2θ
17
The slope at θ =
π
4
equals
cos π2
0
= =0
− sin π2
1
The right answer is (b)
Full Questions :
21. Evaluate the integral
The answer :
=
Z
Z
1 − sin2 x
Z
2
sin2 x cos5 x dx
5
sin x cos x dx =
2
Z
[4 marks]
cos4 x sin2 x cos x dx
sin2 x cos x dx
Put u = sin x then du = cos x dx
Z
Z
Z
2
1 − sin2 x sin2 x cos x dx = (1 − u2 )2 u2 du = (1 − 2u2 + u4 )u2 du
2u5
u7
u3
−
+
+c
3
5
7
=
Z
=
2
1
1
sin3 x − sin5 x + sin7 x + c
3
5
7
(u2 − 2u4 − u6 ) du =
22. Find the surface area generated by revolving y =
the x-axis .
[4 marks]
s
2
Z 4
√
1
√
x 1+
The answer : SA = 2π
dx
2 x
1
r
Z 4
√
1
x 1+
dx
= 2π
4x
1
r
Z 4
√
4x + 1
dx
= 2π
x
4x
1
Z 4√
4x + 1
= 2π
dx
2
1
Z 4
Z
√
1
π 4
=π
4x + 1 dx =
(4x + 1) 2 4dx
4
1
1
4
h
i
3
3
3
π
π 2
(17) 2 − (5) 2
(4x + 1) 2 =
=
4 3
6
1
√
x , 1 ≤ x ≤ 4 about
x3
dx [6 marks]
x2 (x2 + 1)
Z
Z
x
x3
dx =
dx
The answer :
2
2
2
x (x + 1)
x +1
23. Evaluate the integral
Z
18
=
1
2
Z
1
2x
dx = ln(x2 + 1) + c
x2 + 1
2
24. Evaluate the integral
Z
ln x
√ dx
x
Using integration by parts
u = ln x
du =
1
dx
x
1
1
dv = √ = x− 2
x
√
v=2 x
Z
√ 1
√
ln x
√ dx = 2 x ln x − 2 x dx
x
x
Z
Z
√
√
1
1
= 2 x ln x − 2 √ dx = 2 x ln x − 2 x− 2 dx
x
√
√
= 2 x ln x − 4 x + c
Z
25. Sketch the region R bouned by the graph of the equations y = ex , y = e
and y-axis. Find the volume of the solid generated by revolving the region
R around the x-axis. (using Washer method)
[6 marks]
The answer :
3
y =e
e
2
y = ex
1
1
Points of intersection between y = ex and y = e
ex = e ⇒ x = 1.
Using Washer method
Z
Z 1h
i
2
x 2
(e) − (e ) dx = π
V =π
0
1
0
19
e2 − e2x
dx
1
1
1
1
V = π e2 x − e2x = π e2 − e2 − 0 − e0
2
2
2
0
1
π
1
V = π e2 +
= (e2 + 1)
2
2
2
26. Sketch the region R that lies inside both of the graphs of the equations
r = 3 + 3 cos θ and r = 3 − 3 cos θ. Set up (do not evaluate) an integral
that can be used to find its area.
[5 marks]
The answer :
Π
2
2Π
Π
3
3
5Π
Π
6
6
0
Π
0.
1.
2.
3.
4.
5.
6.
7Π
11 Π
6
6
4Π
5Π
3
3
3Π
2
Angles of intersection between r = 3 + 3 cos θ and r = 3 − 3 cos θ
3 + 3 cos θ = 3 − 3 cos θ ⇒ cos θ = 0 ⇒ θ =
Area =
1
2
Z
π
2
0
(3 − 3 cos θ)2 dθ+
1
2
Z
3π
2
π
2
π
2
, θ=
3π
2
(3 + 3 cos θ)2 dθ+
1
2
Z
2π
3π
2
((3 − 3 cos θ)2 dθ
Since the region is symmetric with respect to the polar axis and to the
line θ = π2 then
!
Z π
1 2
(3 − 3 cos θ)2 dθ
Area = 4
2 0
20
M 106 - INTEGRAL CALCULUS
Dr. Tariq A. AlFadhel 4
Solution of the first mid-term exam
Second semester 1432-1433 H
Multiple choice questions (One mark for each question)
25
X
(k 2 + αk) = 0 , then the value of α is equal to :
Question 1. If
k=1
(a) −16
(b) 16
(c) −17
(d) −1
Answer:
25
X
2
(k + αk) = 0 ⇒
k=1
⇒
25
X
k=1
k 2 = −α
25
X
k=1
25
X
k
k=1
25
X
2
2
!
