MAS110 Solutions: Integration
1. Divide the interval [0, X] into N equal parts by using the evenly spaced points
0<
1
2
i−1
i
N −1
N
X < X < ... <
X < X < ... <
X < X = X.
N
N
N
N
N
N
i
Overestimate the area by using the rectangle Ai with base [ i−1
N X, N X] and
height Ni X, i = 1, 2, . . . N .
Figure 1: Diagram for Q1
Now area of Ai is
X
Z
x dx ≈
0
i
NX
× ( Ni X −
i−1
N X)
=
i
X 2.
N2
Hence
N
X
i
X 2 N (N + 1)
X2 1
X2
2
X
=
=
1
+
→
N2
N2
2
2
N
2
as N → ∞.
i=1
2. Divide the interval [0, X] into N equal parts by using the evenly spaced points
0<
1
2
i−1
i
N −1
N
X < X < ... <
X < X < ... <
X < X = X.
N
N
N
N
N
N
i
Under-estimate the area by using the rectangle with base [ i−1
N X, N X] and
2
height (i−1)
X 2 , i = 1, 2, . . . N . This has area X
N ×
N2
The approximation to the area from below is then
N
X
(i − 1)2
i=1
N3
X3 =
(i−1)2 2
X
N2
X3 2
2
2
0
+
1
+
.
.
.
+
(N
−
1)
N3
X 3 (N − 1)N (2N − 1)
N3
6
1 3
1
1 = X 1−
2−
.
6
N
N
=
1
=
(i−1)2 3
X .
N3
Figure 2: Diagram for Q2
Combining with the upper-estimate from Example 7.2, we obtain
Z X
1 1
1 1 1
1 3
x2 dx ≤ X 3 1 +
≤
.
X 1−
2−
2+
N
N}
N
N}
0
|6
{z
|6
{z
→
Z
So
X3
6
→
· 2 when N → ∞
X
x2 dx =
0
X3
6
· 2 when N → ∞
X3
.
3
3. Part (a). We refer to Figure 3 for the graph. In the part over the interval
[an+1 X, an X], n = 0, 1, 2, . . ., the height of the over-estimating rectangle is
(an X)1/2 . So the over-estimate is
∞
X
(an X − an+1 X)(an X)1/2 =
n=0
∞
X
X 3/2 (1 − a)an an/2
n=0
= X 3/2 (1 − a)
∞
X
(a3/2 )n
n=0
=X
2
3/2
1−a
.
1 − a3/2
Figure 3: Diagram for Q3
Part (b). Write
√
a = b; so b → 1 as a → 1. Then
1 − b2
1−a
=
1 − b3
1 − a3/2
(1 − b)(1 + b)
=
(1 − b)(1 + b + b2 )
1+b
=
1 + b + b2
1+1
2
→
=
as b → 1.
1+1+1
3
Hence
RX
0
x1/2 dx = 23 X 3/2 .
Part (c). Using the same partition as in part (a), we obtain the over-estimate
∞
X
m
(ak X − ak+1 X)(ak X)m/n = X n +1
k=0
1−a
.
m
1 − a n +1
1
n
Put b = a . Then
1−a
1 − bn
(1 − b)(1 + b + . . . + bn−1 )
=
=
m
1 − bn+m
(1 − b)(1 + b + . . . + bm+n−1 )
1 − a n +1
n
→
when b (equivalently a) goes to 1.
m+n
RX
m
n
Hence 0 xm/n dx = m+n
X n +1 .
4. Using anti-derivatives:
Z X
Z
n
x dx +
0
0
Xn
xn+1
x1/n dx =
n+1
=
3
X n+1
n+1
X
0
+
"
1
x n +1
+ 1
n +1
nX n+1
n+1
#X n
0
= X n+1 .
Let’s see why this is true geometrically. Consider the rectangle OABC in Figure
4. It has area X · X n i.e. X n+1 . This has to be the sum of the area of the
Figure 4: Diagram for Q4
region OAB and the area of the region OBC.
RX
The area of OAB is 0 xn dx. The area of OBC is the area of the region
enclosed by the graph and the y-axis between y = 0 and y = X n . To calculate
this, we need to express the graph as a function of
then integrate between
R Xynand
n
1/n
limits y = 0 and y = X . Thus area of OBC is 0 y dy, which is the same
R Xn
as 0 x1/n dx (by a relabelling of the variable). Hence we obtain
Z Xn
Z X
n
n+1
x dx +
x1/n dx.
