3
ESTIMATING WITH FINITE SUMS
1
Estimating with finite sums
2
Sigma and limit
3
Definite integral
1
For positive(temporarily) conti function f (x) on [a, b], Let A be the region
under the graph of y = f (x), between x = a, x = b To find area, we divide
[a, b] into small subintervals. ( 4.1). For example, uniform n-subintervals
are {x0 , x1 , . . . , xn }
xi = a + i(b − a)/n,
i = 0, 1, · · · , n
The area A is approximated by
Sn =
n−1
X
i=0
f (xi )(xi+1 − xi )
y
y = f (x)
O
a
b
x
4.1: S6
In general, nonuniform intervals allowed except that maxi |xi+1 − xi | → 0
as n → ∞.
f (xi ) can be replaced by f (ci ). A partition of [a, b] is a set of points
x0 , x1 , . . . , xn such that a = x0 < x1 < · · · < xn−1 < xn = b. P =
{x0 , x1 , . . . , xn } is a partition of [a, b] and f (x) is defined on [a, b]. For any sequence {c1 , c2 , . . . , cn } satisfying xi−1 ≤ ci ≤ xi The Riemann sum R(f, P )
of f (x) w.r.t P .( 4.2)
R(f, P ) =
n
X
i=1
f (ci )(xi+1 − xi )
2
y
(cn , f (cn ))
b
(ck , f (ck ))
b
y = f (x)
c1
O x0 = a x1
c2
b
(c1 , f (c1 ))
b
x2
xk
b
ck
cn
xk+1
b
xn = b
x
b
b
(c2 , f (c2 ))
4.2: R(f, P )
Definition 3.1. kP k: norm of P = {x0 , x1 , . . . , xn } is defined by
kP k = max (xi+1 − xi )
0≤i<n
Suppose the Riemann sum R(f, P ) of f (x) approaches some number I as
the norm kP k of partition P approaches 0 then this number is defined as the
area under the graph.
4
Definite Integral
Definition 4.1 (Definite Integral as limit of Riemann Sum). f (x) is defined
on [a, b] We say a number I is the definite integral of f over [a, b] and f (x)
Rb
is integrable on [a, b] : This number I is denoted by a f (x) dx and called
the definite integral of f (x) on [a, b].
When
lim R(f, P ) = lim
kP k→0
kP k→0
n
X
i=1
f (ci )(xi − xi−1 )
this limit exists We write it as
Z
b
f (x) dx =
a
=
lim R(f, P )
kP k→0
lim
kP k→0
n−1
X
i=0
f (ci )(xi − xi−1 )
4
DEFINITE INTEGRAL
3
Non-integrable function
f (x) =
(
0, x rational number
1, x irrational number
(
0,
g(x) =
1,
Theorem 4.2 (Definite integral).
Z b
g(x) dx
x ∈ [0, 1]
x ∈ (1, 2]
(1)
Z
b
(f (x) + g(x)) dx =
a
a
a
(2)
(3)
Z
Z
b
kf (x) dx = k
a
Z
b
f (x) dx
a
c
f (x) dx =
a
(4)
Z
b
f (x) dx +
a
Z
c
f (x) dx, (a ≤ b ≤ c)
b
[a, b] f (x) ≥ 0
(5) min f (x) · (b − a) ≤
x∈[a,b]
,
Z
Z
b
a
f (x) dx ≥ 0
b
a
f (x) dx ≤ max f (x) · (b − a)
x∈[a,b]
Z b
Z b
f (x) dx ≤
|f (x)| dx
(6) a
a
Definition 4.3.
Z
a
f (x) dx = 0,
a
Example 4.4. Find
Z
b
Ra
0
a
f (x) dx = −
x2 dx by definition.
Z
Z
a
b
f (x) dx.
b
f (x) dx +
4
sol. Suppose P = {x0 , x1 , . . . , xn } is a uniform partition of [0, a]. xi = i a/n
n
X
i=1
x2i (xi − xi−1 ) =
=
=
n X
i·a 2 a
n
n
i=1
n
a3 X
n3
i2
i=1
a3 (n + 1)(2n + 1)
6n2
So
Z
a
x2 dx = lim
n→∞
0
= lim
=
n
X
x2i (xi − xi−1 )
i=1
a3 (n
n→∞
a3
+ 1)(2n
6n2
+ 1)
3
Theorem 4.5 (Mean value theorem for integral). Let a < b. If f (x) is conti.
on closed interval [a, b] then there is a c in [a, b]
Z b
1
f (c) =
f (x) dx
b−a a
This value is called the average of f on [a, b] denoted by av(f )
Proof. Since f (x) is continuous on [a, b] there are min and max;
f (x0 ) = min f (x),
x∈[a,b]
f (x1 ) = max f (x)
x∈[a,b]
for some x0 , x1 in [a, b].
First if x0 = x1 , f (x) is const. This holds for all c ∈ [a, b]. Suppose
x0 < x1 . Then by intermed thm, f (x) assumes all values between f (x0 ) and
f (x1 ) in [x0 , x1 ]. Since
Z b
1
f (x0 ) ≤
f (x) dx ≤ f (x1 )
b−a a
So there is a c such that
1
f (c) =
b−a
c ∈ [x0 , x1 ] ⊂ [a, b]
The case x0 < x1 is the same.
Z
a
b
f (x) dx
5
FUNDAMENTAL THEOREM OF CALCULUS
5
y
y = f (x)
f (c)
O
a
c
b−a
b
x
4.3: MVT for Integral
5
Fundamental theorem of Calculus
If f is integrable on I, the integral
R x from a fixed point a to another point x
defines a new function F (x) = a f (t) dt. If f is conti, this new function is
differentiable, and
Theorem 5.1 (Fundamental theorem of Calculus I,II). f (x) is conti on [a, b]
(1) F (x) =
Rx
a
f (t) dt is differentiable on (a, b) and
d
d
F (x) =
dx
dx
Z
x
f (t) dt = f (x)
a
(2) If F (x) is the anti-derivative of f (x)
Z
b
a
f (x) dx = F (b) − F (a)
Proof. (1)
d
F (x + h) − F (x)
F (x) = lim
h→0
dx
h
F (x + h) − F (x)
h
=
=
1
h
1
h
Z
Z
x+h
a
x+h
f (x) dx −
f (x) dx
x
By MVT there is c between x and x + h s.t.
1
h
Z
x
x+h
f (x) dx = f (c)
Z
x
f (x) dx
a
6
f (c) approaches f (x) as h → 0. Hence
d
F (x) =
dx
F (x + h) − F (x)
h→0
h
Z x+h
1
= lim
f (x) dx
h→0 h x
= f (x)
lim
Rx
(2) Set G(x) = a f (x) dx then by (1) G(x) is anti-derivative of f (x) on
[a, b]. Since F (x) is also an anti-derivative of f (x) we have
G(x) = F (x) + C
But G(a) =
F (a).
Ra
a
f (x) dx = 0, C = −F (a). Hence G(x) =
y
Rx
a
f (x) dx = F (x) −
y = f (t)
A(x) =
O
Rx
a
f (t) dt
a
x
t
4.4: A′ (x) = f (x)
Z x2
d
cos t dt.
Example 5.2. (1)
dx 1
Z
d
1
(2)
dt.
dx 3+x2 1 + et
Example 5.3. Find derivative of A(x) when h(t) is continuous and u(x), v(x)
are differentiable. Find the derivative of
Z v(x)
A(x) =
h(t) dt
u(x)
sol. Let H(t) be an antiderivative of h(t). Then A(x) = H(v(x)) − H(u(x))
and A′ (x) = h(v(x))v ′ (x) − h(u(x))u′ (x).
Z v(x)
d
h(t) dt = h(v(x))v ′ (x) − h(u(x))u′ (x).
dx u(x)
Z b
Example 5.4.
ex dx = eb − ea .
a
Total area.
7
INDEFINITE INTEGRALS AND SUBSTITUTION
6
Indefinite integrals and substitution
d
dx
So we have
Z
Example 6.1.
chain rule
un+1
n+1
= un
du
dx
un+1
+C
n+1
un du =
R√
R
1 + x2 2xdx = u1/2 du =
d
d
d
F (u(x)) =
F (u) u(x)
dx
du
dx
Z
d
d
F (u) u(x) dx = F (u(x)) + C
du
dx
Z
d
F (u) du = F (u) + C
du
Z
Z
d
d
d
F (u) u(x) dx =
F (u) du
du
dx
du
Z
Z
Proposition 6.2.
f (g(x)) · g′ (x) dx = f (u) du.
Proof. Let F (u) be an anti-derivative of f (u).
d
F (g(x)) = f (g(x)) g′ (x)
dx
Z
Z
′
f (g(x)) g (x) dx = f (u) du
Example 6.3.
Z
π/2
sin x
e
cos x dx =
0
7
Z
1
eu du.
0
Area between curves
If f (x) > 0
Rb
a
f (x) dx is the area defined by
y = f (x),
a≤x≤b
In general, when f (x) ≤ g(x) on [a, b], the area defined by
y = f (x),
is
Rb
a (g(x)
− f (x)) dx.
y = g(x),
a≤x≤b
7
8
6
Application of integrals
1
Volume
Volume: Slice method
Definition 1.1. The volume of a solid with known cross-sectional area A(x)
from a to b is given by
Z b
V =
A(x) dx
a
Example 1.2 (Vol of Pyramid). Suppose cross section A(x) = x2 0 ≤ x ≤ 3,
then
Z 3
x3
x2 dx = |30 = 9
V =
3
0
Cavalieri principle
Solid of revolution: Disk method
The solid generated by rotating a plane region about an axis is called Solid
of revolution The cross sectional area is A(x) = πR(x)2 . For the graph of
y = f (x) is rotated about x-axis, the area is π(f (xi ))2 Hence it is approximated
by
n−1
X
π(f (xi ))2 (xi+1 − xi )
i=0
and the volume is
Example 1.3. y =
√
Z
b
πf (x)2 dx
a
x 0 ≤ x ≤ 4 is rotated about x-axis.
sol.