+α
k⇒α=
k
k=0
k=1
k
− k=1
25
X
25
X
=−
(25)(25+1)(2(25)+1)
6
25(25+1)
2
=−
k=1
The right answer is (c)
Question 2. The value of the integral
Z
1
2|x|3 dx is equal to :
−1
(a) 2
(b) 1
(c) 0
(d) −1
Answer: Note that |x| = x if x ≥ 0 , and |x| = −x if x < 0 .
Z 1
Z 0
Z 1
3
3
2|x|3 dx
2|x| dx +
2|x| dx =
0
−1
−1
Z 1
Z 0
Z 1
Z 0
2x3 dx
2x3 dx +
2x3 dx = −
2(−x)3 dx +
=
0
−1
−1
4 0
40 1
x
x
= −2
+2
4 −1
4 0
1
1 1
1
+2
−0 = + =1
= −2 0 −
4
4
2 2
4 E-mail
: [email protected]
21
51
= −17
3
The right answer is (b)
Question 3. The value of the integral
Z
sin(tan(x))
dx is equal to :
cos2 (x)
(a) cos(tan(x)) + c
(b) sin(tan(x)) + c
(c) − cos(tan(x)) + c
(d) − sin(tan(x)) + c
Z
Z
sin(tan(x))
Answer :
dx
=
sin(tan(x)) sec2 (x) dx = − cos(tan(x)) + c
cos2 (x)
The right answer (c)
Question 4. The derivative of the integral
Z
x
0
1+
d tan(t)
dt
(a) 1 + tan x
(b) 1 − tan x
(c) 1 − sec2 x
(d) 1 + sec2 x
Z x
d tan x
d tan(t)
d
dt = 1 +
1+
= 1 + sec2 x
Answer :
dx 0
dt
dx
The right answer is (d)
Question 5. If G(x) =
Z
x2
e
ln(t)
dt then G′ (e) is equal to :
4
(a) 2e
(b) 1
(c) e
(d) 4e
Z x2
d
ln(x2 )
ln(t)
Answer : G (x) =
dt =
(2x)
dx e
4
4
2 ln(x)
4x ln(x)
=
(2x) =
= x ln(x)
4
4
′
G (e) = e ln(e) = e(1) = e
The right answer is (c)
′
22
dt is equal to :
Question 6. If log2
x−1
x
= 1 , then x is equal to :
(a) 1
(b) 2
(c)
1
2
(d) −1
x−1
x−1
=1⇒
= 2 ⇒ x − 1 = 2x ⇒ x = −1
x
x
The right answer is (d)
Answer : log2
Question 7. The value of the integral
(a)
4 ln 5
5
(b)
ln 5
4
(c)
4
ln 5
(d)
5 ln 5
4
Answer :
Z
1
5x dx =
0
The right answer is (c)
5x
ln 5
1
0
=
(b)
1
5x dx is equal to :
0
5
1
4
−
=
ln 5 ln 5
ln 5
Question 8. The value of the integral
(a)
Z
Z
1 2p 2
x x +1+c
2
p
x x2 + 1 dx is equal to :
3
2 2
(x + 1) 2 + c
3
3
2
(c) − (x2 + 1) 2 + c
3
3
1 2
(x + 1) 2 + c
3
Z p
Z
3
1
1
1 (x2 + 1) 2
2
2
2
Answer :
x x + 1 dx =
(x + 1) (2x) dx =
+c
3
2
2
2
3
1
= (x2 + 1) 2 + c
3
The right answer is (d)
(d)
23
Question 9. The value of the integral
(a)
Z
1
0
ex
dx is equal to :
(ex + 1)2
e−1
2(1 + e)
(b) −1
(c) 0
1
(1 + e)2
Z
Z 1
ex
dx = (ex + 1)−2 ex dx
Answer :
x
2
0 (e + 1)
x
1 1
(e + 1)−1
−1
=
= x
−1
e +1 0
0
−1
−1
1
−2 + e + 1
e−1
−1
−
=
+ =
=
=
e+1 1+1
e+1 2
2(e + 1)
2(e + 1)
The right answer is (a).
(d)
Question 10. The value of the integral
cos−1
(a) −
25
x
16
cos−1
(b)
25
x
16
+c
sin−1
(c)
5
5x
4
+c
Z
1
√
dx is equal to :
16 − 25x2
+c
sin−1 5x
4
+c
(d) −
5
Z
Z
1
1
1
5
5x
−1
√
p
dx = sin
dx =
Answer :
+c
2
2
2
5
5
4
16 − 25x
(4) − (5x)
The right answer is (c)
24
Full questions
Question 11. Use Trapizoidal rule to approximate the integral
Z
3
1
p
x2 + 3 dx
with n = 4. [3 marks]
p
Answer : [a, b] = [1, 3] , n = 4 , and (x) = 3 + x2 .
b−a
3−1
2
1
∆x =
=
= = = 0.5
n
4
4
2
x0 = 1 , x1 = 1.5 , x2 = 2 , x3 = 2.5 , x4 = 3 .
