Sample Midterm Test 205
1. (11)
(a) Write in sigma notation the right equidistant Riemann sum Rn for
Z2
2
f (x) = 4 x in [0; 2] and also evaluate f (x) dx as the lim Rn .
n!1
0
0 x
1
Z
d @
(b) Use the fundamental theorem of Integral; Calculus to evaluate
ln t2 + 1 dtA.
dx
e
x
2. (15) Calculate :
Z
3x3
p
dx
(a)
16 x2
Z
x2 + 3
dx
(b)
x2 2x + 5
Z
2x
(c)
dx
22x 4
3. (6) Evaluate the antiderivative F (x) of f (x) =
sec2 x
dx such that
sec2 x + 3
F (0) = 0.
4. (12) Evaluate exactly:
(a)
Z4
p
(b)
Z1
x2 cos ( x) dx
x2
dx
2x + 1
0
0
5. (6) Calculate the area enclosed by the graphs: y = x2 3x+5 an y = 3 x.
Bonus (3) Given that
Z
(f (x) + f 00 (x)) sin xdx = 2 and f ( ) = 1, determine
0
f (0).
1
Solutions
1. (11)
(a) Write in sigma notation the right equidistant Riemann sum Rn for
Z2
2
f (x) = 4 x in [0; 2] and also evaluate f (x) dx as the lim Rn .
n!1
n
2X
2
4
h = : Rn =
n
n i=1
lim
n!1
8
2
n
n
X
i=1
2i
n
2
4
4i
n2
= lim
n!1
0
2
!
0
0
!
2B
@4n
n
Z2
0
f (x) dx =
0
11
2
n/ (2n + 1) (n + 1)
B3
CC
@
AA
n2/
1
n+1
2n + 1
Z2
B
C 16
n
n
C=
4
4 lim B
and
A
n!1 @
3
3
x2 dx =
16
3
0
0 x
1
Z
d @
(b) Use the fundamental theorem of Integral; Calculus to evaluate
ln t2 + 1 dtA =
dx
x
e
ln x2 + 1 + ln e
2x
+1
e
x
.
2. (15) Calculate :
Z
p
3x3
p
(a)
16 x2 x2 + 32 + C either odd or by formula
dx =
2
16 x
!
Z
Z
0
R
x2 2x + 5 + 2x 2
x2 2x + 5
x2 + 3
(b)
dx =
dx =
1+
dx
x2 2x + 5
x2 2x + 5
x2 2x + 5
= x + ln x2 2x + 5 + C
Z
Z
1
dt
2x
x
x
(c)
dx
=
j2
=
t
!
2
ln
2dx
=
dtj
=
=
22x 4
ln 2 t2 4
Z
1
dt
1
=
(ln (t 2) ln (t + 2)) =
ln 2 (t 2) (t + 2)
4 ln 2
1
(ln (2x 2) ln (2x + 2)) + C
4 ln 2
1
A
B
as
=
+
! 1 = A (t + 2) + B (t 2) !
(t 2) (t + 2)
t 2 t+2
1
1
1
t = 2 : 1 = 4B & t = 2 : 1 = 4A !
=
(t 2) (t + 2)
4 t+2
2
1
t
2
3. (6) Evaluate the antiderivative F (x) of f (x) =
sec2 x
dx such that
sec2 x + 3
F (0) = 0.
p
R sec2 x
p
dx = t = sec2 x ! dt = 2 sec2 x 1 sec2 xdx = 2t 1 tdx =
2
sec x + 3
R
R 2udu
p
dt
p
= 1 t = u ! 1 t = u2 ! dt = 2udu =
=
2 (4 u2 )
2 (t + 3) 1 t
1
1
1
ln u2 4 + C = ln jt + 3j + C = ln sec2 x + 3 + C
2
2
2
4. (12) Evaluate exactly:
(a)
Z4
p
p
x2
t2 1
dx = t = 2x + 1 ! t2 = 2x + 1 ! x =
; tdt = dx
2
2x + 1
0
1
4
Z3
t4
2t2 + 1
1
tdt =
t
4
1
(b)
Z1
0
2
t4
2t2 + 1 dt =
124
15
1
x2 cos ( x) dx =
Z1
Z3
u = x2
u0 = 2x
v 0 = cos ( x)
sin ( x)
v=
x2 sin ( x)
=
1
0
x sin ( x) dx
0
v 0 = sin ( x)
cos ( x)
u0 = 1 v =
u=x
=
2
2
[x cos (
1
x)]0
+
2
2
Z1
0
2
2
3
cos ( x) dx =