V =
Z
4
√
π( x)2 dx = 8π
0
9
10
6
y
APPLICATION OF INTEGRALS
Cross section is disk
with area π( radius )2 ,i.e, π(f (x))2
y = f (x
)
a
O
b
x
axis of
rotatio
n
radius:
r = f (x
)
6.1: Slice method
Example 1.4. y =
√
x, 0 ≤ x ≤ 4 is rotated about y = 1.
sol.
V =
Z
4
0
√
π( x − 1)2 dx = 7π/6
Example 1.5. Find the volume of solid obtained when the area {(x, y) | 0 ≤
x ≤ π, 0 ≤ y ≤ sin x} is rotated about x-axis.
Z π
Z π
1 − cos 2x
π2
2
sol. π
sin x dx = π
dx =
.
2
2
0
0
When the graph of x = g(y) on [c, d] is rotated about y-axis the volume is
Z
d
πg(y)2 dy
c
Example 1.6. y = 2/x 1 ≤ y ≤ 4 is rotated about y = 1.
sol.
π
Z
0
4
2
( )2 dy = 3π.
y
2
VOLUME BY CYLINDRICAL SHELLS (SHELL METHOD)
11
Example 1.7. x = y 2 + 1, 1 ≤ x ≤ 3 is rotated about x = 3.
sol.
V
= π
= π
=
1.1
Z
Z
√
2
√
(2 − y 2 )2 dy
√
[4 − 4y 2 + y 4 ]dy
− 2
√
2
− 2
√
64π 2
15
Washer method
When we the disk method, sometimes the cross section is not a circle, instead
a washer of outer radius R and inner radiusr. In this case the volume is
Z b
π(R(x)2 − r(x)2 )dx
V =
a
Example 1.8. The region between y = x2 + 1 and y = −x + 3 is rotated
about x-axis.
sol. Find points of intersection. x2 + 1 = −x + 3, so x = −2, x = 1. Considering graphs,
Z 1
V = π
[(−x + 3)2 − (x2 + 1)2 ]dx
−2
=
2
117π
5
Volume by cylindrical shells (Shell method)
Let a ≥ 0 and f (x) ≥ 0 on [a, b]. The volume of solid by rotating the region
bounded by y = f (x), y = 0, x = a, x = b about y-axis can be obtained by ....
Let P = {x0 , x1 , . . . , xn } be a partition of [a, b]. The volume of shell between
xi , xi+1 is
Vi = Area of bottom cross section -washer × height = π(x2i+1 − x2i )f (xi )
Hence
n−1
X
i=0
π(xi+1 + xi )f (xi )(xi+1 − xi )
12
6
In the limit
Z
a
b
2πxf (x) dx =
Z
APPLICATION OF INTEGRALS
b
a
2π shell radius × shell height = dx.
There is another interpretation: One side the shell has area between 2πxi
and 2πxi+1 . Now the height is f (x̄i ) doe some x̄i ∈ [xi , xi+1 ]. Hence each shell
has volume between 2πxi f (x̄i )∆xi and 2πxi+1 f (x̄i )∆xi . By adding this we get
above relation.
y
y=f
(x )
(nonn
egativ
e)
O
a
∆x i
b
x
6.2: Shell method
√
Example 2.1. {(x, y) | 0 ≤ y ≤ x, 0 ≤ x ≤ 4} about y-axis
Z 4
Z 4
√
sol.
2πx x dx = 2π
x3/2 dx = 128π/5
0
0
Example 2.2. {(x, y) | 0 ≤ y ≤
Z 2
sol.
2πy(4 − y 2 ) dy = 8π.
√
x,
0 ≤ x ≤ 4} is rotated about x-axis
0
Example 2.3. Find the volume of solid rotating the region {(x, y) | 0 ≤ x ≤
π, 0 ≤ y ≤ sin x} about y-axis.
Z π
sol.
2πx sin x dx = 2π [sin x − x cos x]π0 = 2π 2 .
0
3
LENGTH OF PLANE CURVES
13
Similarly if a region x = g(y) c ≤ y ≤ d is rotated about x-axis, then
Z
3
d
2πg(y) dy
c
Length of plane curves
Length of parametric curves
Find the length of a curve given by (x(t), y(t)), [a, b].
y
(x(ti ), y(ti ))
(x(ti+1 ), y(ti+1 ))
x
O
6.3: Sum of line segments
Let P = {t0 , t1 , . . . , tn } be a partition of [a, b]. The lengths is
n−1
Xq
i=0
n−1
X
i=0
s
(x(ti+1 ) − x(ti ))2 + (y(ti+1 ) − y(ti ))2 ,
x(ti+1 ) − x(ti )
ti+1 − ti
2
+
y(ti+1 ) − y(ti )
ti+1 − ti
2
(ti+1 − ti )
(fig 6.3). Thus as kP k → 0
x(ti+1 ) − x(ti )
→ x′ (ti ),
ti+1 − ti
x′ (ti ) → x′ (t),
Hence
y(ti+1 ) − y(ti )
→ y ′ (ti )
ti+1 − ti
y ′ (ti ) → y ′ (t),
Z bp
a
∆ti = ti+1 − ti → dt
(x′ (t))2 + (y ′ (t))2 dt
14
6
APPLICATION OF INTEGRALS
Using Leibniz notation
n q
X
L ≈
(∆xi )2 + (∆yi )2
i=1
n
X
=
i=1
In the limit,
Z
a
b
s
s
dx
dt
Example 3.1. Find the length
sol.
Z
∆xi
∆ti
2
2
+
+
dy
dt
∆yi
∆ti
2
2
∆ti
dt
(x(t), y(t)) = (cos t, sin t), 0 ≤ t ≤ π/2
Z π/2
π/2 p
π
2
2
(− sin t) + (cos t) dt =
dt = .
2
0
0
Example 3.2 (Astroid).
x = cos3 t, y = sin3 t0 ≤ x ≤ 2π
sol. By (6.1)
Z
π/2
0
.
iπ/2 2π
p
2π h
2π cos x 1 + (− sin x)2 dx =
(1 + sin2 x)3/2
=
(23/2 − 1)
3
3
0
Length of a curve y = f (x)
y = f (x)(Fig4.9).
Z bp
a
Example 3.3.
y=
1 + (f ′ (x))2 dx
1 x
(e + e−x ),
2
0≤x≤2
sol. y ′ = 12 (ex + e−x )
Z 2r
0
.
1
1
1 + (ex + e−x )2 dx == (e2 − e−2 )
4
2
4
MOMENT AND CENTER OF MASS
15
y
B
p
A
(∆xk )2 + (∆yk )2
Q
∆yk
y = f (x)
P
O
6.4: P Q =
∆xk
xk+1
xk
a
p
b
x
(∆xk )2 + (∆yk )2 , ∆yk = f (xk+1 ) − f (xk )
Differential formula
ds =
p
dx2 + dy 2
L=
4
Z p
dx2 + dy 2 =
Z r
dy
1 + ( )2 dx =
dx
Z s
1+(
dx 2
) dy
dy
Moment and Center of mass
Moment
Suppose masses m1 , m2 , · · · are placed on a fulcrum at the origin. Each mass
exerts a downward force mk g. This force has tendency of turning the axis
about the origin. This effect is torque. It is measured by multiplying the
mass with the signed distance. The sum is called a system torque
System torque = m1 gx1 + m2 gx2 + m3 gx3 = g(m1 x1 + m2 x2 + m3 x3 )
System torque without g is called moment. We find a special point x̄ about
which the moment is zero:
System torque about x̄ = g
Then
X
(xk − x̄)mk = 0
P
mk xk
x̄ = P
mk
This point is Center of mass
Continuous Mass-Wires or Rods
A thin rod is placed along x-axis. suppose the density is given by δ(x) [a, b].
Find Center of mass x̄
System torque =
X
xk ∆mk
16
6
APPLICATION OF INTEGRALS
Here ∆mk ≈ δ(xk )∆xk Then
P
P
xk δ(xk )∆xk
xk ∆mk
= P
x̄ ≈ P
∆mk
δ(xk )∆xk
In the limit Center of mass x̄ is
Rb
Total mass is M =
Thin flat plates
Rb
a
δ(x) dx
x δ(x) dx
x̄ = Ra b
a δ(x) dx
Density δ center of mass (x̄, ȳ). D ⊂ {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d}
To mimic in one -dim case, we slice the region by a strips parallel to one
of the coordinate axis(y-axis for x̄). Let
x̃ be the center of mass of the vertical strip of mass ∆m
ỹ be the center of mass of the horizontal strip of mass ∆m.
Then M and (x̄, ȳ) are given
M
Z
=
Za
Moments about x-axis Mx =
Z
Moments about y-axis My =
b
ỹ dm
ỹ dm
x̃ dm
My
,
M
x̄ =
ȳ =
Mx
M
Now suppose D is a plane region D Centroid. Then δ is constant. Set
it 1. Then the x̃ is at the mid point over each strip.