Z 3p
3−1
3 + x2 dx ≈
[f (1) + 2f (1.5) + 2f (2) + 2f (2.5) + f (3)]
2(4)
Z1 3 p
1
3 + x2 dx ≈ [2 + 2(2.29129) + 2(2.64575) + 2(3.04138) + 3.4641]
4
Z1 3 p
21.4209
≈ 5.35523
3 + x2 dx ≈
4
1
Question 12. If f (x) = xcosh x , then find f ′ (x). [2 marks]
cosh x
Answer : f (x) = xcosh x ⇒ ln |f (x)| = ln |x|
= cosh x ln |x|
Differentiate both sides
f ′ (x)
1
= sinh x ln |x| + cosh x
f (x)
x cosh x
′
f (x) = f (x) sinh x ln |x| +
x
cosh x
cosh x
′
sinh x ln |x| +
f (x) = x
x
Question 13. Find the number z that satisfies the conclusion
h π i of the Mean
Value Theorem for the function f (x) = cos(2x) where x ∈ 0, . And also find
2
the averahe value fav of f (x) . [3 marks]
Answer :
First : Calculating fav
Z π2
Z π
cos 2x dx
π
1
2 1 2
0
cos(2x) 2 dx = [sin(2x)]02
fav =
=
π
π
2
π
−
0
0
2
1
1
= [sin(π) − sin(0)] = (0) = 0 .
π
π
Second : Calculating the number z that satisfies
conclusion of the MVT
h the
πi
such that
According to MVT there exists a number z ∈ 0,
2
Z π2
cos 2x dx
cos(2z) = 0 π
2 −0
π
π
cos(2z) = 0 ⇒ 2z = ⇒ z =
2
4
25
Question 14. Evaluate the integral J =
Z
cos(x)
q
dx [2 marks]
sin(x) 4 − sin2 (x)
Answer
:
Z
Z
cos(x)
cos(x)
q
p
J=
dx =
dx
2
sin(x) (2)2 − (sin(x))2
sin(x) 4 − sin (x)
1
sin x
−1
+c
= − sech
2
2
26
M 106 - INTEGRAL CALCULUS
Dr. Tariq A. AlFadhel 5
Solution of the second mid-term exam
Second semester 1432-1433 H
Multiple choice questions (One mark for each question)
2x − 3x
is equal to
Question 1. lim
x→0
x
(a) ∞
(b) ln 23
(c) ln 32
(d) −1
0
2x − 3x
Answer: lim
x→0
x
0
Apply L’Hôpital’s rule
2x − 3x
2x ln 2 − 3x ln 3
2
lim
= lim
= 20 ln 2 − 30 ln 3 = ln 2 − ln 3 = ln
x→0
x→0
x
1
3
The right answer is (b)
Question 2. The partial fraction decomposition of
(a)
x2 + 2
(x4 − 1)(x − 1)
Cx + D
E
Ax + B
+ 2
+
x2 + 1
x −1
x−1
(b)
C
D
E
Ax + B
+
+
+
2
2
x +1
x − 1 (x − 1)
x+1
(c)
Ax + B
Cx + D
E
+
+
2
2
x +1
(x − 1)
x+1
(d)
C
E
Ax + B
+
+
2
2
x +1
(x − 1)
x+1
x2 + 2
x2 + 2
x2 + 2
=
=
(x4 − 1)(x − 1)
(x2 + 1)(x2 − 1)(x − 1)
(x2 + 1)(x − 1)2 (x + 1)
Ax + B
C
D
E
x2 + 2
= 2
+
+
+
(x4 − 1)(x − 1)
x +1
x − 1 (x − 1)2
x+1
The right answer is (b)
Answer:
Question 3. The integral
(a) ln |x2 − 4x + 3| + c
x + 1
+c
(b) ln x + 3
5 E-mail
Z
2 dx
is equal to
x2 − 4x + 3
: [email protected]
27
x − 3
+c
(c) ln x − 1
x − 1
+c
(d) ln x − 3
Answer : Using partial fractions
2
A
B
2
=
=
+
x2 − 4x + 3
(x − 1)(x − 3)
x−1 x−3