X
=
0
5.
d
dx
Z
x
0
f (t) dt = f (x). Sub u = h(x) to obtain
0
d
dx
Z
h(x)
0
d Z u
du
f (t) dt =
f (t) dt
= f (u)u0 = f (h(x))h0 (x).
du 0
dx
Last part:
d
dx
Z
h(x)
g(x)
Z
d
f (t) dt =
dx
d
=
dx
h(x)
f (t) dt −
0
Z
h(x)
0
!
g(x)
Z
f (t) dt
0
d
f (t) dt −
dx
Z
g(x)
f (t) dt
0
= f (h(x))h0 (x) − f (g(x))g 0 (x).
4
6. Take u(x) = v(x) = 1, for example.
7.
2π
Z
x sin x dx =
0
2π
Z
Z
+
0
2π
cos x dx = −2π + [sin x]2π
0 = −2π.
2π Z 2π
1 2
x cos(2x) dx =
x sin(2x)
x sin(2x) dx
−
2
0
0
2π
Z
1 2π
1
cos(2x) dx = π.
x cos(2x)
−
=
2
2 0
0
2
0
[−x cos x]2π
0
π
Z
cos2 x dx. Then
8. Set I =
0
Z
π
[cos x sin x]π0
Z
π
cos x cos x dx =
−
(− sin x) sin x dx
0
Z0 π
Z π
=
sin2 x dx =
(1 − cos2 x) dx = π − I.
I=
0
0
So 2I = π i.e. I = π/2.
9. We have identities
2 cos(mx) sin(nx) = sin(m + n)x − sin(m − n)x;
2 cos(mx) cos(nx) = cos(m + n)x + cos(m − n)x;
2 sin(mx) sin(nx) = cos(m − n)x − cos(m + n)x.
Z 2π
Z 2π
Also
cos(ax) dx =
sin(ax) dx = 0 for all 0 6= a ∈ Z. Hence
0
Z
0
2π
Z
2π
2π
sin(m + n)x dx −
2 cos(mx) sin(nx) dx =
0
Z 2π
Z
0
Z
sin(m − n)x dx = 0;
0
2π
Z
2π
cos(m − n)x dx
0
(0
0,
m 6= n,
=
2π, m = n
Z 2π
Z 2π
Z 2π
2 sin(mx) sin(nx) dx =
cos(m − n)x dx −
cos(m + n)x dx
0
0
(0
0,
m 6= n or m = n = 0,
=
2π, m = n 6= 0
2 cos(mx) cos(nx) dx =
cos(m + n)x dx +
0
for m, n ∈ Z.
5
10.
i1
h
x2 dx = 1 ex2
= 12 (e − 1).
xe
2
0
0
h
2 i1
R1
R 1 3 x2
R1 2
2
x2 dx = x2 · ex
− 0 xex dx = 21 e− 12 (e−1) = 12 .
x
e
dx
=
(x
)
xe
2
0
0
d
dx
ex
2
2
= 2xex . So
R1
0
11.
x2
0 1+x3
R 1 x2
0 1+x2
dx =
dx =
1
3 1
3 ln(1 + x ) 0 = 3
R1
1
0 1 − 1+x2 dx =
x3
0 1+x2
dx =
R1
R1
R1
12.
R
1
0
ln 2.
1
x − tan−1 x 0 = 1 − π4 .
h 2
i1
ln(1+x2 )
x
x
x − 1+x
= 12 − 12 ln 2.
−
2 dx =
2
2
0
tan−1 x dx = x tan−1 x dx −
R
x
1+x2
dx = x tan−1 x − 12 ln(1 + x2 ) + C.
13. We have
Z
π
cos x]π0
Z
π
ex sin x
Z π
π
π
x
ex cos x dx
= −e − 1 + [e sin x]0 −
0
Z π
π
x
= −e − 1 −
e cos x dx.
x
e cos x dx = [e
0
x
+
0
0
Hence 2
Rπ
0
ex cos x dx
=
−eπ
− 1 i.e.
Rπ
0
ex cos x dx = − 21 (1 + eπ ).