Example 4.1. Region surrounded by y = 0, x = 1 and y = x2 . Find Centroid
of D
sol. Here along vertical strip dm = f (x)dx = x2 dx, while along horizontal
√
strip dm = (1 − y)dy. So
My =
Z
1
3
x dx,
Mx =
0
0
M=
x̄ =
R1
0
Z
x3 dx
3
= ,
M
4
Z
1
1
y(1 −
√
y) dy
1
x2 dx = ,
3
0
R1
√
3
0 y(1 − y) dy
ȳ =
= .
M
10
4
MOMENT AND CENTER OF MASS
y
17
mass ∆m = ρ f (x′ )∆x
Center of mass
ȳ
b
O
x̄
(x̄, ȳ)
x
x′
6.5: Vertical strip of mass ∆m
b
(x̃, ỹ)
dm = (1 −
b
√
y)dy =
(x̃, ỹ)
dm = x2 dx
6.6: Horizontal and Vertical strip of mass ∆m
Example 4.2 (Constant density wire). A wire with constant density can be
regarded as a curve. Find center of mass a wire on the following quarter circle.
(x(t), y(t)) = (a cos t, a sin t),
0 ≤ t ≤ π/2
18
6
APPLICATION OF INTEGRALS
sol. If density is 1, the length is same as the mass. Length is
ds = adθ
dm = ds = adθ
ỹ = sin θ
M =a
Z
0
(x̄, ȳ) ,
My =
Z
π/2
π/2 p
2
a cos t
0
p
Z
π/2
dt =
0
(− sin t)2
+
(cos t)2 dt,
Mx =
Z
π/2
0
a2
My
x̄ =
=
M
a2
Mx
=
ȳ =
M
5
(− sin t)2 + (cos t)2 dt = a
R π/2
0
R π/2
0
cos t
sin t
p
p
πa
2
p
a2 sin t (− sin t)2 + (cos t)2 dt
(− sin t)2 + (cos t)2 dt
2a
=
,
aπ/2
π
(− sin t)2 + (cos t)2 dt
2a
=
.
aπ/2
π
Surface area of revolution and Pappus theorem
The area of surface obtained when we rotate the graph of y = f (x) about
x-axis. Let P = {x0 , x1 , . . . , xn } be a partition of [a, b]. The average height is
f (xi )+f (xi−1 )
, slant length is
2
q
So
2π
f (xi ) + f (xi−1 )
2
π (f (xi ) + f (xi−1 ))
Hence
∆x2i + ∆yi2
Z
a
b
s
1+
q
∆x2i + ∆yi2 ,
f (xi ) − f (xi−1 )
xi − xi−1
p
2πf (x) 1 + (f ′ (x))2 dx
2
∆xi
(6.1)
5
SURFACE AREA OF REVOLUTION AND PAPPUS THEOREM 19
On the other hand, if the x = g(y) is rotated about y-axis, the surface area is
Z
b
a
p
2πg(y) 1 + (g′ (y))2 dy
(6.2)
Example 5.1. The line segment x = 1, 0 ≤ y ≤ 1 is revolved about y-axis.
Find the area.
sol. x = 1 − y So dx/dy = −1.
S = 2π
p
Z
1
0
1 + (g′ (y))2 =
√
2 and the area is
√
√
2π(1 − y( y dy = π 2
P
Length
p of P Q:
L = (∆xi )2 + (∆y
y = f (x)
|∆yi |
Q
yi = f (xi )
yi+1 = f
x
x
6.7: Surface area of revolution ∆xi = xi+1 − xi , ∆yi = yi+1 − yi
Example 5.2. Surface area of revolution
y = cos x,
0≤x≤
π
2
is revolved about x-axis
sol. By (6.1)
Z
π/2
0
.
iπ/2 2π
p
2π h
2π cos x 1 + (− sin x)2 dx =
(1 + sin2 x)3/2
=
(23/2 − 1)
3
3
0
20
6
5.1
APPLICATION OF INTEGRALS
Parameterized curve
The length differential of a parameterized curve (f (t), g(t)) is
p
[f ′ (x)]2
+
=
s
dx
dt
2
2πy
s
dx
dt
2
dx
dt
2
[g′ (x)]2 dx
+
dy
dt
2
+
dy
dt
2
dt
(6.4)
+
dy
dt
2
dt
(6.5)
(6.3)
Hence surface area of revolution is
about x-axis S =
Z
b
Z
b
a
about y-axis S =
2πx
a
s
Example 5.3. The circle of radius 1 at (0, 1) is
x = cos t,
0, y = 1 + sin t ≤ t ≤ 2π
Rotated it about x-axis
ds = 1 So
S =
Z
2π
2πy
0
=
Z
2π
dy
dx
Z
b
s
dx
dt
2
+
y 2
dt
dt
2π(1 + sin t)dt = 4π 2
0
Differential form
S=
Z
b
2πy
a
s
1+
2
dx and
Z
b
Z
b
2πx
a
s
1+
dx
dy
2
dy
are often written as
S=
a
or
S=
Z
2πy ds and
2πx ds
a
b
a
2π radius × band width
Z
b
2πρ ds
a
Example 5.4. Find area of surface revolving y = x2 , 0 ≤ x ≤ 1/2.
5
SURFACE AREA OF REVOLUTION AND PAPPUS THEOREM 21
sol. Differential form S =
Rb
a
2πρ ds
dy = 3x2 , ds =
Hence
S =
Z
1/2
2πy ds =
0
=
Z
1/2
0
=
61π
1728
2πx2
p
p
Z
1 + 9x4 dx
1/2
2πy
0
p
1 + 9x4 dx
1 + 9x4 dx
Pappus’s theorem
Theorem 5.5. If a plane region is revolved about a line that does not cut
through the region’s interior, the the vol generated is
2πρA
where A is read of the region, ρ is the distance from the centroid to the line L.
Proof. The idea is to express the volume by shell method. Assume the region
D is between y = c and y = d(d ≥ c ≥ 0). For simplicity, assume x-axis is the
axis of rotation. Then the y-coord of centroid is ȳ = ρ and f (y) is the length
of segment intersecting with D. Then
Rd
yf (y) dy
ρ= c
A
While the volume using the shell method is
Z d
Z d
V =
2πyf (y) dy = 2π
yf (y) dy = 2πρA
c
c
Example 5.6. The area of volume obtained rotating the circle {(x, y) | (x −
b)2 + y 2 = a2 } about y-axis (0 < a < b)
sol. A = πa2 ρ = b. So volume is (2πb) × (πa2 ) = 2π 2 ba2 .
22
6
APPLICATION OF INTEGRALS
y
area A
Centroid (x̄, ȳ)
b
ȳ
x
O
6.8: Volume of solid of revolution about x-axis is 2π ȳA .
Example 5.7 (Centroid√of circular region). Find y-coord of centroid of the
region of semi-circle y = a2 − x2 .
sol. ȳ =
V
2πA
=
4
3π a.
Theorem 5.8 (Pappus’ theorem for surface area). If an arc of a smooth
curve is revolved about a line not cutting through arc, then area of revolution
generated by the
arc is length × the distance traveled by the centroid = 2πρL.
Here ρ is the distance from centroid to the axis.
Proof. Assume the x-axis is the axis of rotation. Then surface area is
Z
S = 2 2πy ds
But the y-coord of centroid of arc is
ȳ =
Hence
So
Z
R
y ds
L
y ds = ȳL
S = 2π ȳL = 2πρL
Example 5.9. The area of surface obtained rotating the circle {(x, y) | (x −
b)2 + y 2 = a2 } about y-axis (0 < a < b)
sol. Length is 2πa centroid is (b, 0). Distance from centroid to axis is b. So
the surface area is (2πb) × (2πa) = 4abπ 2 .
7
Integrals and transcendental
function
In chapter 2 we introduced log function as an inverse of exponential function
ex , where the number e was chosen to satisfy certain slope condition. In this
chapter we introduce an alternative theory for exponential and log. function
1
Logarithm defined as integral
Definition 1.1.
ln x =
Z
x
1
1
dt, (x > 0)
t
From this, we see that
(ln x)′ = 1/x(ln x)′ = 1/ and (ln x)′ = 1/x
Definition 1.2. The number e is defined as the number satisfying
Z e
1
ln e = 1, or
dt = 1
1 t
Derivative of ln x
d
d
ln x =
dx
dx
Hence
By substitution
Z
Z
Z
1
x
1
1
dt =
t
x
1
dt = ln |x| + C
t
f ′ (x)
dx = ln |f (x)| + C
f (x)
23
24
7
(2)
π/2
−π/2
Z
2
2x
dx = ln |u|−1
−5
2−5
x
0
Z 5
4 cos θ
2
dθ =
du
3 + 2 sin θ
1 u
Example 1.3.
Z
INTEGRALS AND TRANSCENDENTAL FUNCTION
(1)
Exponential function as inverse of log function
We define the inverse of ln x by exp(x). Then since ln(e) = 1 we have exp(1) =
e. Since er is also an inverse of ln x, we see er = exp(r).
Exponential Growth and decay
The decay of radioactive material or money earning interests in bank account,
temperature between a cup of hot water and room air it sits, etc follows the
law of exponential change Suppose y(t) denotes some quantity which changes
according to the exponential law: The rate of change of y is proportional to
y.
dy
= ky
dt
with I.C. Then y = Aekt .
Example 1.4. Assume a disease is spreading ”Entero virus”, ”A.I” Let y be
the number of people infected by disease. Assume we cure people as much
as possible. Then dy/dt is proportional to y.(The more people, the more
infected, the more cured) Suppose for each year the number is reduced by
20% and 10,000 people infected today, how many years will it take to reduce
to 1, 000?
sol. y = Aekt , A = 10, 000 Since it is reduced by 0.2 each year, we see
0.8 = ek·1 → k = ln 0.8 < 0
So we have y = 10, 000e(ln 0.8)t we want 10, 000e(ln 0.8)t = 1, 000. So
1
e(ln 0.8)t = 10
. ln(0.8)t = ln(0.1). t = ln(0.1)
ln(0.8) ≈ 10.32 yrs.