2 = A(x − 3) + B(x − 1)
Put x = 1 then 2 = A(1 − 3) ⇒ 2 = −2A ⇒ A = −1
Put
⇒B=1
Z Z x = 3 then 2 = 2B
−1
1
2
dx
dx =
+
2
x−1 x−3
Z x − 4x + 3
x − 3
2
+c
dx = − ln |x − 1| + ln |x − 3| + c = ln x2 − 4x + 3
x − 1
The right answer (c)
Question 4. The value of the integral
(a)
1
6
sin6 x −
1
8
sin8 x + c
(b)
1
5
sin5 x −
1
3
sin3 x + c
(c)
1
3
sin5 x −
1
2
sin2 x + c
Z
sin5 x cos3 x dx is equal to
sin5 x − 18 sin8 x + c
Z
Z
5
3
Answer :
sin x cos x dx = sin5 x cos2 x cos x dx
Z
= sin5 x(1 − sin2 x) cos x dx
(d)
1
3
Put
Z u = sin x ⇒ du = cos x dx Z
sin5 x(1 − sin2 x) cos x dx = u5 (1 − u2 ) du
Z
u8
u6
−
+c
= (u5 − u7 ) du =
6
8
1
1
= sin6 x − sin8 x + c
6
8
The right answer is (a)
Question 5. The substitution u = tan
x
2
into
(a)
Z
u2
2
du
− 2u + 5
(b)
Z
u2
2
du
− 2u + 3
(c)
Z
2
du
u2 + 2u + 5
28
transforms the integral
Z
1
dx
3 − sin x + 2 cos x
(d)
Z
2
du
u2 + 2u + 3
Answer : Using the half angle substitution u = tan x2
1 − u2
2u
2
cos x =
, sin x =
and dx =
du
2
2
1
+
u
1
+
u
1
+
u2
Z
Z
2
1
1
dx =
du
2u
1−u2
3 − sin x + 2 cos x
1
+
u2
3 − 1+u2 + 2 1+u2
Z
Z
1 + u2
2
2
=
du
=
du
2 ) − 2u + 2(1 − u2 ) 1 + u2
2 − 2u + 2 − 2u2
3(1
+
u
3
+
3u
Z
2
=
du
u2 − 2u + 5
The right answer is (a)
Z
Question 6. To evaluate the integral
dx
√
, we use the substitution
x2 x2 − 25
(a) x = 5 sec θ
(b) x = sec5 θ
(c) x = 5 tan θ
(d) x = tan5 θ
p
p
Answer : x2 − 25 = (x)2 − (5)2
So, we use the substitution x = 5 sec θ
The right answer is (a)
Question 7. The improper integral
Z
∞
x2
0
1
dx
+1
(a) converges to π
(b) diverges
(c) converges to
π
2
(d) converges to +∞
Z t
Z ∞
1
1
dx
=
lim
dx
Answer :
2
2
t→∞
x +1
x +1
0
−1
t
−1 0
π
π
= lim tan (x) 0 = lim tan (t) − tan−1 (0) = − 0 =
t→∞
t→∞
2
2
The right answer is (c)
Question 8. The value of the integral
Z
(a) x − ln |x + 1| + c
(b) x − ln (ex + 1) + c
29
1
dx is equal to
1 + ex
x2
− ln (ex + 1) + c
2
2
x
− ln |x + 1| + c
(d) ln
2
Z
Z
1
(1 + ex ) − ex
Answer :
dx
=
dx
x
1 + ex Z 1+ e
Z x
x
x
e
e
1+e
1−
dx =
dx
−
=
1 + ex
1 + ex
1 + ex
x
= x − ln (e + 1) + c
The right answer is (b)
(c)
Question 9. The value of the integral
(a)
ln |x|
−x+c
x
Z
ln |x| dx is equal to
(b) x ln |x| − ln |x| + c
(c)
ln2 |x|
+c
2
(d) x ln |x| − x + c
Answer : Using integration by parts
u = ln |x|
dv = dx
1
du = dx v = x
x
Z
Z
1
ln |x| dx = x ln |x| − x dx
x
R
= x ln |x| − dx = x ln |x| − x + c
The right answer is (d).
