Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 5 pdf
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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
sin x
dx.
Hint, Solution
Exercise 4.5 (mathematica/calculus/integral/fundamental.nb)
Evaluate
x
2
x
3
−5
dx.
Hint, Solution
4.6.2 The Definite Integral
Exercise 4.6 (mathematica/calculus/integral/definite.nb)
Use the result
b
a
f(x) dx = lim
N→∞
N−1
n=0
f(x
n
)∆x
134
where ∆x =
b−a
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) − h(g(x))g
(x).
(Don’t use the limit definition of differentiation, use the Fundamental Theorem of Integral Calculus.)
Hint, Solution
Exercise 4.9 (mathematica/calculus/integral/definite.nb)
Let A
n
be the area between the curves x and x
n
on the interval [0 . . . 1]. What is lim
n→∞
A
n
? Explai n this result
geometrically.
Hint, Solution
Exercise 4.10 (mathematica/calculus/integral/taylor.nb)
a. Show that
f(x) = f(0) +
x
0
f
(x − ξ) dξ.
b. From the above identity show that
f(x) = f(0) + xf
(0) +
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!
ξ
n
f
(n+1)
(x − ξ) dξ.
Hint, Solution
Exercise 4.11
Find a function f(x) whose arc length from 0 to x is 2x.
Hint, Solution
Exercise 4.12
Consider a curve C, bounded by −1 and 1, on the interval (−1 . . . 1). Can the length of C be unbounded? What if we
change to the closed interval [−1 . . . 1]?
Hint, Solution
4.6.3 The Fundamental Theorem of Integration
4.6.4 Techniques of Integration
Exercise 4.13 (mathematica/calculus/integral/parts.nb)
Evaluate
x sin x dx.
Hint, Solution
Exercise 4.14 (mathematica/calculus/integral/parts.nb)
Evaluate
x
3
e
2x
dx.
Hint, Solution
Exercise 4.15 (mathematica/calculus/integral/partial.nb)
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.
Hint, Solution
Exercise 4.18 (mathematica/calculus/integral/improper.nb)
Evaluate
1
0
1
√
x
dx.
Hint, Solution
Exercise 4.19 (mathematica/calculus/integral/improper.nb)
Evaluate
∞
0
1
x
2
+4
dx.
Hint, Solution
137
4.7 Hints
Hint 4.1
Make the change of variables u = 2x + 3.
Hint 4.2
Make the change of variables u = ln x.
Hint 4.3
Make the change of variables u = x
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
0
sin
2
x dx.
Hint 4.8
Let H
(x) = h(x) and evaluate the integral in terms of H(x).
138
Hint 4.9
CONTINUE
Hint 4.10
a. Evaluate the integral.
b. Use integration by parts to evaluate the integral.
c. Use integration by parts with u = f
(n+1)
(x − ξ) and dv =
1
n!
ξ
n
.
Hint 4.11
The arc length from 0 to x is
x
0
1 + (f
(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
(x + 2)
1 = a(x + 2) + b(x −2)
139
Set x = 2 and x = −2 to solve for a and b.
Hint 4.16
Expanding the integral in partial fractions,
x + 1
x
3
+ 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)
Set x = 0, x = 2 and x = −3 to solve for a, b and c.
Hint 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
dx
Hint 4.18
1
0
1
√
x
dx = lim
→0
+
1
1
√
x
dx
Hint 4.19
1
x
2
+ a
2
dx =
1
a
arctan
x
a
140
4.8 Solutions
Solution 4.1
(2x + 3)
10
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
Solution 4.2
(ln x)
2
x
dx =
(ln x)
2
d(ln x)
dx
dx
=
(ln x)
3
3
Solution 4.3
x
√
x
2
+ 3 dx =
√
x
2
+ 3
1
2
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
dx
= ln |sin x|
Solution 4.5
x
2
x
3
− 5
dx =
1
x
3
− 5
1
3
d(x
3
)
dx
dx
=
1
3
ln |x
3
− 5|
Solution 4.6
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
= lim
N→∞
N(N − 1)
2N
2
=
1
2
142
Solution 4.7
Let u = sin x and dv = sin x dx. Then du = cos x dx and v = −cos x.