Example 1.5 (Half life of a radioactive material). y0 e−kt = 12 y0 . so t = ln 2/k.
Example 1.6 (Carbon 14). It is estimated the half life of Carbon 14 is
5700 yrs. AS wooden artifact was found from an ancient site. This contains carbon 14 about 10% less than the living tree. How old is the site?
k = ln 2/Half lif e = n2/5700. y = y0 e−kt = 0.9y0 So e−kt = 0.9 or
t = −5700 lnln0.9
2 = 866 yrs.
2
RELATIVE RATE OF GROWTH
25
Example 1.7 (Law of Cooling). If H is the temperature of an object and Hs
the surrounding temperature. Then the rate of change(cooling) is proportional
to the temperature difference. Thus
dH
= −k(H − Hs )
dt
Solving
H − Hs = (H0 − Hs )e−kt .
A boiled egg at 98o is put in the sink of 18o to cool down. In 5 min, the egg
was 38o . how much longer will it take to reach 20o ?
sol.
H − 18 = (98 − 18)e−kt ,
H = 18 + 80e−kt
Set H = 18, t = 5. e−5k = 1/4, k = 1/5 ln 1/4 = 0.2 ln 4 = 0.28
H = 18 + 80e−(0.2 ln 4)t
Solving t ≈ 13 min.
2
Relative Rate of Growth
Definition 2.1. Suppose f (x), g(x) are positive for sufficiently large x.
(1) f grows faster than g as x → ∞ if
f (x)
=∞
x→=∞ g(x)
lim
(2) f grows at the same rate as g as x → ∞ if
f (x)
= L, for some postive finite number L
x→=∞ g(x)
lim
Example 2.2.
(1) ex grows faster than x3 as x → ∞
(2) 3x grows faster than 2x as x → ∞
(3) x grows faster than ln x as x → ∞
Definition 2.3. (1) f (x) is smaller order than g(x) as x → ∞ if limx→=∞
0 and write f = o(g)
f (x)
g(x)
=
26
7
INTEGRALS AND TRANSCENDENTAL FUNCTION
(2) f grows at the same order as g as x → ∞ if
lim
x→=∞
f (x)
≤ L, for some postive finite number L
g(x)
then we write f = O(g).
Example 2.4.
(1) ln x = o(x) as x → ∞
(2) x2 = 0(x3 ) g as x → ∞
(3) x + sin x = O(x)
(4) x = O(ex )
3
Hyperbolic function
hyperbolic function
Any f (x) can be written as even part and odd part
f (x) =
f (x) + f (−x) f (x) − f (−x)
+
2
2
|
{z
} |
{z
}
even part
odd part
Hence ex can be written as
ex =
ex + e−x ex − e−x
+
2
2
Definition 3.1 (hyperbolic function).
1
ex + e−x
,
2
ex − e−x
sine sinh x =
,
2
sinh x
ex − e−x
tangent tanh x =
= x
,
cosh x
e + e−x
1
ex + e−x
cotangent coth x =
= x
,
tanh x
e − e−x
1
2
secant sech x =
= x
,
cosh x
e + e−x
1
2
cosecant csch x =
= x
.
sinh x
e − e−x
hyperbolic cosine cosh x =
hyperbolic
hyperbolic
hyperbolic
hyperbolic
hyperbolic
(7.1)
3
HYPERBOLIC FUNCTION
27
y
y = cosh x
y
y = sinh x
y=
ex
2
1
y=
0
−1
−1
y = coth x
e−x
2
y=1
x
0
1
−x
y = −e2
y = tanh x
x
y = −1
y = coth x
y
y
y = cosh x
y = sinh x
y=1
x
0
y = csch x
7.1: hyperbolic functions
Some identities of hyperbolic functions:
Proposition 3.2.
(1) sinh 2x = 2 sinh x cosh x
(2) cosh 2x = cosh2 x + sinh2 x
(3) sinh2 x =
cosh 2x − 1
2
(4) cosh2 x =
cosh 2x + 1
2
(5) cosh2 x − sinh2 x = 1
(6) tanh2 x = 1 − sech2 x
1
sin x cos x
eiθ = cos θ + i sin θ
sin θ cos θ
.
.
e−iθ = cos θ − i sin θ
cos θ =
eiθ + e−iθ
,
2
x
0
y = sech x
sin θ =
eiθ − e−iθ
2i
28
7
INTEGRALS AND TRANSCENDENTAL FUNCTION
(7) coth2 x = 1 + csch2 x
Proposition 3.3.
(1)
du
d
(sinh u) = cosh u
dx
dx
(2)
d
du
(coth u) = − csch2 u
dx
dx
(3)
d
du
(cosh u) = sinh u
dx
dx
(4)
d
du
(sech u) = − sech u tanh u
dx
dx
(5)
d
du
(tanh u) = sech2 u
dx
dx
(6)
d
du
(csch u) = − csch u coth u
dx
dx
Proposition 3.4.
Z
(1)
sinh u du = cosh u + C
(2)
(3)
(4)
(5)
(6)
Z
Z
Z
Z
Z
cosh u du = sinh u + C
sech2 u du = tanh u + C
csch2 u du = − coth u + C
sech u tanh u du = − sech u + C
csch u coth udu = − csch u + C
Example 3.5. (1) The indefinite integral of sinh2 x can be computed just
as that of sin2 x.
Z 1
Z 1
cosh 2x − 1
2
sinh x dx =
dx
2
0
0
1
1 sinh 2x
=
−x
2
2
0
sinh 2 1
=
− .
4
2
3
HYPERBOLIC FUNCTION
29
(2) Using the definition of sinh x
Z ln 2
Z ln 2
Z ln 2
ex − e−x
4ex sinh x dx =
4ex
dx =
(2e2x − 2) dx
2
0
0
0
2x
ln 2
= e − 2x 0
= 3 − 2 ln 2.
Inverse hyperbolic function
The function y = sinh x is defined on (−∞, ∞) having values in (−∞, ∞). So
(inverse hyperbolic sine ) y = sinh−1 x is defined on (−∞, ∞).
The function y = cosh x restricted to x ≥ 0 is 1-1 to [1, ∞). So inverse
y = cosh−1 inverse hyperbolic cosine is defined on [1, ∞).
y = sech x restricted to x ≥ 0 is one-to-one. Hence its inverse y = sech−1 x
is defined on (0, 1]. Meanwhile y = tanh x, y = coth x, y = csch x are oneto-one on (−∞, ∞). Hence their inverses y = tanh−1 x, y = coth−1 x, y =
csch−1 x are defined accordingly. The graphs are as in figure ??
y
y
y = cosh x
y = sinh x
y=x
y=x
y = sinh−1 x
y = cosh−1 x
x
x
y
y
y = sech−1 x
y=x
y = csch x
y=x
x
y = csch−1 x
y = sech x, x ≥ 0
x
y y = tanh−1 x
y=x
y
y = coth x
y = tanh x
x
y=x
x
y = coth−1 x
7.2: Inverse hyperbolic functions
Inverse hyperbolic functions can be represented using log functions.
Proposition 3.6.
(1) sinh−1 x = ln x +
p
x2 + 1 ,
−∞ < x < ∞
30
7
INTEGRALS AND TRANSCENDENTAL FUNCTION
(2) cosh−1 x = ln x +
p
x2 − 1 ,
x≥1
1 1+x
ln
, |x| < 1
2 1−x
√
1 + 1 − x2
−1
(4) sech x = ln
, 0<x≤1
x
√
1
1 + x2
−1
(5) csch x = ln
+
, x 6= 0
x
|x|
(3) tanh−1 x =
(6) coth−1 x =
1 x+1
ln
,
2 x−1
|x| > 1
Proof. We prove the formula for sinh−1 x.
y = sinh x =
ex − e−x
ex − e−x
,
2
= 2y,
e2x − 2yex − 1 = 0,
p
p
ex = y + y 2 + 1. (Since y − y 2 + 1 is negative, we drop it.)
p
√
Hence x = ln(y + y 2 + 1), or y = ln(x + x2 + 1) is the sinh−1 x.
Proposition 3.7.
(1) sech−1 x = cosh−1
1
x
(2) csch−1 x = sinh−1
1
x
(3) coth−1 x = tanh−1
1
x
Derivatives of inverse hyperbolic functions
Proposition 3.8.
(1)
d(sinh−1 u)
1
du
=√
2
dx
1 + u dx
(2)
d(cosh−1 u)
1
du
=√
,
2
dx
u − 1 dx
(3)
(4)
d(tanh−1 u)
1 du
=
,
dx
1 − u2 dx
d(coth−1 u)
1 du
=
,
dx
1 − u2 dx
u>1
|u| < 1
|u| > 1
3
(5)
(6)
HYPERBOLIC FUNCTION
d(sech−1 u)
−du/dx
,
= √
dx
u 1 − u2
31
0<u<1
d(csch−1 u)
−du/dx
√
=
,
dx
|u| 1 + u2
u 6= 0
Proposition 3.9.
Z
du
√
(1)
= sinh−1 u + C
2
1+u
Z
du
√
(2)
= cosh−1 u + C, u > 1
u2 − 1
(
Z
tanh−1 u + C, if |u| < 1,
du
(3)
=
1 − u2
coth−1 u + C, if |u| > 1
Z
du
1
−1
−1
√
= − sech |u| + C = − cosh
+C
(4)
2
|u|
u 1−u
Z
du
1
−1
−1
√
(5)
= − csch |u| + C = − sinh
+C
2
|u|
u 1+u
Example 3.10.