Question 10. The area of the region bounded by the graphs of y = x , y = −x
and y = 1 is equal to
(a) 1
(b) 0
(c) 2
(d)
1
2
Answer :
y =1
1
y =x
y = -x
1
-1
30
y = x intersects
y = 1 at x =Z1 and y = −x intersects y = 1 at x = −1
Z 0
1
(1 − x) dx
(1 − (−x)) dx +
Area =
0
−1
0
1
x2
x2
+ x−
Area = x +
2
2
−1 0
1
1
1 1
Area = 0 − −1 +
+ 1−
−0 = + =1
2
2
2 2
The right answer is (a)
Full questions
Z
Question 11. Evaluate
sin5 5x dx . [3 marks]
Z
Z
Answer :
sin5 5x dx = sin4 5x sin 5x dx
Z
Z
2
2
2
1 − cos2 5x sin 5x dx
sin 5x
sin 5x dx =
=
1
Put
Z dx ⇒ − 5 du = sin 5x dx
Z u = cos 5x then du = −5 sin 5x
2
2
1
1 − u2 du
1 − cos2 5x sin 5x dx = −
5
Z
2u3
u5
1
1
2
4
+c
+
(1 − 2u + u ) du = − u −
=−
5
5
3
5
cos 5x 2 cos3 5x cos5 5x
=−
+
−
+c
5
15
25
Z
x+1
Question 12. Evaluate the integral
dx [2 marks]
x2 + x − 2
Answer : Using partial fractions
x+1
A
B
x+1
=
=
+
2
x +x−2
(x + 2)(x − 1)
x−1 x+2
x + 1 = A(x + 2) + B(x − 1)
Put x = 1 then 2 = 3A ⇒ A = 23
Put x = −2 then −1 = −3B ⇒ B = 31
2
1
x+1
= 3 + 3
2
x − 1 Zx + 2
Zx + x − 2
Z
x+1
1
1
2
1
dx
=
dx
+
dx
2
x +x−2
3
x−1
3
x+2
1
2
= ln |x − 1| + ln |x + 2| + c
3
3
Z
Question 13. Evaluate the integral
tan x sec3 x dx [3 marks]
Z
Z
Answer :
tan x sec3 x dx = sec2 x sec x tan x dx
Z
sec3 x
+c
= (sec x)2 sec x tan x dx =
3
Question 14. Sketch the region bounded by the graphs of y = − π2 x + 1 ,
y = π2 x − 1 , x = 0 and x = π . Find its area. [2 marks]
Answer :
Points of intersection between y = − π2 x + 1 and y = π2 x − 1
31
2
2
4
π
x−1=− x+1⇒ x=2⇒x=
π
π
π
2
1
y =
-2
x +1
Π
Π
Π
2
y =
-1
2
x -1
Π
x =Π
Z π
2
2
2
2
− x+1−
x−1
dx +
x−1− − x+1
dx
π
π
π
π
π
0
2
Z π2 Z π
4
4
Area =
− x + 2 dx +
x − 2 dx
π
π
π
0
2
π2 π
2
2 2
Area = − x2 + 2x +
x − 2x
π
π
π
2
0 i h
π
i π π
h π
−π = + =π
Area = − + π − 0 + (2π − 2π) −
2
2
2
2
Area =
Z
π
2
32
M 106 - INTEGRAL CALCULUS
Dr. Tariq A. AlFadhel 6
Solution of the Final Exam
Second semester 1432-1433 H
Multiple choice questions (one mark for each question)
√
1. The average value of 3 x + 1 0n [−2, 0] is equal to
(a) −
1
2
(b)
1
2
(c) 0
Z
The answer : fav =
fav =
3
4
0
(d) 1
√
3
x + 1 dx
−2
=
0 − (−2)
4
4
(0 + 1) 3 − 43 (−2 + 1) 3
=
2
3
4
h
3
4
4
(x + 1) 3
(1) −
2
2
3
4
(1)
i0
−2
=0
The right answer is (c) .
2. The integral
Z
sin 5x [cos 5x]−3 dx is equal to
[cos 5x]−2
[cos 5x]−2
+ c (b) −
+c
10 −2
10 −2
(c) [cos 5x] + c (d) −[cos 5x] + c
Z
Z
1
[cos 5x]−3 (− sin 5x) 5 dx
The answer :
sin 5x [cos 5x]−3 dx = −
5
(a)
=−
1 [cos 5x]−2
[cos 5x]−2
+c=
+c
5
−2
10
The right answer is (a) .
3. The integral
Z
ln
√
x dx is equal to
1
x(ln x − 1) + c (b) 2x(ln x − 1) + c
2
1
(c) x ln x − 1 + c
(d) x ln x − x + c
2
Z
Z
Z
√
1
1
ln x dx
The answer :
ln x dx = ln(x) 2 dx =
2
(a)
Using integration by parts
u = ln x
1
du = dx
x
6 E-mail
dv = dx
v=x
: [email protected]
33
Z
1
1
x ln x − x dx
ln x dx =
2
x
Z
1
1
1
x ln x − dx = (x ln x − x) + c = x (ln x − 1) + c
=
2
2
2
1
2
Z
The right answer is (a) .
