MATH 101, Section 212 (CSP)
Week 1: Marked Homework Solutions
2011 Jan 13
1. [2] We use n subintervals of equal length ∆x = n2 , for i = 1, . . . , n. Then x0 = 0,
for i = 0, 1, . . . , n. The ith subinterval is
x1 = n2 , x2 = n4 , x3 = n6 , etc. In general xi = 2i
n
2(i−1) 2i
[xi−1 , xi ] = [ n , n ].
(a) [6] The right endpoint of the ith subinterval is xi =
i) The Riemann sum using right endpoints is
Rn =
n
X
2i
,
n
i = 1, . . . , n.
f (xi ) ∆x
i=1
=
=
n
X
(xi − x2i )∆x
i=1
n X
2i
n
−
4i2
n2
2
n
=
i=1
Pn
4
2
i=1
n
4 n(n+1)
n2
2
−
=
2(n+1)
n
4(n+1)(2n+1)
3n2
=
−
= 2 1+
1
n
i−
8
n3
Pn
i2
i=1
8 n(n+1)(2n+1)
n3
6
−
4
3
1+
22
ii) if n = 10 then R10 = − 25
= −0.8800.
1717
iii) if n = 100 then R100 = − 2500
= −0.6868.
iv) limn→∞ Rn = 2(1) − 34 (1)(2) = − 23 ≈ −0.6667.
1
n
(b) [6] The left endpoint of the ith subinterval is xi−1 =
i) The Riemann sum using left endpoints is
Ln =
=
n
X
.
2(i−1)
,
n
i = 1, . . . , n.
f (xi−1 ) ∆x
(xi−1 − x2i−1 )∆x
n h
X
2(i−1)
n
=
i=1
Pn
4
i=1 (i
n2
=
4
n2
=
1
n
i=1
n
X
i=1
=
2+
4
n2
Pn
−
4(i−1)2
n2
− 1) −
i
8
n3
2
n
Pn
i=1 (i
− 1)2
P
− 1) − n83 ni=1 (i2 − 2i + 1)
P
P
P
P
P
[ ni=1 i − i=1 1] − n83 [ ni=1 i2 − 2 ni=1 i + ni=1 1]
i=1 (i
1
h
n(n+1)
2
=
4
n2
=
2(n+1)
n
4
n
−
= 2 1+
1
n
i
−n −
−
−
8
n3
h
n(n+1)(2n+1)
6
4(n+1)(2n+1)
3n2
4
n
4
3
−
1+
+
1
n
− 2 n(n+1)
+n
2
8(n+1)
n2
− n82
2 + n1 + n8 1 + n1 −
12
ii) if n = 10 then L10 = − 25
= −0.4800.
1617
iii) if n = 100 then L100 = − 2500
= −0.6468.
4
iv) limn→∞ Ln = 2(1) − 0 − 3 (1)(2) + 0(1) − 0 = − 23 ≈ 0.6667.
(c) [6] The midpoint of the ith subinterval is x̄i = 21 (xi−1 + xi ) =
i) The Riemann sum using midpoints is
Mn =
=
n
X
i=1
n
X
i=1
=
i = 1, . . . , n.
(x̄i − x̄2i )∆x
n h
X
2i−1
n
(2i−1)2
n2
−
=
=
2
n2
=
2
n2
=
2(n+1)
n
2
n2
2i−1
,
n
8
.
n2
f (x̄i ) ∆x
i=1
Pn
2
i=1 (2i
n2
=
i
Pn
i
− 1) −
2
n
2
n3
Pn
i=1 (2i
− 1)2
P
− 1) − n23 ni=1 (4i2 − 4i + 1)
P
P
P
P
P
[2 ni=1 i − i=1 1] − n23 [4 ni=1 i2 − 4 ni=1 i + ni=1 1]
h
i
h
i
n(n+1)(2n+1)
n(n+1)
2
2 n(n+1)
−
n
−
4
−
4
+
n
2
n3
6
2
i=1 (2i
= 2 1+
−
1
n
2
n
−
−
4(n+1)(2n+1)
3n2
2
n
−
4
3
1+
+
1
n
4(n+1)
n2
− n22
2 + n1 + n4 1 + n1 −
2
.
n2
33
= −0.6600.
ii) if n = 10 then M10 = − 50
3333
iii) if n = 100 then M100 = − 5000
= −0.6666.
4
iv) limn→∞ Mn = 2(1) − 0 − 3 (1)(2) + 0(1) − 0 = − 23 ≈ −0.6667.
(Notice that for a fixed finite value of n, using midpoints gives a more accurate result than
using either right or left endpoints, but all three Riemann sums approach the same limiting
value as n → ∞.)