Z
2 dx
√
3 + 4x2
=
=
=
d2 y
dx2
Z
√
du
√
, u = 2x, a = 3
a2 + u2
u
sinh−1 ( ) + C
a
2x
sinh−1 ( √ ) + C
3
r
dy 2
=a 1+
dx
(7.2)
32
7
INTEGRALS AND TRANSCENDENTAL FUNCTION
8
Techniques of Integration
1
Basic integration formulas
Basic integration formulas
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
xn dx =
xn+1
+C
n+1
(n 6= −1)
1
dx = ln x + C
x
cos x dx = sin x + C
sin x dx = − cos x + C
tan x dx = − ln | cos x| + C
sec2 x dx = tan x + C
csc2 x dx = − cot x + C
sec x tan x dx = sec x + C
csc x cot x dx = − csc x + C
√
1
dx = sin−1 x + C
2
1−x
1
dx = tan−1 x + C
1 + x2
33
34
(12)
(13)
(14)
(15)
(16)
(17)
(18)
8
Z
Z
Z
Z
Z
Z
Z
TECHNIQUES OF INTEGRATION
ex dx = ex + C
ax dx =
ax
+C
ln a
(a > 0, a 6= 1)
cosh x dx = sinh x + C
sinh x dx = cosh x + C
tanh x dx = ln cosh x + C
coth x dx = ln | sinh x| + C
sech x dx = tan−1 sinh x + C
Z
x csch x dx = ln tanh + C
2
Z
2x − 9
√
Example 1.1. (1)
dx
2
x − 9x + 1
Z
dx
√
(2)
dx
8x − x2
(19)
sol. 1.
Z
2x − 9
√
dx =
2
x − 9x + 1
Z
du
√
u
2.
Z
Example 1.2. Find
dx
√
dx =
8 − x2
Z
dx
p
16 − (x − 4)2
Z
du
√
=
2
a − u2
u
= sin−1 ( ) + C
a
Z
(sec x + tan x)2 dx
dx
1
BASIC INTEGRATION FORMULAS
35
sol.
(sec x+tan x)2 = sec2 x+2 sec x tan x+tan2 x = 2 sec2 x−1+2 sec x tan x
Hence
Z
Z
2
(sec x + tan x) dx =
= 2 sec2 x − 1 + 2 sec x tan x dx
= 2 tan +2 sec x − x + C.
Example 1.3.
Z
π/4 √
1 + cos 4x dx
0
Using
cos2 θ =
Example 1.4. Find
Z
The idea is to multiply sec x + tan x
Z
Z
sec x dx =
Z
=
Z
=
1 + cos 2θ
2
sec x dx
both the numerator and denominator:
sec x + tan x
dx
sec x + tan x
sec2 x + secx tan x
dx
sec x + tan x
du
u
= ln | sec x + tan x| + C
Similarly, we obtain
Z
sec x ·
csc x dx = − ln | csc x + cot x| + C
Integral tables
(1)
Z
u
1
1
du = tan−1
a2 + u2
a
a
(a > 0)
(2)
Z
√
1
u
du = sin−1
2
a
−u
(a > 0)
a2
36
8
Example 1.5. For
3dx = du, and
R
Z
2
TECHNIQUES OF INTEGRATION
1/(4 + 9x2 ) dx, use substitution first. Let 3x = u then
Z
1
1
1
dx =
du
2
2
4 + 9x
3
2 + u2
1 1
−1 u
=
tan
+C
3 2
2
1
3
= tan−1 x + C
6
2
Integration by Parts
integral by parts
d
dv
du
(uv) = u
+v
dx
dx
dx
Integrating w.r.t x
Z
Z
dv
du
uv = u dx + v
dx
dx
dx
Z
Z
= u du + v dv
Thus
Proposition 2.1 (Integration by Parts I).
Z
Z
u dv = uv − v du
(8.1)
Proposition 2.2 (Integration by Parts II).
Z
Z
f (x)g′ (x) dx = f (x)g(x) − f ′ (x)g(x) dx
Proposition 2.3 (Definite integral).
Z
a
b
′
f (x)g (x) dx =
[f (x)g(x)]ba
Example 2.4. Find the following
Z π
x sin x dx
(1)
0
(2)
Z
ln x dx
−
Z
a
b
f ′ (x)g(x) dx
(8.2)
2
INTEGRATION BY PARTS
37
sol. (1) Let u = x, dv = sin x dx. Then du = dx, v = − cos x. (Fig ??)
Z
π
0
x sin x dx = [x(− cos x)]π0 −
= π + [sin x]π0
Z
π
(− cos x) dx
0
= π.
(2) Let u = ln x, dv = dx. Then we have du = (1/x)dx, v = x.
Z
Z
1
ln x dx = (ln x)x − x · dx
x
= x ln x − x + C.
y
y = x sin x
1
π x
0
8.1:
Repeated integration by parts
f and its derivative
x2
2x
2
0
Example 2.5. Find
Z
x2 sin x dx..
(+)
(−)
(+)
g and its integral
ex
ex
ex
ex
38
8
TECHNIQUES OF INTEGRATION
sol. Let u = x2 , dv = sin x dx. Then du = 2xdx, v = − cos x and hence
Z
Z
x2 sin x dx = x2 (− cos x) − (− cos x)2x dx
Z
= −x2 cos x + 2x cos x dx
Again, set u = 2x, dv = cos x dx. Then du = 2 dx, v = sin x.
Z
Z
2
2
x sin x dx = −x cos x + 2x sin x − 2 sin x dx
= −x2 cos x + 2x sin x + 2 cos x + C.
Example 2.6. Find
Z
x2 ex dx.
sol. f (x) = x2 , g(x) = ex
f and its derivative
x3
3x2
6x
6
0
Example 2.7. Find
Z
(+)
(−)
(+)
(-)
g and its integral
sin x
− cos x
− sin x
cos x
sin x
x3 sin x dx.
Hence
Z
x3 sin x dx = −x3 cos x + 3x2 sin x + 6x cos x − 6 sin x + C
Example 2.8. Find
Z
ex sin x dx.
2
INTEGRATION BY PARTS
39
sol. If u = ex , dv = sin x dx, then du = ex dx, v = − cos x.
Z
Z
ex sin x dx = ex (− cos x) − ex (− cos x) dx
Z
= −ex cos x + ex cos x dx
Again let u = ex , dv = cos x dx so that du = ex dx, v = sin x.
Z
Z
x
x
e sin x dx = −e cos x + ex cos x dx
Z
x
x
= −e cos x + e sin x − ex sin x dx.
Solving this for
R
ex sin x dx we obtain
Z
1
ex sin x dx = ex (sin x − cos x) + C
2
Reduction formula
Example 2.9. Express
sol.
Z
cos
n−1
R
cosn x dx in terms of low power of cos x.
x cos x dx = cos
n−1
sin2 x cosn−2 x dx
Z
(1 − cos2 x) cosn−2 x dx
Z
Z
n−1
2
= cos
sin x + (n − 1) cos x dx − (n − 1) cosn x dx
= cos
n−1
sin x + (n − 1)
Z
sin x + (n − 1)
So
n
Z
cos
n−1
x cos x dx = cos
n−1
sin x + (n − 1)
Z
cos2 x dx
Example 2.10. Prove
Z
x(a2 ± x2 )
2na2
(a ± x ) dx =
+
2n + 1
2n + 1
2
2 n
Z
(a2 ± x2 )n−1 dx
1
(n 6= − )
2
40
8
TECHNIQUES OF INTEGRATION
sol. Integration by parts
Z
Z
(a2 ± x2 )n dx = x(a2 ± x2 )n − x · n(a2 ± x2 )n−1 (±2x) dx
Z
= x(a2 ± x2 )n − 2n(a2 ± x2 )n−1 (a2 ± x2 − a2 ) dx
Z
= x(a2 ± x2 )n − 2n (a2 ± x2 )n dx
Z
2
(a2 ± x2 )n−1 dx.
+ 2na
If n 6= −1/2,
Z
Z
x(a2 ± x2 )n
2na2
2
2 n
+
(a2 ± x2 )n−1 dx
(a ± x ) dx =
2n + 1
2n + 1
3
Integration of Rational functions by partial
fraction
When p(x), q(x) are rational functions
p(x)
r(x)
= Q(x) +
q(x)
q(x)
deg of r(x) less than deg of q(x) Rhus, the fraction must be proper;
Distinct linear factors
α1 , . . . , αr are distinct and p(x) is polynomial of degree of is less than r. Then
p(x)
A1
Ar
=
+ ··· +
(x − α1 ) · · · (x − αr )
x − α1
x − αr
Here Ai are obtained by method of undetermined coefficients.
Z
r
X
dx
=
Ai ln |x − αi | + C
(x − α1 ) · · · (x − αr )
Example 3.1. Find
i=1
Z
x+1
dx.
x(x + 2)
(8.3)
3
INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTION41
sol.
x+1
1
1
=
+
x(x + 2)
2x 2(x + 2)
Z 1
1
1
x+1
dx =
+
dx
x(x + 2)
2
x x+2
1
= ln |x(x + 2)| + C.
2
Z
Example 3.2. Find
Z
2x + 1
dx.
x3 − x
sol. Since x3 − x = x(x − 1)(x + 1)
2x + 1
A
B
C
= +
+
x3 − x
x
x−1 x+1
Solving for A, B, C we getA = −1, B = 3/2, C = −1/2. Hence
Z
Z 2x + 1
−1
3/2
−1/2
dx =
+
+
dx
x3 − x
x
x−1 x+1
1
3
= − ln |x| + ln |x − 1| − ln |x + 1| + C.