4. lim
x→0
(a)
tan x − x
is equal to
x2
1
3
(b) 0
(c)
3
2
(d)
2
3
0
0
tan x − x
x→0
x2
The answer : lim
Apply L’Hôpital’s rule
tan x − x
sec2 x − 1
=
lim
x→0
x→0
x2
2x
0
0
lim
Apply L’Hôpital’s rule
sec2 x − 1
2 sec2 x tan x
0
= lim
= =0
x→0
x→0
2x
2
2
lim
The right answer is (b)
5. If exp(x2 ) = exp(4x − 4) then the value of x is equal to
(a) −2
(b) 2
(c) 1
(d) −1
The answer : exp(x2 ) = exp(4x − 4) ⇒ x2 = 4x − 4 ⇒ x2 − 4x + 4 = 0
⇒ (x − 2)2 = 0 ⇒ x = 2
The right answer is (b)
x2 + 1
is equal to
x→∞ ln(x + 1)
6. lim
(a) ln 2
(b) ∞
(c) 1
(d)
x2 + 1
x→∞ ln(x + 1)
The answer : lim
Apply L’Hôpital’s rule
1
ln 2
∞
∞
x2 + 1
2x
= lim 1 = lim 2x(x + 1) = ∞
x→∞ ln(x + 1)
x→∞
x→∞
x+1
lim
The right answer is (b)
34
7. To evaluate the integral
(a) u = cos x
Z
sin3 x dx , we use the substitution
(b) u = sin x
(c) u =
cos x
3
(d) u =
sin x
3
The answer : Since the power of the sine function is odd then we use
u = cos x
The right answer is (a) .
8. The partial fraction decomposition of
A
Bx + C
+ 2
+1
2x + 2
Ax + B
C
(c)
+
2x2 + 1
2x2 + 2
(a)
2x2
The answer :
2x2 + 2x + 10
takes the form
8x4 + 12x2 + 4
A
B
+ 2
+ 1 2x + 2
Ax + B
Cx + D
(d)
+
2x2 + 1
2x2 + 2
(b)
2x2
x2 + x + 5
x2 + x + 5
2x2 + 2x + 10
= 2
=
4
2
2
8x + 12x + 4
4x + 6x + 2
(2x + 1)(2x2 + 2)
Ax + B
Cx + D
x2 + x + 5
=
+
+ 1)(2x2 + 2)
2x2 + 1
2x2 + 2
(2x2
Note that 2x2 + 1 and 2x2 + 2 are irreducible quadratics .
The right answer is (d) .
9. If F (x) = 2tan x , then F ′ (x) is equal to
(a) (ln 2) (sec2 x) 2tan x
(c) (ln 2) (sec x tan x) 2tan x
(b) (sec2 x) 2tan x
(d) (ln 2) (sec2 x)
The answer : F ′ (x) = 2tan x sec2 x ln 2 = (ln 2) (sec2 x) 2tan x
The right answer is (a)
10.
1
dx is equal to
x2 − 6x + 13
x−3
1
+ c (b) log |x2 − 6x + 13| + c
(a) tan−1
2
2
1
x−3
(c) tan
+c
(d) log |x − 3| + c
3
2
Z
Z
1
1
The answer :
dx
=
dx
x2 − 6x + 13
(x2 − 6x + 9) + 4
Z
1
x−3
1
−1
+c
dx
=
tan
=
(x − 3)2 + (2)2
2
2
Z
The right answer is (a)
35
11. The area of the region bounded by the graphs of y = x2 and y = |x| is
equal to
(a)
1
3
(b) 1
(c)
2
3
(d)
4
3
The answer :
1
x ¤
x2
1
-1
Points of intersection between y = x2 and y = |x|
x2 = |x| ⇒ x2 = ±x ⇒ x2 ± x = 0 ⇒ x(x ± 1) = 0 ⇒ x = 0 , x = ±1
Since the region is symmetric with respect to the y-axis then
Z 1
Z 1
x − x2 dx
|x| − x2 dx =
Area = 2
0
0
Area = 2
3 1
2
x
x
−
2
3
=2
0
1 1
−
2 3
1
1
− (0 − 0) = 2 =
6
3
The right answer is (a)
12. The improper integral
Z
∞
−∞
1
dx
x2 + 9
(a) converges to 0
(b) converges to
(c) diverges
(d) converges tp
The answer :
= lim
t→−∞
Z
0
t
0
Z
∞
−∞
1
x2 + 9
1
dx =
2
x +9
Z
dx + lim
k→∞
Z
0
−∞
k
0
π
3π
2
1
dx +
2
x +9
Z
∞
0
x2
1
dx
+9
1
dx
x2 + 9
k
1
1
dx
+
lim
dx
2
2
2
t→−∞ t x + 3
k→∞ 0 x + 32
0
k
1
1
−1 x
−1 x
= lim
tan
+ lim
tan
t→−∞ 3
3 t k→∞ 3
3 0
1
1
1
1
t
k
−1
−1
−1
−1
= lim
+ lim
− tan (0)
tan (0) − tan
tan
t→−∞ 3
k→∞ 3
3
3
3
3
= lim
Z
Z
36
1 π
1 π π π
1 π
−
=
+
−0 =
+
= 0−
3
2
3 2
3 2
2
3
The right answer is (b) .
