Math 121: Homework 1 solutions
1. (a)
2
2
2
2
2 − 3 + 4 − 5 + · · · − 99
2
99
∑ (−1)i i2
=
i =2
49
∑ [(2i)2 − (2i + 1)2 ] =
=
i =1
49
= − ∑ [4i + 1] = −4
i =1
49
∑ [4i2 − 4i2 − 4i − 1]
i =1
49 × 50
− 49
2
= −4949.
(b)
n
4
n
i
1 2 3
+ + +
+ ··· + n = ∑ i.
2 4 8 16
2
i =1 2
Let s =
1
2
4
+ 24 + 38 + 16
+ · · · + 2nn . Then
s
1 2
3
4
n
= + +
+
+ · · · + n +1 .
2
4 8 16 32
2
Subtracting these two sums, we get
s
1 1 1
1
n
=
+ + + · · · + n − n +1
2
2 4 8
2
2
1 1 − (1/2n )
n
=
− n +1
2 1 − 1/2
2
n+2
= 1 − n +1 .
2
Thus,
s = 2−
n+2
.
2n
2. The formula ∑in=1 i3 = n2 (n + 1)2 /4 holds for n = 1, since it says 1 = 1 in this case.
Now assume that it holds for n = k ≥ 1; that is, ∑ik=1 i3 = k2 (k + 1)2 /4. Then for
n = k + 1, we have
k +1
∑
i3 =
i =1
=
k
∑ i 3 + ( k + 1)3
i =1
k 2 ( k + 1)2
+ ( k + 1)3 =
( k + 1)2 2
[k + 4(k + 1)]
4
4
( k + 1)2
=
( k + 2)2 .
4
Thus the formula also holds for n = k + 1. By induction, it holds for all positive
integers n.
1
Figure 1: 3a
3. (a) The required area is (see the figure 3a)
3
3
3
[(−1 + )2 + 2(−1 + ) + 3
n→∞ n
n
n
6
6 2
+(−1 + ) + 2(−1 + ) + 3
n
n
3n 2
3n
+ · · · + (−1 + ) + 2(−1 + ) + 3]
n
n
2
6
3
6
3
= lim [(1 − + 2 − 2 + + 3)
n→∞ n
n n
n
12 62
12
+(1 −
+ 2 −2+
+ 3)
n
n
n
6n 9n2
6n
+ · · · + (1 −
+ 2 −2+
+ 3)]
n
n
n
27 n(n + 1)(2n + 1)
= lim (6 + 3
)
n→∞
6
n
= 6 + 9 = 15sq.units.
A =
lim
(b) The height of the region at position x is 0 − ( x2 − 2x ) = 2x − x2 . The base is an
interval of length 2, so we approximate using n rectangles of width 2/n. The
shaded area is
n
2 2i 4i2
A = lim ∑ (2 − 2 )
n→∞
n n
n
i =1
n
=
8i
∑ ( n2 −
n→∞
lim
i =1
8i2
)
n3
8 n ( n + 1)
8 n(n + 1)(2n + 1)
− 3
)
2
2
6
n
n
8
4
= 4 − = sq.units.
3
3
=
lim (
n→∞
2
Figure 2: 3b
4.
P = { x0 < x1 < · · · < x n },
P 0 = { x 0 < x 1 < · · · < x j −1 < x 0 < x j < · · · < x n } .
Let m j and M j be, respectively, the minimum and maximum values of f ( x ) on the
interval [ xi−1 , xi ], for 1 ≤ i ≤ n. Then
n
L( f , P) =
∑ m i ( x i − x i −1 ),
i =1
n
U ( f , P) =
∑ Mi ( x i − x i − 1 ) .
i =1
If m0j and M0j are the minimum and maximum values of f ( x ) on [ x j−1 , x 0 ], and if m00j
and M00j are the corresponding values for [ x 0 , x j ], then
m0j ≥ m j ,
m00j ≥ m j ,
M0j ≤ M j ,
M00j ≤ M j .
Therefore we have
m j ( x j − x j−1 ) ≤ m0j ( x 0 − x j−1 ) + m00j ( x j − x 0 ),
M j ( x j − x j−1 ) ≥ M0j ( x 0 − x j−1 ) + M00j ( x j − x 0 ),
Hence L( f , P) ≤ L( f , P0 ) and U ( f , P) ≥ U ( f , P0 ). If P00 is any refinement of P we
can add the new points in P00 to those in P one at a time, and thus obtain
L( f , P) ≤ L( f , P00 ),
5. (a)
n
U ( f , P) ≥ U ( f , P00 ).
2n + 3i
=
∑
2
n→∞
n
i =1
lim
3
Z 1
0
7
(2 + 3x )dx = .
2
Figure 3: 5c
Figure 4: 5d
(b)
Z 3
−3
(2 + t )
p
9 − t2 dt
= 2
Z 3 p
−3
9 − t2 dt +
Z 3 p
−3
t
9 − t2 dt
1
= 2( π32 ) + 0 = 9π.
2
(c)
Z 2
−1
sgn( x )dx = 2 − 1 = 1.
(d)
Z 4
−3
(| x + 1| − | x − 1| + | x + 2|)dx
= area( A1 ) − area( A2 )
5+8
1+2
1+2
11
41
15
(5) +
(3) −
(1) −
(1) −
(1) = .
=
23
2
2
2
23
2
4
Figure 5: 5e
(e)
Z 3 2
x −x
dx
| x − 1|
= area( A1 ) − area( A2 )
1+3
1
7
=
(2) − (1)(1) =
2
2
2
0
(f)
Z b
a
=
Z b
a
=
Z b
a
( f ( x ) − k)2 dx
2
( f ( x )) dx − 2k
Z b
a
f ( x )dx + k
2
Z b
a
dx
( f ( x ))2 dx − 2k(b − a) f¯ + k2 (b − a)
= (b − a)(k − f¯)2 +
This is minimum if k = f¯.
5
Z b
a
( f ( x ))2 dx − (b − a)( f¯)2