Homework #2 Solutions
Math 2423 Spring 2012
4:1#8 f (x) = 1 + x2 ; 1 x 1: length of subinterval = 4x = 2=n: We …rst …nd the
upper and lower sums by taking n = 3: For n = 3; 4x = 2=3: From the …gure below (left),
upper sum is the sum of the area of rectangles I; II and III: Therefore,
1
[f ( 1) + f ( ) + f (1)]:4x
3
10
2
92
= [2 +
+ 2]: = :
9
3
27
upper sum
=
The lower sum is (see …gure below right),
lower sum
Next, for n = 4; 4x =
1
1
) + f (0) + f ( )]:4x
3
3
10
10 2
58
= [ + 1 + ]: = :
9
9 3
27
=
[f (
1
2
= : (You can make a similar diagram for this case). Then
4
2
1
1
) + f ( ) + f (1)]:4x
2
2
5 5
1
13
= [2 + + + 2]: =
4 4
2
4
upper sum
=
[f ( 1) + f (
and
1
1
) + f (0) + f (0) + f ( )]:4x
2
2
5
5 1
9
= [ + 1 + 1 + ]: = :
4
4 2
4
lower sum
4:1#20 Here f (x) = x2 +
4x =
b
a
n
=
p
[f (
1 + 2x; 4
=
7
4
n
=
x
7: We compute 4x and xi as follows
3
3i
; and xi = a + i4x = 4 + :
n
n
1
Using the limit formula for area, we obtain
A =
lim Rn = lim
n !1
=
lim
n !1
=
lim
n !1
n !1
n
X
i=1
n
X
i=1
n
X
i=1
3i
n
f
4+
"
3i
4+
n
Z4
g(x)dx
a
3
n
2
+
s
3i
1+2 4+
n
#
:
3
n
( 2)
6
= = 1:
2
6
6
(a) We have to …nd R6 : Right end-points are 1; 0; 1; 2; 3; 4: Therefore
4:2#6 Here n = 6 and 4x =
b
f (xi )4x
=
4
R6 = [g( 1) + g(0) + g(1) + g(2) + g(3) + g(4)]4x
2
3 1
1
3
+ 0 + + + ( 1) + ]:1
2
2 2
2
= [
= 0:
(b) We have to …nd L6 : Left end-points are
Z4
g(x)dx
2; 1; 0; 1; 2; 3: Therefore
L6 = [g( 2) + g( 1) + g(0) + g(1) + g(2) + g(3)]4x
2
= [0 + (
3
3 1
) + 0 + + + ( 1)]:1
2
2 2
1
:
2
=
(c) We have to …nd M6 : Mid points are
Z4
g(x)dx
M6 = [g(
3=2; 1=2; 1=2; 3=2; 5=2; 7=2: Therefore
3
1
1
3
5
7
) + g( ) + g( ) + g( ) + g( ) + g( )]4x
2
2
2
2
2
2
2
= [ 1 + ( 1) + 1 + 1 + 0 + (
=
1
)]:1
2
1
2
4:2#18 We have given the limit
lim
n !1
n
X
cos xi
i=1
xi
4x on [ ; 2 ]:
2
and we have to …nd the de…nite integral corresponding to this expression. The idea is to
compare this expression with the de…nition of de…nite integral,
Z
b
f (x)dx = lim
n !1
a
n
X
f (xi )4x:
i=1
In our case, a is ; and b is 2 : Next we have to …nd f (x): From the given expression, we
cos xi
cos x
have f (xi ) =
: This gives f (x) =
: Therefore the de…nite integral for the given
xi
x
expression is
Z 2
cos x
dx:
x
4:2#34 (a)
Z
2
g(x)dx is the area of triangle. This area of triangle is the half of the area
0
of rectangle of height 4 and base 2: Therefore
Z 2
g(x)dx = area of triangle
0
1
2
1
=
2
area of rectangle
=
(b)
Z
(4
2) = 4:
6
g(x)dx is the negative of the area of semicircle. The area of semicircle is the half
2
of the area of circle. From the given graph, radius of semicircle (r) = 2: Therefore
Z 6
g(x)dx = negative of the area of semicircle
2
=
=
=
(c)
Z
1
2
1
2
1
2
area of circle
r2
(2)2 =
2 :
7
g(x)dx is the area of triangle. By the similar method what we did in (a); we have
6
Z
7
g(x)dx = area of triangle
6
1
2
1
=
2
area of rectangle
=
(1
3
1
1) = :
2
Finally, total area is given by
Z 7
g(x)dx =
0
Z
2
Z
6
Z
7
g(x) dx
+
6
2
0
Z 6
Z 7
Z 2
g(x)dx +
g(x)dx +
g(x)dx
=
+
0
2
1
= 4 2 +
2
= 4 2 + 0:5
= 4:5 2 :
4
6