13. To evaluate the integral
(a) u =
√
x
√
3
x
Z
dx , we use the substitution
2
x + x3
1
1
2
(c) u = x 5
(d) u = x 5
√
Z
3
x
1
3
The answer : Since the integrand in the integral
2 dx involves x
x + x3
2
1
and x 3 then we use the substitution x = u3 ⇒ u = x 3
(b) u = x 3
(note that l.c.m(3, 3) = 3) .
The right answer is (b)
14. The substitution u = tan
(a)
2
Z
du
Z −u2 − 1
2
(c)
du
2
−u − 2u + 1
x
2
Z
transforms the integral
1
dx into
2 + cos x
1
du
2 − 2u + 1
−u
Z
2
(d)
du
2
u +3
(b)
Z
The answer : Using half-angle substitution
u = tan
x
2
⇒ cos x =
2
1 − u2
, dx =
du
1 + u2
1 + u2
1
Z
1
dx =
2 + cos x
=
Z
1 + u2
2
du =
2
2
2(1 + u ) + 1 − u 1 + u2
=
Z
u2
Z
2+
1−u2
1+u2
2
du
1 + u2
Z
2+
2u2
2
du
+ 1 − u2
2
du
+3
The right answer is (d)
15. The parametric curve C : x = t + 1 , y = 2t + 1 , 0 ≤ t ≤ 2 represents
(a) a line segment
(b) a parabola
(c) a circle
(d) an ellipse
The answer : x = t + 1 ⇒ t = x − 1
y = 2t + 1 ⇒ y = 2(x − 1) + 1 = 2x − 2 + 1 = 2x − 1
The parametric curve represents a line segment from (1, 1) to (3, 5)
The right answer is (a)
37
16. The volume of the solid obtained by revolving the region bounded by the
1
, y = 0 , x = 0 and x = 1 about the x-axis is equal to
graphs of y =
x+1
π
π
(b)
(c) π (d) 2π
(a)
4
2
The answer :
x =0
x =1
1
1
y =
x +1
1
Using the disk method
2
Z 1
Z
1
V =π
dx = π (x + 1)−2 dx
x+1
0
1
1
1
π
(x + 1)−1
−1
V =π
= π − − (−1) =
=π
−1
x+1 0
2
2
0
The right answer is (b)
17. If (x, y) = (0, −2) then its polar coordinates (r, θ) equals to
(c) 2, π2
(d) None of these
(a) (−2, π) (b) 2, 3π
2
38
The answer :
Hr, Θ L = H2 , 3 Π  2 L
r=
p
(0)2 + (−2)2 =
√
Hx , y L = H0 , - 2 L
4=2
3π
(see the above figure)
2
(r, θ) = 2, 3π
2
θ=
The right answer is (b)
18. The point on C : x = t , y = t2 ; − 1 ≤ t ≤ 1 at which the tangent line
is horizontal is
1
(c) u = − 21 , 41
(d) 41 , 16
(a) (0, 0) (b) 21 , 41
The answer : The slope of the tangent line is m =
dy
dx
= 2t and
=1
dt
dt
The tangent line is horizontal when
dy
dt
dx
dt
dy
dx
= 0 and
6= 0
dt
dt
dy
= 0 ⇒ 2t = 0 ⇒ t = 0
dt
Note that
dx
6= 0 at t = 0
dt
When t = 0 : x = 0 and y = 02 = 0
The point on C at which the tangent line is horizontal is (0, 0)
The right answer is (a)
39
19. The arc length of the polar curve r = 4 sin θ ; 0 ≤ θ ≤
(a)
π
2
(b) π
The answer : L =
L=
Z
L=
Z
π
2
0
π
2
0
p
(c) 2π
Z
π
2
0
s
π
2
equals to
(d) 4π
[r(θ)]2 +
dr
dθ
16 sin2 θ + 16 cos2 θ dθ =
π
4 dθ = [4θ]02 = 4
The right answer is (c)
π
2
2
Z
π
2
0
dθ =
q
Z
π
2
0
q
2
[4 sin θ]2 + [4 cos θ] dθ
16 sin2 θ + cos2 θ dθ
− 0 = 2π
Note: The polar curve r = 4 sin θ ; 0 ≤ θ ≤ π2 represents half a circle of
center (0, 2) and radius equals 2 , therefore the arc length is half of the
circuference of the circle which equals to 12 2π(2) = 2π
20. The polar curve r =
(a) a straight line
The answer : r =
3
represents
sin θ
(b) a circle
(c) a cardioid
(d) a rose curve
3
⇒ r sin θ = 3 ⇒ y = 3
sin θ
3
represents a straight line parallel to the polar
The polar curve r =
sin θ
axis .