2
2
Repeated linear factor
deg of p(x) is less than r
p(x)
A1
A2
Ar
=
+
+ ··· +
r
2
(x − α)
x − α (x − α)
(x − α)r
Find A1 , A2 , . . . , Ar
Z
Z p(x)
A1
A2
Ar
dx
dx =
+
+ ··· +
(x − α)r
x − α (x − α)2
(x − α)r
A2
(r − 1)Ar
= A1 ln |x − α| −
− ··· −
+C
x−α
(x − α)r−1
Z
x2
Example 3.3. Find
dx.
(x − 2)3
42
8
TECHNIQUES OF INTEGRATION
sol. Since x2 = (x − 2)2 + 4(x − 2) + 4,
1
4
4
x2
=
+
+
3
2
(x − 2)
x − 2 (x − 2)
(x − 2)3
Hence
Z
Z 1
4
4
+
+
dx
x − 2 (x − 2)2 (x − 2)3
4
8
= ln |x − 2| −
−
+ C.
x − 2 (x − 2)2
x2
dx =
(x − 2)3
irreducible quadratic factor
Suppose x2 + β1 x + γ1 , . . . , x2 + βr x + γr are distinct quadratic factor without
having real roots. Suppose p(x) is polynomial of degree less than 2r. Then we
can set
r
X Bi x + Ci
p(x)
=
(x2 + β1 x + γ1 ) · · · (x2 + βr x + γr )
x2 + βi x + γi
i=1
for some B1 , . . . , Br , C1 , . . . , Cr . Set
Z
r
X
p(x)
dx
=
(x2 + β1 x + γ1 ) · · · (x2 + βr x + γr )
i=1
Z
Bi x + Ci
dx
+ βi x + γi
x2
Di = Ci − Bi βi /2
Bi
(2x + βi ) + Di
2
Bi 2
=
(x + βi x + γi )′ + Di
2
Bi x + Ci =
So that
Z
Bi (x2 + βi x + γi )′
Di
+ 2
dx
2 x2 + βi x + γi
x + βi x + γi
Z
Bi
Di
=
ln(x2 + βi x + γi ) +
dx
2
2
x + βi x + γi
Bi x + Ci
dx =
2
x + βi x + γi
Z Use Di /(x2 + βi x + γi )
Z
u2
du
1
u
= tan−1 + C
2
+a
a
a
3
INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTION43
Example 3.4. Find
Z
x4
2x
dx
+ x2 + 1
sol. Since x4 + x2 + 1 = (x2 − x + 1)(x2 + x + 1), we obtain
x4
2x
B1 x + C1
B2 x + C2
= 2
+ 2
2
+x +1
x −x+1 x +x+1
2
2
with
√ B12 = B2 = 0, C1 = 1, C2 = −1. Since x ± x + 1 = (x ± 1/2) +
( 3/2) we see
Z
2x
dx
4
x + x2 + 1
Z 1
1
√
√
dx
=
−
(x − 1/2)2 + ( 3/2)2
(x + 1/2)2 + ( 3/2)2
2
2x − 1
2x + 1
=√
tan−1 √
− tan−1 √
+ C.
3
3
3
Repeated irreducible quadratic factor
Suppose p(x) is polynomial of degree less than 2r, and x2 + βx + γ does not
have real roots. Then we set
(x2
p(x)
B1 x + C1
B2 x + C2
Br x + Cr
= 2
+ 2
+ ··· + 2
r
2
+ βx + γ)
x + βx + γ (x + βx + γ)
(x + βx + γ)r
for some B1 , B2 , . . . , Br , C1 , C2 , . . . , Cr . Then
Z
p(x)
dx
(x2 + βx + γ)r
Z B1 x + C1
B2 x + C2
Br x + Cr
=
+
+ ··· + 2
dx
x2 + βx + γ (x2 + βx + γ)2
(x + βx + γ)r
with Di = Ci − Bi β/2
Bi
(2x + β) + Di
2
Bi 2
=
(x + βx + γ)′ + Di
2
Bi x + Ci =
we see
Z
Z Bi x + Ci
Bi (x2 + βx + γ)′
Di
dx
=
+
dx
(x2 + βx + γ)i
2 (x2 + βx + γ)i (x2 + βx + γ)i
Z
Bi
Di
=−
+
dx
2
i−1
2
2(i − 1)(x + βx + γ)
(x + βx + γ)i
44
8
TECHNIQUES OF INTEGRATION
For the integral of Di /(x2 + βx + γ)i (i ≥ 2) Use the recurrence relation
Z
Z
u
2i − 3
du
du
=
+
2
2
i
2
2
2
i−1
2
2
(u + a )
a (2i − 2)(u + a )
a (2i − 2)
(u + a2 )i−1
Z 4
x + 2x3 + 5x2 + 6
Example 3.5. Find
dx.
(x2 + 2)3
sol.
A1 x + B1 A2 x + B2 A3 x + B3
x4 + 2x3 + 5x2 + 6
=
+ 2
+ 2
2
3
(x + 2)
x2 + 2
(x + 2)2
(x + 2)3
Multiply (x2 + 2)3 to see
x4 + 2x3 + 5x2 + 6 = A1 x5 + B1 x4 + (4A1 + A2 )x3 + (4B1 + B2 )x2
+ (4A1 + 2A2 + A3 )x + 4B1 + 2B2 + B3
Solving we get A1 = 0, A2 = 2, A3 = −2, B1 = 1, B2 = 1, B3 = 0.
Hence the integral is
x4 + 2x3 + 5x2 + 6
1
2x + 1
−4x
= 2
+ 2
+ 2
2
3
2
(x + 2)
x + 2 (x + 2)
(x + 2)3
Z
x4 + 2x3 + 5x2 + 6
dx
(x2 + 2)3
Z
Z
Z
Z
dx
2x
1
−4x
=
+
dx +
dx +
dx
x2 + 2
(x2 + 2)2
(x2 + 2)2
(x2 + 2)3
Z
x
1
x
1
1
1
1
+
+
dx + 2
= √ tan−1 √ − 2
2
2
x +2
(x + 2)2
2
2 x + 2 4(x + 2) 4
5
x
1
x−4
= √ tan−1 √ +
+ 2
+ C.
2
4 2
2 4(x + 2) (x + 2)2
Heaviside cover up method for linear factors
Example 3.6.
x2 + 1
A
B
C
=
+
+
(x − 1)(x − 2)(x − 3)
x−1 x−2 x−3
Here
A=
(1)2 + 1
(x − 1)(1 − 2)(1 − 3)
| {z }
cover
3
INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTION45
B=
(2)2 + 1
(2 − 1) (x − 2)
B=
cover
(2 − 3)
=
(3)2 + 1
(3 − 1)(3 − 2) (x − 3)
5
= −5
(1)(−1)
=
10
=5
(2)(1)
cover
Example 3.7. Do the same with
Z
x+4
x(x − 2)(x + 5)
sol. Note
x+4
A
B
C
= +
+
x(x − 2)(x + 5)
x
x−2 x+5
A
B
C
x+4
= +
+
A =
x(x − 2)(x + 5)
x
x−2 x+5
B =
C =
0+4
x (0 − 2)(0 + 5)
2+4
2 (x − 2) (2 + 5)
(−5)(−5 − 2) (x + 5)
Using differentiation-repeated factors
Example 3.8.
x−1
A
B
C
=
+
+
3
2
(x + 1)
x + 1 (x + 1)
(x + 1)3
Write
x − 1 = A(x + 1)2 + B(x + 1) + C
Substitute x = −1. Then take derivative
1 = 2A(x + 1) + B
and Substitute x = −1. to get B etc.
Example 3.9. Do the same with
Z
x+4
x(x − 2)(x + 5)
−5 + 4
=−
1
35
46
8
sol. Note
TECHNIQUES OF INTEGRATION
A
B
C
x+4
= +
+
x(x − 2)(x + 5)
x
x−2 x+5
x+4
A
B
C
= +
+
A =
x(x − 2)(x + 5)
x
x−2 x+5
B =
C =
4
0+4
x (0 − 2)(0 + 5)
2+4
2 (x − 2) (2 + 5)
−5 + 4
(−5)(−5 − 2) (x + 5)
=−
Integration of Trigonometric function
Products of powers of Sines and Cosines
Integral of sinm x cosn x
(1) If m is odd, then set m = 2k + 1 and use sin2 x = 1 − cos2 x sin x dx =
−d(cos x) to transform it to
Z
Z
sin2k+1 x cosn x dx = − (1 − cos2 x)k cosn x d(cos x)
(2) If n is odd n = 2k + 1, use cos2 x = 1 − sin2 x cos x dx = d(sin x) to
obtain
Z
Z
m
2k+1
sin x cos
x dx = sinm x(1 − sin2 x)k d(sin x)
(3) If both m, n are even, use sin2 x = (1 − cos 2x)/2, cos2 x = (1 + cos 2x)/2
lower the degree and repeat the previous technique
Z
Example 4.1. Find
sin5 x dx.
sol.
Z
sin5 x dx = −
Z
(1 − cos2 x)2 d(cos x)
Z
= − (1 − 2 cos2 x + cos4 x) d(cos x)
1
2
= − cos5 x + cos3 x − cos x + C.
5
3
1
35
4
INTEGRATION OF TRIGONOMETRIC FUNCTION
Example 4.2. Find
sol.
