Kuwait University
Dept. of Mathematics
Math 102
Second Exam
April 19, 2014
Duration: 90 min
Calculators are not allowed. All communication devices must be switched off.
Answer all questions. Total points: 100.
1. (10 pt) Evaluate:
2x + 4 x
lim
x→∞ 5x − 2x
2. (10 pt) Find the value of a for which lim
x→0
and evaluate the limit for this value of a.
tan x + ax
exists
x2
3. (15 pt each) Evaluate the following:
Z
(a)
(arcsin x)2 dx
(b)
Zπ/2
cos x sin2(2x) dx
0
Z
5
dx
x4 + 2x3 + 5x2
Z
tan (x/2)
dx
sin x − cos x − 1
(c)
(d)
4. (20 pt) Determine whether the following integral is convergent
or divergent. Find its value if it is convergent.
Z∞
x+1
dx
x5/2 − x1/2
2
Solution
—————————————————————
x
∞
∞
since 5x − 2x = 5x 1 − 52 → ∞(1 − 0) → ∞. L’Hospital rule can be applied but it
is not useful. Instead we have
2x + 4x
0
(2/5)x + (4/5)x
−→
=
= 0.
=
x
x→∞
5x − 2x
1
1 − 25
2
∗
2. 00 , so L = lim tan xx+ax
= lim sec 2xx+a = 1+a
.
2
0
1.
x→0
x→0
∗
(a) If a = −1, then L = lim 2 sec
x→0
2
x tan x
2
= 0. (exists)
(b) If a 6= −1, then L = ±∞ (Does not exist)
3. (a)
Z
−1
(sin
2
Z
x
dx
(sin−1 x) √
1 − x2
√
= x(sin−1 x)2 + 2(sin−1 x) 1 − x2 − 2x + C
−1
x) dx = x(sin
2
x) − 2
(b)
Zπ/2
I = 4 sin2 x cos3 xdx
0
Z1
=
u=sin x
4
u2 (1 − u2 )du =
8
15
0
(c)
A B
Cx + D
5
= + 2+ 2
+ 2x + 5)
x x
x + 2x + 5
2
2
This implies 5 = (Ax+B)(x +2x+5)+(Cx+D)x . Now we obtain, A = −2/5, B =
1, C = 2/5, and D = −1/5.
Z
Z
Z
Z
−2
1
2x − 1
5
=
+
+
4
3
2
2
2
x + 2x + 5x
5x
x
5x + 10x + 25
−2 ln |x| 1
4 3
2
x+1
−1
=
− −
ln √
−
tan
+C
5
x 10 x2 + 2x + 5 10
2
x2 (x2
(d) If we let t = tan( x2 ), then
Z
Z
2t dt
t dt
x
x
I=
=
= t+ln |t−1|+C = tan( )+ln | tan( )−1|+C
2
2
2t − (1 − t ) − (1 + t )
t−1
2
2
4. If we let u = x1/2 , then
Z 2
Z
Z Z
u − 1
u +1
2
x+1
1
1
+C
dx = 2
du =
du =
−
= ln x5/2 − x1/2
u4 − 1
u2 − 1
u−1 u+1
u + 1
√
√
√
Zt
2 − 1
2 − 1
t − 1
− ln √
I = lim
= lim ln √
= − ln √
.
t→∞
t→∞
t+1
2+1
2 + 1
2
√ Therefore, it is convergent and it converges to − ln √2−1
.
2+1