The right answer is (a)
Full Questions :
21. Approximate the integral
Z
regular partition with n=4
The answer : f (x) =
2
0
1
dx bu using Simpson’s rule and
+1
[4 marks]
3x3
2−0
1
, [a, b] = [0, 2] and ∆x =
= 0.5
3x3 + 1
4
x0 = 0 , x1 = 0.5 , x2 = 1 , x3 = 1.5 and x4 = 2 .
Z 2
1
2−0
dx ≈
[f (0) + 4f (0.5) + 2f (1) + 4f (1.5) + f (2)]
3
3 (4)
0 3x + 1
Z 2
1
1
dx ≈ [1 + 4(0.727273) + 2(0.25) + 4(0.0898876) + 0.04]
3+1
3x
6
0
Z 2
1
1
dx ≈ [1 + 2.90909 + 0.5 + 0.359551 + 0.04]
3
6
0 3x + 1
40
Z
2
0
1
1
dx ≈ [4.80864] ≈ 0.80144
+1
6
3x3
22. Evaluate
Z
x2
1
√
dx .
x2 + 16
[6 marks]
The answer : Using trigonometric substitutions
Put x = 4 tan θ ⇒ tan θ =
x
4
dx = 4 sec2 θ dθ
Z
Z
1
4 sec2 θ dθ
p
√
dx =
x2 x2 + 16
(4 tan θ)2 (4 tan θ)2 + 16
Z
Z
4 sec2 θ dθ
4 sec2 θ dθ
p
=
=
16 tan2 θ (4 sec θ)
16 tan2 θ 16(tan2 θ + 1)
Z
Z
Z
1
1
1
sec θ dθ
cos2 θ 1
dθ
=
=
sin θ)−2 cos θ dθ
=
16
16
16
tan2 θ
sin2 θ cos θ
=
1
1 (sin θ)−1
+ c = − csc θ + c
16
−1
16
x 2 + 16
4
Θ
Z
1
1
√
dx = −
2
2
16
x x + 16
x
√
x2 + 16
+c
x
1
√
dx
4
x x − 16
Z
Z
1
1
√
dx =
The answer :
2
x x4 − 16
23. Evaluate the integral
Z
[5 marks]
2x
q
2
x2 (x2 ) − (4)2
2
2
1 1
1
x
x
=
+c=
+c
sec−1
sec−1
2 4
4
8
4
41
dx
24. Let R be the region bounded by the graph of the equations y = e2x−1 ,
y = 0 , x = 0 and x = 1 .
Sketch the region R and evaluate the volume of the solid obtained by
revolving the region R about the x-axis . [5 marks]
The answer :
3
2
y = e 2 x -1
1
1
x =1
Note that y = e2x−1 is a positive increasing exponential function .
Using the disk method
Z
Z 1
2x−1 2
dx = π
e
V =π
0
V =
1
e
4x−2
0
π 4x−2 1
π 2
e
e − e−2
=
0
4
4
42
π
dx =
4
Z
1
e4x−2 4 dx
0
25. Sketch the region inside the polar curve r = 2 + 2 cos θ and outside the
polar curve r = 1 , and find its area
[5 marks]
The answer :
Π
2
2Π
Π
3
3
5Π
Π
6
6
0
Π
0.
1.
2.
3.
4.
7Π
11 Π
6
6
4Π
5Π
3
3
3Π
2
Angle of intersection between r = 2 + 2 cos θ and r = 1
2 + 2 cos θ = 1 ⇒ cos θ = − 12 ⇒ θ =
2π
3
, θ=
4π
3
Since the region is symmetric with respect to the polar axis then
!
Z 2π
Z 2π
3
3
1
1
(2 + 2 cos θ)2 dθ −
(1)2 dθ
Area = 2
2 0
2 0
Area =
Z
Area =
Z
Area =
Z
2π
3
0
[(4 + 8 cos θ + 4 cos2 θ) − 1] dθ
2π
3
[3 + 8 cos θ + 2(1 + cos 2θ)] dθ
0
2π
3
[5 + 8 cos θ + 2 cos 2θ] dθ
0
2π
3
Area = [5θ + 8 sin θ + sin 2θ]0
26. Evaluate
Z
√
10π 7 3
+
=
3
2
π
4
sin x ln cos x dx
[5 marks]
0
The answer : Using integration by parts
u = ln cos x
− sin x
du =
dx
cos x
dv = sin x dx
v = − cos x
43
Z
π
4
0
π
4
sin x ln cos x dx = [− cos x ln cos x]0 −
π
4
= [− cos x ln cos x]0 −
π
Z
π
4
sin x dx
0
π
= [− cos x ln cos x]04 − [cos x]04
#
# "√
"
√ !
√
2
2
2
ln
−1
−0 +
=
−
2
2
2
44
Z
π
4
0
sin x
−
cos x
(− cos x) dx