Z
Z
sin2 x cos3 x dx.
sin2 x cos3 x dx =
Z
sin2 x(1 − sin2 x) d(sin x)
1
1
= − sin5 x + sin3 x + C.
5
3
Example 4.3. Find
sol.
Z
47
Z
sin4 x cos2 x dx..
Z 1 − cos 2x 2 1 + cos 2x
sin4 x cos2 x dx =
dx
2
2
Z
1
=
1 − 2 cos 2x + cos2 2x (1 + cos 2x) dx
8Z
1
=
1 − cos 2x − cos2 2x + cos3 2x dx
8Z 1
1 + cos 4x
2
=
1 − cos 2x −
+ (1 − sin 2x) cos 2x dx
8 Z
2
1
=
1 − cos 4x − sin2 2x · 2 cos 2x dx
16 1
1
1
3
=
x − sin 4x − sin 2x + C.
16
4
3
Integral of
√
1 ± sin ax,
√
1 ± cos ax
Use the double angle formula.
sin 2A = 2 sin A cos A
cos 2A = 2 cos2 A − 1
= 1 − 2 sin2 A
Change the form 1 ± sin ax, 1 ± cos ax to complete swquare.
Example 4.4. Find
Z
0
π
√
1 − sin x dx.
48
8
TECHNIQUES OF INTEGRATION
sol. Since 1 − sin x = 1 − 2 sin(x/2) cos(x/2) = (sin(x/2) − cos(x/2))2
Z π
Z π
√
x
x 1 − sin x dx =
sin − cos dx
2
2
0
0
Z π Z π/2 x
x
x
x
+
dx
=
cos − sin
sin − cos
2
2
2
2
0
π/2
h
x
x iπ/2 h
x iπ
x
= 2 sin + 2 cos
+ −2 cos − 2 sin
2
2 0
2
2 π/2
√
√
√
√
= ( 2 + 2 − 2) + (−2 + 2 + 2)
√
= 4( 2 − 1).
Example 4.5. Find
Z
π/2 √
1 + cos 2x dx.
0
sol. 1 + cos 2x = 2 cos2 x ,
Z
0
π/2 √
√ Z π/2
1 + cos 2x = 2
|cos x| dx
0
√
π/2
= 2 [sin x]0
√
= 2.
Tangent and secant
Recall 1 + tan2 x = sec2 x d/dx tan x sec2 x, d/dx sec x = sec x tan x.
Z
Example 4.6.
sec x dx.
sol. Multiply sec x + tan x.
Z
Z
sec x(sec x + tan x)
sec x dx =
dx
sec x + tan x
Z
(sec x + tan x)′
=
dx
sec x + tan x
= ln | sec x + tan x| + C.
4
INTEGRATION OF TRIGONOMETRIC FUNCTION
Example 4.7.
Z
49
tan2 x sec x dx.
sol. RSince tan2 x sec x = (sec2 x − 1) sec x = sec3 x − sec x, we can find
sec3 x dx. Let u = sec x, dv = sec2 x dx then v = tan x, du =
sec x tan x dx, we have
Z
Z
3
sec x dx = sec x tan x − (tan x) sec x tan x dx
Z
= sec x tan x − (sec2 x − 1) sec x dx
Z
Z
= sec x tan x − sec3 x dx + sec x dx
we obtain
Hence
Z
Example 4.8.
Z
1
1
sec x dx = sec x tan x +
2
2
3
tan2 x sec x dx =
Z
sec3 x dx −
Z
Z
sec x dx
sec x dx
Z
1
1
= sec x tan x −
sec x dx
2
2
1
1
= sec x tan x − ln | sec x + tan x| + C.
2
2
Z
tan6 x dx.
50
8
TECHNIQUES OF INTEGRATION
sol. Since tan2 x = sec2 x − 1
Z
Z
6
tan x dx = tan4 x(sec2 x − 1) dx
Z
Z
4
2
= tan x sec x dx − tan4 x dx
Z
Z
4
2
= tan x sec x dx − tan2 x(sec2 x − 1) dx
Z
Z
Z
= tan4 x sec2 x dx − tan2 x sec2 x dx + tan2 x dx
Z
Z
Z
= tan4 x sec2 x dx − tan2 x sec2 x dx + (sec2 x − 1) dx
=
1
1
tan5 x − tan3 x + tan x − x + C.
5
3
Remark 4.9. For cot x or csc x, use 1 + cot2 x = csc2 x, d/dx cot x = − csc2 x,
d/dx csc x = − csc x cot x.
sin mx sin nx, sin mx cos nx, cos mx cos nx
Addition formula:
sin(A + B) = sin A cos B + cos A sin B
sin(A − B) = sin A cos B − cos A sin B
cos(A + B) = cos A cos B − sin A sin B
cos(A − B) = cos A cos B + sin A sin B
From these we get(with A = mx, B = nx)
1
[cos(m − n)x − cos(m + n)x]
2
1
sin mx cos nx = [sin(m − n)x + sin(m + n)x]
2
1
cos mx cos nx = [cos(m − n)x + cos(m + n)x]
2
sin mx sin nx =
Example 4.10.
Z
π/6
sin 4x sin 3x dx.
0
5
TRIG SUBSTITUTION
√
a2 + x2
51
a
x
x
x
√
a
x2 − a2
a
a2 − x2
x = a sin θ
x = a tan θ
√
x = a sec θ
8.2: trig substitution
sol. Figure ??
Z
5
π/6
0
1
sin 4x sin 3x dx =
2
Z
π/6
(cos x − cos 7x) dx
0
π/6
2
1
1
= .
sin x − sin 7x
=
2
7
7
0
Trig Substitution
Quadratic term
a2 − u2 , a2 + u2 u2 − a2 .
Substitute by u = a sin θ, u = a tan θ, u = a sec θ
a2 − u2 = a2 − a2 sin2 θ = a2 (1 − sin2 θ) = a2 cos2 θ
2
2
2
2
2
2
2
2
2
2
2
(8.5)
2
(8.6)
a + u = a + a tan θ = a (1 + tan θ) = a sec θ
2
2
2
2
2
(8.4)
2
u − a = a sec θ − a = a (sec θ − 1) = a tan θ
(1) u = a sin θ is defined on −π/2 ≤ θ ≤ π/2.
(2) u = a tan θ θ = tan−1 (u/a) on −π/2 < θ < π/2 .
(3) u = a sec θ θ = sec−1 (u/a) Since |u| ≥ a 0 ≤ θ < π/2 (if u ≥ a), or
π/2 < θ ≤ π (if u ≤ −a )
Example 5.1.
Z
a2
du
.
+ u2
52
8
TECHNIQUES OF INTEGRATION
sol. Use substitution u = a tan θ, du = a sec2 θ dθ
Z
Z
a sec2 θ dθ
du
=
a2 + u2
a2 sec2 θ
Z
dθ
=
a
1
= ·θ+C
a
1
u
= tan−1 + C.
a
a
Example 5.2. Find
Z p
a2 − u2 du. (a > 0)
sol. Use u = a sin θ, du = a cos θ dθ
Z p
Z
2
2
a − u du = a cos θ · a cos θ dθ
Z
a2
=
(1 + cos 2θ) dθ
2
a2
sin 2θ
=
θ+
+C
2
2
a2
= (θ + sin θ cos θ) + C
2
!
r
2
a2
u
u
u
=
sin−1 +
1− 2 +C
2
a a
a
u 1 p
a2
sin−1 + u a2 − u2 + C.
=
2
a 2
Example 5.3. Find
Z
√
du
. (|u| > a > 0)
u2 − a2
5
TRIG SUBSTITUTION
53
sol. Let u = a sec θ
u2 − a2 = a2 (sec2 θ − 1)
= a2 tan2 θ,
du = a sec θ tan θ dθ
Then
Z
du
√
=
2
u − a2
=
=
=
Z













a sec θ tan θ dθ
a| tan θ|
Z
sec θ dθ (0 < θ < π/2)
Z
− sec θ dθ (π/2 < θ < π)
ln | sec θ + tan θ| + C (0 < θ < π/2)
− ln | sec θ + tan θ| + C (π/2 < θ < π)
u √u2 − a2 ln +
+ C (u > a)
a
a


u √u2 − a2 



+ C (u < −a).
 − ln a −
a
On the other hand,
u √u2 − a2 p
ln +
= ln u + u2 − a2 − ln a
a
a
Hence
u √u2 − a2 a
√
− ln −
= ln a
a
u − u2 − a2 √
a(u + u2 − a2 )
√
√
= ln (u − u2 − a2 )(u + u2 − a2 ) a(u + √u2 − a2 ) = ln a2
u + √u2 − a2 = ln a
p
= ln u + u2 − a2 − ln a
Z
√
p
du
= ln u + u2 − a2 + C ′
2
2
u −a
54
8
Example 5.4.
Z
√
TECHNIQUES OF INTEGRATION
dx
.
x2 + 9
sol. Let x = 3 tan θ (−π/2 < θ < π/2), dx = 3 sec2 θ dθ,
Z
Z
3 sec2 θ
dx
√
=
dθ
x2 + 9 Z 3 sec θ
=
sec θ dθ
= ln |sec θ + tan θ| + C
r
x 2
x
= ln +1+ +C
3
3
p
= ln x + x2 + 9 + C.
Involving ax2 + bx + c– Completing square
px2 + qx + r (p, q 6= 0) Use u = x + q/(2p) to get = px2 + qx + rp(u2 ± a2 )
Z p
Example 5.5. Find
2x − x2 dx.
sol. Since 2x − x2 = 1 − (x − 1)2 u = x − 1 we have ?? a = 1
Z p
Example 5.6.
