Math 121: Homework 2 solutions
1.
2 f (x) + 1 = 3
Z 1
x
f (t)dt
2 f 0 ( x ) = −3 f ( x )
f ( x ) = Ce−3x/2 ,
since 2 f ( x ) + 1 = 0, f (1) = − 21 = Ce−3/2 , C = − 12 e3/2 .
1
f ( x ) = − e(3/2)(1− x) .
2
2. (a) The curves y = x42 and y = 5 − x2 intersect where x4 − 5x2 + 4 = 0. Thus the
intersections are at x = ±1 and x = ±2. And the region is symmetric about
y−axis. We have
Z 2
4
)dx
x2
1
2
x3 4 4
= 2 (5x −
+ ) = sq.units.
3
x 1
3
Area = 2
(5 − x 2 −
(b)
Looparea = 2
Let
u2
Z 0
−2
√
x2 2 + x,
= 2 + x, 2udu = dx. Thus we have
Area = 2
Z √2
0
Z √2
(u2 − 2)2 u(2u)du
(u6 − 4u4 + 4u2 )du
√
1 7 4 5 4 3 2
= 4 ( u − u + u )
7
5
3
0
√
256 2
=
sq.units.
105
= 4
0
3. (a)
I=
Z 4p
0
9t2
+ t4 dt
=
Z 4 p
0
t
let u = 9 + t2 , du = 2tdt,
Z
1 25 √
I =
udu
2 9
1 3/2 25
=
u 3
9
98
=
.
3
1
9 + t2 dt,
(b)
t
t
cos2 ( ) sin2 ( )dt
5
5
Z
1
2t
=
sin2 ( )dt
4
5
Z
1
4t
=
(1 − cos( ))dt
8
5
1
5
4t
= (t − sin( )) + C.
8
4
5
Z
I =
(c)
Z
4
cos xdx =
Z
[1 + cos(2x )]2
dx
4
1
[1 + 2 cos(2x ) + cos2 (2x )]dx
4
Z
x sin(2x ) 1
=
+
+
1 + cos(4x )dx
4
4
8
x sin(2x ) x sin(4x )
+
+ +
+C
=
4
4
8
32
3x sin(2x ) sin(4x )
=
+
+
+C
8
4
32
Z
=
(d)
I=
Z
dx
=
ex + 1
Z
e− x dx
,
1 + e− x
let u = 1 + e− x , du = −e− x dx. Thus we have
du
u
= − ln |u| + C
= − ln(1 + e−x ) + C.
I = −
Z
(e)
Area =
Let u =
x2 ,
Z 2
0
xdx
.
x4 + 16
du = 2xdx. Thus, we have
1 4 du
2 0 u2 + 16
1
u 4
=
tan−1 8
4 0
π
=
sq.units.
32
Area =
2
Z
(f)
I=
Z π/2 q
0
1 − sin(θ )dθ =
Z π/2 q
0
1 − cos(π/2 − x )dx.
Let u = π/2 − x, du = −dx.
I = −
Z 0 √
1 − cos udu
π/2
Z π/2 r
√
u
u π/2
=
2 sin du = 2 (−2 cos )
2
2 0
0
√
√
= −2 + 2 2 = 2( 2 − 1).
2
4. We want to prove that for each positive integer k,
n
∑
jk =
j =1
n k +1
nk
+
+ Pk−1 (n),
k+1
2
where Pk−1 is a polynomial of degree at most k − 1. First check the case k = 1:
n
∑
j=
j =1
n 1+1
n
n ( n + 1)
=
+ + P0 (n),
2
1+1 2
where P0 (n) = 0 has degree ≤ 0. Now assume that the formula above holds for
k = 1, 2, 3, · · · , m, we will show that it also holds for k = m + 1. TO this end, sum
the formula
( j + 1 ) m +2 − j m +2 = ( m + 2 ) j m +1 +
(m + 2)(m + 1) m
j + · · · + 1,
2
for j = 1, 2, · · · , n. The left side telescopes, we get
n
( n + 1 ) m +2 − 1 = ( m + 2 ) ∑ j m +1
j =1
+
(m + 2)(m + 1)
2
n
n
j =1
j =1
∑ jm + · · · + ∑ 1.
Expanding the binomial power on the left and using the induction hypothesis on
the other terms we get
n
m +2
+ ( m + 2) n
m +1
n
+ · · · = ( m + 2 ) ∑ j m +1
j =1
+
3
(m + 2)(m + 1) nm+1
+··· ,
2
m+1
where the · · · represent terms of degree m or lower in the variable n. Solving for the
remain sum, we get
n
∑ j m +1
=
m + 2 m +1
1
( n m +2 + ( m + 2 ) n m +1 + · · · −
n
−···)
m+2
2
=
n m +2
n m +1
+
+··· ,
m+2
2
j =1
so that the formula is also correct for k = m + 1. Hence it is true for all positive
integer k by induction.
5.
F(x) =
Note that 0 <
1
1+ t2
Z 2x− x2
0
cos(
1
)dt.
1 + t2
≤ 1 for all t, and hence
0 < cos(1) ≤ cos(
1
) ≤ 1.
1 + t2
The integrand is continuous for all t, so F ( x ) is defined and differentiable for all x.
Since limx→±∞ (2x − x2 ) = −∞, therefore limx→±∞ F ( x ) = −∞. Now
1
)=0
1 + (2x − x2 )2
F 0 ( x ) = (2 − 2x ) cos(
only at x = 1. Therefore F must have a maximum value at x = 1, and no minimum
value.
6. (a) If m and n are integers, and m 6= n, then
1 π
cos(mx ) cos(nx )dx =
(cos(m − n) x + cos(m + n) x )dx
2 −π
−π
1 sin(m − n) x sin(m + n) x π
(
+
)
=
2
m−n
m+n
−π
= 0,
Z π
Z
1 π
sin(mx ) sin(nx )dx =
(cos(m − n) x − cos(m + n) x )dx
2 −π
−π
1 sin(m − n) x sin(m + n) x π
=
(
−
)
2
m−n
m+n
−π
= 0,
Z π
Z
1
sin(mx ) cos(nx )dx =
2
−π
Z π
Z π
(sin(m + n) x + sin(m − n) x )dx
1 cos(m + n) x cos(m − n) x π
= − (
+
)
2
m+n
m−n
−π
= 0,
4
−π
If m = n 6= 0 then
Z π
1
2
sin(mx ) cos(mx )dx =
Z π
(sin(2mx ))dx
π
1
(cos(2mx ))
= −
4m
−π
= 0,
−π
−π
(b) If 1 ≤ m ≤ k, we have
Z π
−π
a0
f ( x ) cos(mx )dx =
2
Z π
−π
k
∑
+
an
n =1
k
∑ bn
+
cos(mx )dx
Z π
−π
Z π
−π
n =1
cos(nx ) cos(mx )dx
sin(nx ) cos(mx )dx.
By the previous part, all the integrals on the right side are zero except the one
in the first sum having n = m. Thus the whole right side reduces to
am
Z π
−π
2
cos (mx )dx = am
=
Thus
1
am =
π
Z π
1
bm =
π
Z π
−π
Z π
1 + cos(2mx )
−π
2
am
(2π + 0) = πam .
2
f ( x ) cos(mx )dx.
Similarly, we have
−π
f ( x ) sin(mx )dx.
For m = 0 we have
Z π
−π
f ( x ) cos(mx )dx =
=
so the formula for am also holds for m = 0.
5
Z π
−π
a0 π,
f ( x )dx
dx