Math 21 B, Homework 2 Solutions
2
X
6k
Exercise 5.2, (1). Write the sum
without sigma notation, then evaluate.
k+1
k=1
Solution.
2
X
6(1)
6(2)
6 12
6k
=
+
= +
= 7.
k+1
(1) + 1 (2) + 1
2
3
k=1
Exercise 5.2, (2). Write the sum
3
X
k−1
k=1
Solution.
3
X
k−1
k=1
k
k
without sigma notation, then evaluate.
(1) − 1 (2) − 1 (3) − 1
0 1 2
9
3
+
+
= + + = = .
(1)
(2)
(3)
1 2 3
6
2
=
Exercise 5.2, (10). Which formula is not equivalent to the other two?
a.
4
X
(k − 1)2
b.
3
X
(k + 1)2
−1
X
k2 .
k=−3
k=−1
k=1
c.
Solution.
a.
4
X
(k − 1)2 = ((1) − 1)2 + ((2) − 1)2 + ((3) − 1)2 + ((4) − 1)2
k=1
= 02 + 12 + 22 + 32 = 14;
b.
3
X
(k + 1)2 = ((−1) + 1)2 + ((0) + 1)2 + ((1) + 1)2 + ((2) + 1)2 + ((3) + 1)2
k=−1
= 02 + 12 + 22 + 32 + 42 = 30;
c.
−1
X
k 2 = (−3)2 + (−2)2 + (−1)2
k=−3
= 12 + 22 + 32 = 14.
When we evaluate the sums we see that a = c. However, note that the expressions for a and c themselves
are not equivalent since the sum involved in a has 4 terms whereas the sum involved in c has only 3
terms. (Starting at 2 instead of 1 in a would make the expressions equivalent; or ending at 0 instead of
−1 in c.)
Exercise 5.2, (12). Express the sum 1 + 4 + 9 + 16 in sigma notation.
Solution. One possibility is
4
X
3
1
X
X
2
k . Another could be
(j + 1) , or
4i − 102 .
2
j=0
k=1
1
i=10
Exercise 5.2, (16). Express the sum − 15 +
2
5
3
5
−
+
4
5
−
5
5
in sigma notation.
4
5
X
X
k+1
kk
or
.
Solution. Two possibilities are
(−1)k−1
(−1)
5
5
k=0
k=1
7
X
Exercise 5.2, (21). Evaluate the sum
(−2k).
k=1
Solution.
7
X
(−2k) = −2
k=1
7
X
k = −2
k=1
6
X
Exercise 5.2, (24). Evaluate the sum
7
2
(7 + 1)
= −56.
(k 2 − 5).
k=1
Solution.
6
X
(k 2 − 5) =
k=1
6
X
k2 −
k=1
6
X
5=
k−1
6(6 + 1)(2(6) + 1)
− 5(6) = 61.
6
Exercise 5.2, (40). For f (x) = 2x, find a formula for the Riemann sum obtained by dividing the interval
[0, 3] into n equal subintervals and using the right-hand endpoint for each ck . Then take a limit of these
sums as n → ∞ to calculate the area under the curver over [0, 3].
3−0
3
Solution. For the length of subinterval this gives ∆x =
= , and for right-hand endpoints ci this
n
n
3i
gives ci = i∆x = . Then the right-hand sum is
n
X
n
n
n
n
n
X
3
18 X
3i 3 X 6i 3 X 18
2ci
=
=
· =
i
=
2
i,
n
n n
n n
n2
n2
i=1
i=1
i=1
i=1
i=1
where 18
n is able to be factored out since it is constant (not a part of the summation that changes from
term to term as i increases). Using the summation equations from this section,
n
18 X
18 n(n + 1)
9n2 + 9n
i
=
=
.
n2
n2
2
n2
i=1
Then, taking limits as n → ∞,
lim
n
X
6i
n→∞
Exercise 5.3, (1). Express
i=1
lim
kP k→0
9n2 = 9n
9
= lim
= lim 9 +
= 9.
n→∞
n→∞
n
n2
n
n
X
c2k ∆xk as a definite integral, where P is a partition of [0, 2].
k=1
Solution.
Z
2
x2 dx.
0
2
Exercise 5.3, (3). Express lim
kP k→0
n
X
(c2k − 3ck )∆xk as a definite integral, where P partitions [−7, 5].
k=1
Solution.
Z
5
(x2 − 3)dx.
−7
R2
Exercise 5.3, (11). Suppose that 1 F (x)dx = 5. Find
Z 2
Z 2√
Z
a.
f (u)du
b.
3f (z)dz
c.
1
1
Solution.
Z 2
a.
f (u)du = 5
2√
b.
√
3f (z)dz = 5 3
f (t)dt
Z
d.
[−f (x)]dx.
1
1
[−f (x)]dx = −5.
d.
2
Exercise 5.3, (13). Suppose that f is integrable and that
Z 3
Z 4
f (t)dt.
f (z)dz
b.
a.
2
Z
f (t)dt = −5
c.
1
2
Z
2
Z
1
1
1
R3
0
f (z)dz = 3 and
R4
0
f (z)dz = 7. Find
4
3
Solution.
Z 4
a.
f (z)dz = 4
3
Z
f (t)dt − 4.
b.
3
4
√
Z
2
Exercise 5.3, (29). Use the results of Equations (2) and (4) to evaluate the integral
xdx.
1
Solution.
√
Z
1
2
√
( 2)2 (1)2
2 1
1
xdx =
−
= − = .
2
2
2 2
2
Z
2π
Exercise 5.3, (31). Use the results of Equations (2) and (4) to evaluate the integral
θdθ.
π
Solution.
Z
2π
θdθ =
π
(2π)2 (π)2
4π 2 π 2
3π 2
−
=
−
=
.
2
2
2
2
2
Exercise 5.3, (43). Use the rules in Table 5.6 and Equations (2) and (4) to evaluate the integral
Z 2
(2t − 3)dt.
0
Solution.
Z
2
Z
(2t − 3)dt = 2
0
2
Z
tdt −
0
2
3dt = 2
0
22 02
−
2
2
3
− 3(2 − 0) = 2(2) − 3(2) = −2.