Z
2x
− x2 dx
Z p
=
1 − u2 du
1
1 p
sin−1 u + u 1 − u2 + C
2
2
p
1
1
−1
= sin (x − 1) + (x − 1) 2x − x2 + C.
2
2
=
x2
dx
.
+x+1
8
TABLES
55
√
sol. x2 + x + 1 = (x + 1/2)2 + 3/4 u = x + 1/2 a = 3/2
Z
Z
du
dx
=
x2 + x + 1
u2 + 3/4
2
2u
= √ tan−1 √ + C
3
3
2
2x
+1
= √ tan−1 √
+ C.
3
3
6
Tables
7
Numerical Integration
8
Improper Integral
Improper Integral
√
Example 8.1. Find the area surrounded by y = 1/ x, x-axis, y-axis, x =
1(fig ??).
y
y = √1
x
0
1
8.3:
56
8
TECHNIQUES OF INTEGRATION
√
sol. Function 1/ x is not defined at x = 0. So one cannot write the region
as
Z 1
dx
√
x
0
But we can use limit such as
( )
Z
= lim
ε→0+
1
ε
h
dx
√
x
i1
= lim 2x1/2
ε
ε→0+
= lim 2 − 2ε1/2
ε→0+
= 2.
Definition 8.2 (Improper integral).
(1)
(2) When a = −∞ or b = ∞, the integral
Z
b
f (x) dx improper integral
a
Computation of Improper integral
y
y
y = f (x)
0
a
y = f (x)
ub
x
0
a
u
x
8.4: Improper integral on [a, b)
Definition 8.3 (Convergence of Improper integral).
(1) Suppose f (x) is integrable on all closed subinterval of [a, b) and we have
either lim f (x) = ±∞ or b = ∞. If the limite
x→b−
L = lim
u→b−
exists
Z
Z
u
f (x) dx
a
b
f (x) dx
a
(8.7)
8
IMPROPER INTEGRAL
57
lim f (x) = ±∞ a = −∞
x→a+
L = lim
ℓ→a+
Z
b
f (x) dx
(8.8)
ℓ
we say the Improper integral
Z
b
f (x) dx
a
converges. Otherwise diverges.
y
y
y = f (x)
y = f (x)
x
al
0
x
b
l
8.5: Improper integral on (a, b]
Example 8.4.
Z
1
−1
√
1
dx.
1 − x2
y
1
x
−1
0
8.6:
1
0
b
58
8
TECHNIQUES OF INTEGRATION
sol. We distinguish two case: (−1, 0] and [0, 1).
Z
0
−1
1
√
dx = lim
ℓ→−1+
1 − x2
Z
1
dx
1 − x2
ℓ
0
= lim sin−1 x ℓ
ℓ→−1+
−1
Z
= − sin
π
=
2
1
0
0
√
(−1)
Z u
1
1
√
√
dx = lim
dx
2
−
u→1
1−x
1 − x2
0 −1 u
= lim sin x 0
u→1−
−1
= sin
π
=
2
1
Hence
Z
1
1
√
dx =
1 − x2
−1
Example 8.5.
Z
2
Z
0
−1
1
√
dx +
1 − x2
Z
1
√
0
1
dx = π.
1 − x2
dx
.
(x − 1)4/3
0
sol. The funciton 1/(x − 1)4/3 is not defined at x = 1. Hence we separate
Z
2
0
dx
=
(x − 1)4/3
Z
1
0
Z
1
0
dx
+
(x − 1)4/3
dx
= lim
(x − 1)4/3
u→1−
= lim
u→1−
Z
h
u
0
Z
2
1
dx
(x − 1)4/3
dx
(x − 1)4/3
−3(x − 1)−1/3
iu
0
3
= lim −
−3
−
u→1
(u − 1)1/3
=∞
Since
Z 2
1
Z
1
0
dx
(x − 1)4/3
dx
.
(x − 1)4/3
diverges the integral diverges regardless of
8
IMPROPER INTEGRAL
59
y
y=
1
1
4
(x − 1) 3
x
0
1
2
8.7:
The function 1/xp
The integral of 1/xp on (0, 1] or [1, ∞) depends on the value of p. In particular
P
the integral on [1, ∞) is used to decide the convergence of the series
1/np
On [1, ∞)
Example 8.6. Find
Z
1
∞
dx
. ( p > 0.)
xp
y
0
1
x
8.8: Improper integral on [1, ∞)
60
sol.
8
TECHNIQUES OF INTEGRATION
(1) For 0 < p < 1,
Z
∞
1
Z
dx
= lim
u→∞
xp
(2) For p = 1
Z ∞
u
1
dx
= lim
u→∞
xp
1
1−p u
x
dx
u1−p − 1
= ∞,
=
lim
=
lim
u→∞ 1 − p
u→∞ 1 − p
xp
1
Z
u
1
dx
= lim [ln x]u1 = lim ln u = ∞,
u→∞
u→∞
x
(3) For p > 1
Z
1
∞
dx
= lim
u→∞
xp
Z
u
1
1−p u
dx
x
u1−p − 1
1
=
lim
=
lim
=
.
p
u→∞
u→∞
x
1−p 1
1−p
p−1
On (0, 1]
Example 8.7. Find
Z
0
1
dx
. ( p > 0.)
xp
y
0
1
8.9: On (0, 1]
8
IMPROPER INTEGRAL
sol.
61
(1) For 0 < p < 1
Z
1
0
dx
= lim
xp
ℓ→0+
Z
1−p 1
dx
x
1 − ℓ1−p
1
=
lim
=
lim
=
,
xp
1−p
ℓ→0+ 1 − p ℓ
ℓ→0+ 1 − p
1
ℓ
(2) For p = 1
Z
1
0
Z
dx
= lim
xp
ℓ→0+
u
1
dx
= lim [ln x]1ℓ = lim (− ln ℓ) = ∞,
x
ℓ→0+
ℓ→0+
(3) For p > 1
Z
1
0
dx
= lim
xp
ℓ→0+
Z
1
ℓ
1−p 1
dx
x
1 − ℓ1−p
=
lim
=
lim
= ∞.
xp
ℓ→0+ 1 − p ℓ
ℓ→0+ 1 − p
Test for Convergence
Theorem 8.8 (Comparison test). Let 0 ≤ f (x) ≤ g(x) for all x > a. Then
Z ∞
Z ∞
(1) If
g(x) dx converges, then
f (x) dx also converges.
a
(2) If
Z
a
∞
f (x) dx diverges, then
a
Example 8.9. Test whether
Z
0
∞
Z
∞
g(x) dx also diverges.
a
dx
converges or not?
1 + x3
sol. We Rsee, for all x ≥ 1, 1/(1 + x3 ) ≤ 1/x3 holds. ByR example ?? we
∞
∞
see 1 1/x3 dx = 1/2. Hence by Comparison test 1 1/(1 + x3 ) dx
R1
converges. On the other hand, the integra 0 1/(1+x3 ) dx is well defined
R∞
R1
on [0, 1]. RHence 0 1/(1 + x3 ) dx converges and the value is 0 1/(1 +
∞
x3 ) dx + 1 1/(1 + x3 ) dx. Fig ??
Theorem 8.10 (Limit Comparison Test). Assume f (x), g(x) are positive on
[a, ∞) and suppose
f (x)
lim
=L>0
x→∞ g(x)
Z ∞
Z ∞
Then the two integra
f (x) dx and
g(x) dx both converge or both dia
a
verge.
62
8
TECHNIQUES OF INTEGRATION
R∞
Proof. (1) Suppose a g(x) dx converges: Then there is N > a such that
f (x)/g(x) ≤ L+1 holds for all x R≥ N . So we have 0 ≤ f (x) ≤ R(L+1)g(x)
∞
∞
and by Limit Comparison Test, N f (x) dx converge. Hence a f (x) dx
R∞
RN
converges to a f (x) dx + N f (x) dx.
R∞
(2) Suppose a g(x) dx diverges:There exists N > a s.t. for all x ≥ N ,
f (x)/g(x) ≥ L − L/2 = RL/2 holds. Hence f (x) ≥ (L/2)g(x)
R ∞ ≥ 0 and by
∞
Limit Comparison Test N f (x) dx diverges. So does a f (x) dx.
y
y=
1
1 + x3
y=
1
1 + x2
y = 13
x
0
1
x
2
8.10:
Example 8.11. Test whether
Z
0
∞
dx
converges or not?
1 + ex
sol. Let f (x) = 1/(1 + ex ), g(x) = 1/ex . Then
f (x)
ex
= lim
=1
x→∞ g(x)
x→∞ 1 + ex
lim
and
Z
∞
dx
= lim
u→∞
ex
Z
u
u
dx
= lim −e−x 0 = lim −e−u + 1 = 1
x
u→∞
u→∞
0
0 e
R∞
Hence by Limit Comparison Test, 0 1/(1 + ex ) dx converges.
Example 8.12. Test for convergence
Z
2
∞r
x2
x
dx.
−1
8
IMPROPER INTEGRAL
63
y
y=
1
2
1
1 + ex
x
0
8.11:
sol. Set f (x) =
Then
r
x2
x
1
, g(x) = √
−1
x
f (x)
= lim
lim
x→∞ g(x)
x→∞
Z
r
x2
=1
−1
x2
√
√ √ u
dx
√ = lim 2 x 2 = lim 2 u − 2 2 = ∞
u→∞
x u→∞
2
Z ∞r
x
By Limit Comparison Test
dx converges.
2
x −1
2
∞
y
y=
0
1
2
8.12:
r
x
x2 − 1
x