4.6 Exercises
4.6.1 The Indefinite Integral
Exercise 4.1 (mathematica/calculus/integral/fundamental.nb)
Evaluate

(2x + 3)
10
dx.
Hint, Solution
Exercise 4.2 (mathematica/calculus/integral/fundamental.nb)
Evaluate

(ln x)
2
x
dx.
Hint, Solution
Exercise 4.3 (mathematica/calculus/integral/fundamental.nb)
Evaluate

x
√
x
2
+ 3 dx.
Hint, Solution
Exercise 4.4 (mathematica/calculus/integral/fundamental.nb)
Evaluate

cos x

N
and x
n
= a + n∆x, to show that

1
0
x dx =
1
2
.
Hint, Solution
Exercise 4.7 (mathematica/calculus/integral/definite.nb)
Evaluate the following integral using integration by parts and the Pythagorean identity.

π
0
sin
2
x dx
Hint, Solution
Exercise 4.8 (mathematica/calculus/integral/definite.nb)
Prove that
d
dx

f(x)
g(x)
h(ξ) dξ = h(f(x))f



x
0
ξf

(x − ξ) dξ.
135
c. Using induction, show that
f(x) = f(0) + xf

(0) +
1
2
x
2
f

(0) + ··· +
1
n!
x
n
f
(n)
(0) +

x
0
1
n!

Evaluate

1
x
2
−4
dx.
Hint, Solution
136
Exercise 4.16 (mathematica/calculus/integral/partial.nb)
Evaluate

x+1
x
3
+x
2
−6x
dx.
Hint, Solution
4.6.5 Improper Integrals
Exercise 4.17 (mathematica/calculus/integral/improper.nb)
Evaluate

4
0
1
(x−1)
2
dx.

2
+ 3.
Hint 4.4
Make the change of variables u = sin x.
Hint 4.5
Make the change of variables u = x
3
− 5.
Hint 4.6

1
0
x dx = lim
N→∞
N−1

n=0
x
n
∆x
= lim
N→∞
N−1

n=0
(n∆x)∆x
Hint 4.7
Let u = sin x and dv = sin x dx. Integration by parts will give you an equation for

π

(ξ))
2
dξ (4.3)
First show that the arc length of f(x) from a to b is 2(b −a). Then conclude that the integrand in Equation 4.3 must
everywhere be 2.
Hint 4.12
CONTINUE
Hint 4.13
Let u = x, and dv = sin x dx.
Hint 4.14
Perform integration by parts three succes sive times. For the first one let u = x
3
and dv =
e
2x
dx.
Hint 4.15
Expanding the integrand in partial fractions,
1
x
2
− 4
=
1
(x − 2)(x + 2)
=
a
(x − 2)
+
b

1
(x − 1)
2
dx = lim
δ→0
+

1−δ
0
1
(x − 1)
2
dx + lim
→0
+

4
1+
1
(x − 1)
2
dx
Hint 4.18

1
0
1
√
x
dx = lim

dx
Let u = 2x + 3, g(u) = x =
u−3
2
, g

(u) =
1
2
.

(2x + 3)
10
dx =

u
10
1
2
du
=
u
11
11
1
2
=
(2x + 3)
11
22

d(x
2
)
dx
dx
=
1
2
(x
2
+ 3)
3/2
3/2
=
(x
2
+ 3)
3/2
3
141
Solution 4.4

cos x
sin x
dx =

1
sin x
d(sin x)
dx

1
0
x dx = lim
N→∞
N−1

n=0
x
n
∆x
= lim
N→∞
N−1

n=0
(n∆x)∆x
= lim
N→∞
∆x
2
N−1

n=0
n
= lim
N→∞
∆x
2
N(N − 1)
2


π
0
cos
2
x dx
=

π
0
(1 − sin
2
x) dx
= π −

π
0
sin
2
x dx
2

π
0
sin
2
x dx = π

π
0


(x)
Solution 4.9
First we compute the area for positive integer n.
A
n
=

1
0
(x − x
n
) dx =

x
2
2
−
x
n+1
n + 1

1
0
=
1
2
−
1
n + 1

on the interval [0 . . . 1] tends to
the function

0 0 ≤ x < 1
1 x = 1
Thus the area tends to the area of the right triangle with unit base and height.
0.2
0.4 0.6
0.8
1
0.2
0.4
0.6
0.8
1
Figure 4.3: Plots of x
1
, x
2
, x
4
, x
8
, . . . , x
1024
.
144
Solution 4.10
1.
f(0) +

−f

(x − ξ) dξ
= f(0) + xf

(0) − xf

(0) − [f(x −ξ)]
x
0
= f(0) − f(0) + f(x)
= f(x)
3. Above we showed that the hypothesis holds for n = 0 and n = 1. Assume that it holds for some n = m ≥ 0.
f(x) = f(0) + xf

(0) +
1
2
x
2
f

(0) + ··· +
1
n!
x
n
f
(n)
(0) +

ξ
n+1
f
(n+1)
(x − ξ)

x
0
−

x
0
−
1
(n + 1)!
ξ
n+1
f
(n+2)
(x − ξ) dξ
= f(0) + xf

(0) +
1
2
x
2
f

(0) + ··· +

b
a

1 + (f

(x))
2
dx =

b
0

1 + (f

(x))
2
dx −

a
0

1 + (f

(x))
2
dx = 2b −2a
Since a and b are arbitrary, we conclude that the integrand must everywhere be 2.