= n1 , and let x0 = 0, x1 = n1 , x2 = n2 ,. . . ,xn = nn = 1;
2. (a) [4] i) Let a = 0, b = 1, ∆x = b−a
n
i
in general xi = n for i = 0, 1, . . . , n. The P
right endpoint
of
[xi−1 , xi ] is
the ith subinterval
P
xi = ni P
for i = 1, . . . , n. Then limn→∞ n12 ni=1 i ni + 1 = limn→∞ ni=1 ni ni + 1 n1 =
limn→∞ ni=1 xi (xi + 1) ∆x =
Z 1
x(x + 1) dx.
0
2
ii)
(b) [4] i) Let a = 0, b = 1, ∆x = b−a
= n1 , and let x0 = 0, x1 = m1 , x2 = m2 ,. . . ,xm = m
; in
n
m
k
general xk = m for k = 0, 1, . . . , m. The right endpoint
[xk−1 , xk ] is
2 of the kth subinterval
P
P
m
m
k
k
k 2 1
k
for k = 1, . . . , m. Then limm→∞ k=1 mk2 cos m
= limm→∞ k=1 m
cos m
=
xk = m
2
m
Pm
2
limm→∞ k=1 xk cos(xk ) ∆x =
Z 1
x cos(x2 ) dx.
0
ii)
R2 √
3. (a) [3] −2 4 − x2 dx = 21 π22 = 2π is the area of the top half of the circle of radius 2
centred at the origin (x2 + y 2 = 4)
R3
R 3/2
R3
(b) [3] −1 (3 − 2u) du = −1 (3 − 2u) du + 3/2 (3 − 2u) du
= [area of triangle of height 5 and base 23 − (−1) = 25 ]
− [area of triangle of height 3 and base 3 − 23 = 23 ]
= 12 5 25 − 12 3 23 = 4
3
R 10
R5
R 10
(c) [3] 0 |y − 5| dy = 0 |y − 5| dy + 5 |y − 5| dy
= [area of triangle of height 5 and base 5 − 0 = 5 ]
+ [area of triangle of height 5 and base 10 − 5 = 5 ]
= 21 5 · 5 + 21 5 · 5 = 25
√
√
4. (a) [4] If f (t) = 1 + t2 , then f 0 (t) = t/ 1 + t2 exists for all t in (0, 1) and f 0 (t) > 0
for all t√in (0, 1), f (t) is increasing on [0, 1]. So f (0) = 1 is the absolute minimum value,
f (1) = 2 is the absolute maximum value on [0, 1], and
√
√
1 ≤ 1 + t2 ≤ 2 for all t in [0, 1].
Therefore
1(1 − 0) ≤
Z
1
0
√
√
1 + t2 dt ≤ 2(1 − 0).
(b) [4] If f (y) = cos y, then f 0 (y) = − sin y exists for all y in (−π/4, π/4) and f 0 (y) = 0 iff
y = 0 in (−π/4, π/4). Evaluating f at the end points and critical point y = 0, we find that
4
f (−π/4) = f (π/4) =
maximum value, so
√
2/2 is the absolute minimum value, and f (0) = 1 is the absolute
√
2/2 ≤ cos y ≤ 1 for all y in [−π/4, π/4].
Therefore
√ !h
Z π/4
h π π i
2
π π i
≤
,
cos y dy ≤ 1
− −
− −
2
4
4
4
4
−π/4
√
R π/4
or simplifying, π 2/4 ≤ −π/4 cos y dy ≤ π/2.
(c) [4] If f (x) = x1 , then f 0 (x) = − x12 < 0 for all x in (4, 6), f is decreasing on [4, 6], so its
absolute maximum value is f (4) = 41 and
1
1
≤
x
4
for all x in [4, 6].
(1)
1
1
On the other hand, if g(x) = 8−x
, then g 0 (x) = (8−x)
2 > 0 for all x in (4, 6), g is increasing
1
on [4, 6], so its absolute minimum value is g(4) = 4 and
1
1
≤
4
8−x
for all x in [4, 6].
(2)
Therefore from equations (1) and (2) we have
1
1
≤
x
8−x
and
Z
6
4
for all x in [4, 6],
1
dx ≤
x
Z
6
4
1
dx.
8−x
1
2v
0
(d) [4] If f (v) = 1+v
2 , then f (v) = − (1+v 2 )2 < 0 for all v in (0, 2), f is decreasing on [0, 2],
so its absolute maximum value is f (0) = 1 and its absolute minimum value is f (2) = 15 ,
therefore
1
1
≤
≤ 1 for all v in [0, 2],
5
1 + v2
and
Z 2
1
1
(2 − 0) ≤
dv ≤ 1(2 − 0).
2
5
0 1+v
5