14. In the range x ∈ (− π2 , π2 ), we have 1 + sin x > 0 and cos x > 0. Hence
sec x + tan x = (1 + sin x)/ cos x > 0 when x ∈ (− π2 , π2 ).
d 1
1
sec x = dx
cos x = − cos2 x (− sin x) = sec x tan x. So if u = sec x + tan x
2
then du
dx = sec x tan x + sec x = u sec x. Hence
Z
Z
du
= ln u + C = ln(sec x + tan x) + C.
sec x dx =
u
cos x cos x−sin x(− sin x)
d sin x
d
15. We have dx
(tan x) = dx
= cos12 x = sec2 x.
cos x =
cos2 x
d
dx
For the integral:
Z
Z
sec3 x dx = (sec x)(sec2 x) dx
Z
= sec x tan x − sec x tan x tan x dx
Z
= sec x tan x − sec x(sec2 x − 1) dx
Z
Z
= sec x tan x − sec3 x dx + sec x dx.
R
Hence sec3 x dx =
tan x) + C.
1
2
sec x tan x +
1
2
R
6
sec x dx =
1
2
sec x tan x +
1
2
ln(sec x +
16. Put t = tan θ. So dt = sec2 θ dθ and
Z θ=tan−1 x
Z x
Z tan−1 x
sec2 θdθ
dt
√
=
sec θ dθ
=
sec θ
1 + t2
θ=0
0
0
−1
= [ln(sec θ + tan θ)]0tan x
h
itan−1 x
p
= ln(tan θ + 1 + tan2 θ)
0
p
p
= ln(x + 1 + x2 ) − ln(0 + 1) = ln(x + 1 + x2 ).
Rx 1
df
1
dt. Then dx
by the
Alternative method: Let f (x) := 0 √1+t
= √1+x
2
2
Fundamental Theorem of Calculus. Now
p
d
2x
1
√
× 1+ √
ln(x + 1 + x2 ) =
dx
x + 1 + x2
2 1 + x2
√
1
x + 1 + x2
1
√
=
× √
=√
.
2
2
x+ 1+x
1+x
1 + x2
√
Hence f (x) = ln(x + 1 + x2 )√
+ c for some real constant
R x 1 c. Since f (0) = 0
2
we obtain c = 0. Hence ln(x + 1 + x ) = f (x) = 0 √1+t2 dt.
17.
R2
18.
R1
dx
1 x2 +2x
=
R 2 1
1
x
−
1
x+2
dx
2
1
2
=
[ln x − ln(x + 2)]21 =
1
2
ln( 32 ).
R1
dx = 0 √ 1 2 dx. Now substitute x + 1 = sec θ. Thus dx =
(x+1) −1
p
sec θ tan θ dθ, (x + 1)2 − 1 = tan θ and so
Z 1
Z π/3
√ 1
π/3
√
dx =
sec θ dθ = [ln(sec θ + tan θ)]0 = ln 2 + 3 .
x2 + 2x
0
0
0
√ 1
x2 +2x
19. (a)
Z ∞
3 −x4
x e
0
Z
dx = lim
X→∞ 0
X
3 −x4
x e
dx = lim
X→∞
1
4
− e−x
4
X
1 1
1
4
− e−X = .
X→∞ 4 4
4
= lim
0
(b) Sub u = ln(x), so du = x1 dx and x = eu . Thus
Z ∞
Z ∞
ln(x)
−u
−u
−u ∞
dx
=
ue
du
=
−ue
−
e
= 1.
0
x2
1
0
(Note that limu→∞ ue−u = limx→∞
(c) For n ≥ 1 we have
Z X
Z
X
xn e−x dx = −xn e−x 0 +
0
X
ln x
x
= 0.)
Z
nxn−1 e−x dx = −X n e−X +
0
0
7
X
nxn−1 e−x dx.
n
n−1
Now limX→∞ X n e−X = limX→∞ X
= n limX→∞ XeX . Applying L’hopital
eX
repeatedly, we see that limX→∞ X n e−X = 0. Hence
Z ∞
Z ∞
n −x
x e dx = n
xn−1 e−x dx,
0
0
and repeating the process we obtain
Z
Z ∞
n −x
x e dx = n!
0
0
8
∞
e−x dx = n!.