1 + (f


e
2x
dx. Then du = 3x
2
dx and v =
1
2
e
2x
.

x
3
e
2x
dx =
1
2
x
3
e
2x
−
3
2

x
2
e
2x

2
x
2
e
2x
−

x
e
2x
dx


x
3
e
2x
dx =
1
2
x
3
e
2x
−
3
4
x
2
e

−
3
4
x
2
e
2x
+
3
2

1
2
x
e
2x
−
1
2

e
2x
dx


x
3
e
2x
dx =

− 4
=
1
(x − 2)(x + 2)
=
A
(x − 2)
+
B
(x + 2)
1 = A(x + 2) + B(x − 2)
Setting x = 2 yields A =
1
4
. Setting x = −2 yields B = −
1
4
. Now we can do the integral.

1
x
2
− 4
dx =


1
4(x − 2)
−
1

+ x
2
− 6x
=
x + 1
x(x − 2)(x + 3)
=
A
x
+
B
x − 2
+
C
x + 3
x + 1 = A(x −2)(x + 3) + Bx(x + 3) + Cx(x − 2)
Setting x = 0 yields A = −
1
6
. Setting x = 2 yields B =
3
10
. Setting x = −3 yields C = −
2
15
.

x + 1
x
3

|x|
1/6
|x + 3|
2/15
+ C
Solution 4.17

4
0
1
(x − 1)
2
dx = lim
δ→0
+

1−δ
0
1
(x − 1)
2
dx + lim
→0
+

4
1+
1
(x − 1)
2

→0
+

−
1
3
+
1


= ∞ + ∞
The integral diverges.
148
Solution 4.18

1
0
1
√
x
dx = lim
→0
+

1

1
√
x
dx

1
x
2
+ 4
dx
= lim
α→∞

1
2
arctan

x
2


α
0
=
1
2

π
2
− 0

=
π
4
149

2
(x+1)
3
dx.
Solution
Problem 4.5
State the integral mean value theorem.
Solution
Problem 4.6
What is the partial fraction expansion of
1
x(x−1)(x−2)(x−3)
?
Solution
150
4.10 Quiz Solutions
Solution 4.1
Let a = x
0
< x
1
< ··· < x
n−1
< x
n
= b be a partition of the interval (a b). We define ∆x
i
= x
i+1
− x

|x|dx =

0
−1
√
−x dx +

2
0
√
x dx
=

1
0
√
x dx +

2
0
√
x dx
=

2
3
x
3/2

1

x
f(ξ) dξ = f(x
2
)
d
dx
(x
2
) − f(x)
d
dx
(x)
= 2xf(x
2
) − f(x)
151
Solution 4.4
First we expand the integrand in partial fractions.
1 + x + x
2
(x + 1)
3
=
a
(x + 1)
3
+
b
(x + 1)
2

c =
1
2!

d
2
dx
2
(1 + x + x
2
)





x=−1
=
1
2
(2)


x=−1
= 1
Then we can do the integration.

1 + x + x
2
(x + 1)

Let f(x) be continuous. Then

b
a
f(x) dx = (b − a)f(ξ),
for some ξ ∈ [a b].
Solution 4.6
1
x(x − 1)(x − 2)(x − 3)
=
a
x
+
b
x − 1
+
c
x − 2
+
d
x − 3
152
a =
1
(0 − 1)(0 − 2)(0 − 3)
= −
1
6
b =
1

153
Chapter 5
Vector Calculus
5.1 Vector Functions
Vector-valued Functions. A vector-valued function, r(t), is a mapping r : R → R
n
that assigns a vector to each
value of t.
r(t) = r
1
(t)e
1
+ ··· + r
n
(t)e
n
.
An example of a vector-valued function is the position of an object in space as a function of time. The function is
continous at a point t = τ if
lim
t→τ
r(t) = r(τ).
This occurs if and only if the component functions are continu ous. The function is differentiable i f
dr
dt
≡ lim
∆t→0
r(t + ∆t) − r(t)
∆t
exists. This occurs if and only if the component functions are differentiable.

(af) = a

f + af

5.2 Gradient, Divergence and Curl
Scalar and Vector Fields. A scalar field is a function of position u(x) that assigns a scalar to each point in space.
A function that gives the temperature of a material is an example of a scalar field. In two di men sions , you can graph a
scalar field as a surface plot, (Figure 5.1), with the vertical axis for the value of the function.
A vector fie ld is a function of position u(x) that assigns a vector to each point in space. Examples of vectors fields
are functions that give the acceleration due to gravity or the velocity of a fluid. You can graph a vector field in two or
three dimension by drawing vectors at regularly spaced points. (See Figure 5.1 for a vector field in two dimensions.)
Partial Derivatives of Scalar Fields. Consider a scalar field u(x). The partial derivative of u with respect to
x
k
is the derivative of u in which x
k
is considered to be a variable and the remaining arguments are considered to be
parameters. The partial derivative is denoted
∂
∂x
k
u(x),
∂u
∂x
k
or u
x
k
and is defined
∂u

-0.5
0
0.5
1
0
2
4
6
Figure 5.1: A Scalar Field and a Vector Field
Consider a scalar field in R
3
, u(x, y, z). Higher derivatives of u are denoted:
u
xx
≡
∂
2
u
∂x
2
≡
∂
∂x
∂u
∂x
,
u
xy
≡
∂

xy
and u
yx
are continuous, then
∂
2
u
∂x∂y
=
∂
2
u
∂y∂x
.
This is referred to as the equality of mixed partial derivatives.
Partial Derivatives of Vector Fields. Consider a vector field u(x). The partial derivative of u with respect to
x
k
is denoted
∂
∂x
k
u(x),
∂u
∂x
k
or u
x
k
and is defined

∂x
n
e
n
,
which is known as del or nabla. In R
3
it is
∇ ≡
∂
∂x
i +
∂
∂y
j +
∂
∂z
k.
Let u(x) be a differential scalar field. The gradient of u is,
∇u ≡
∂u
∂x
1
e
1
+ ··· +
∂u
∂x
n
e