π
0
sin
2
x dx =
− sin x cos x
π
0
+
π
0
cos
2
x dx
=
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
sin
2
x dx =
π
2
Solution 4.8
Let H
(x) = h(x).
d
dx
f(x)
g(x)
h(ξ) dξ =
d
dx
(H(f(x)) −H(g(x)))
= H
(f(x))f
(x) − H
(g(x))g
(x)
= h(f(x))f
(x) − h(g(x))g
(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
143
Then we consider the area in the limit as n → ∞.
lim
n→∞
A
n
= lim
n→∞
1
2
−
1
n + 1
=
1
2
In Figure 4.3 we plot the functions x
1
, x
2
, x
4
, x
8
, . . . , x
1024
. In the limit as n → ∞, x
n
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) +
x
0
f
(x − ξ) dξ = f(0) + [−f(x − ξ)]
x
0
= f(0) − f(0) + f(x)
= f(x)
2.
f(0) + xf
(0) +
x
0
ξf
(x − ξ) dξ = f(0) + xf
(0) + [−ξf
(x − ξ)]
x
0
−
x
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) +
x
0
1
n!
ξ
n
f
(n+1)
(x − ξ) dξ
= f(0) + xf
(0) +
1
2
x
2
f
(0) + ··· +
1
n!
x
n
f
(n)
(0) +
1
(n + 1)!
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) + ··· +
1
n!
x
n
f
(n)
(0) +
1
(n + 1)!
x
n+1
f
(n+1)
(0)
+
x
0
1
(n + 1)!
ξ
n+1
f
(n+2)
(x − ξ) dξ
This shows that the hypothesis holds for n = m + 1. By indu ction, the hypothesis hold for all n ≥ 0.
145
Solution 4.11
First note that the arc length from a to b is 2(b − a).
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
(x))
2
= 2
f
(x) = ±
√
3
f(x) is a continuous, piecewise differentiable function which satisfies f
(x) = ±
√
3 at the points where it is differentiable.
One example is
f(x) =
√
3x
Solution 4.12
CONTINUE
Solution 4.13
Let u = x, and dv = sin x dx. Then du = dx and v = −cos x.
x sin x dx = −x cos x +
cos x dx
= −x cos x + sin x + C
Solution 4.14
Let u = x
3
and dv =
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
dx
146
Let u = x
2
and dv =
e
2x
dx. Then du = 2x dx and v =
1
2
e
2x
.
x
3
e
2x
dx =
1
2
x
3
e
2x
−
3
2
1
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
2x
+
3
2
x
e
2x
dx
Let u = x and dv =
e
2x
dx. Then du = dx and v =
1
2
e
2x
.
x
3
e
2x
dx =
1
2
x
3
e
2x
−
3
4
x
2
e
2x
+
3
2
1
2
x
e
2x
−
1
2
e
2x
dx
x
3
e
2x
dx =
1
2
x
3
e
2x
−
3
4
x
2
e
2x
+
3
4
x
e
2x
−
3
8
e
2x
+C
Solution 4.15
Expanding the integrand in partial fractions,
1
x
2
− 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
4(x + 2)
dx
=
1
4
ln |x − 2| −
1
4
ln |x + 2| + C
=
1
4
x − 2
x + 2
+ C
147
Solution 4.16
Expanding the integral in partial fractions,
x + 1
x
3
+ 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
2
− 6x
dx =
−
1
6x
+
3
10(x − 2)
−
2
15(x + 3)
dx
= −
1
6
ln |x| +
3
10
ln |x − 2| −
2
15
ln |x + 3| + C
= ln
|x − 2|
3/10
|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
dx
= lim
δ→0
+
−
1
x − 1
1−δ
0
+ lim
→0
+
−
1
x − 1
4
1+
= lim
δ→0
+
1
δ
− 1
+ lim
→0
+
−
1
3
+
1
= ∞ + ∞
The integral diverges.
148
Solution 4.18
1
0
1
√
x
dx = lim
→0
+
1
1
√
x
dx
= lim
→0
+
2
√
x
1
= lim
→0
+
2(1 −
√
)
= 2
Solution 4.19
∞
0
1
x
2
+ 4
dx = lim
α→∞
α
0
1
x
2
+ 4
dx
= lim
α→∞
1
2
arctan
x
2
α
0
=
1
2
π
2
− 0
=
π
4
149
4.9 Quiz
Problem 4.1
Write the limit-sum de finition of
b
a
f(x) dx.
Solution
Problem 4.2
Evaluate
2
−1
|x|dx.
Solution
Problem 4.3
Evaluate
d
dx
x
2
x
f(ξ) dξ.
Solution
Problem 4.4
Evaluate
1+x+x
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
i
and
∆x = max
i
∆x
i
and choose ξ
i
∈ [x
i
x
i+1
].
b
a
f(x) dx = lim
∆x→0
n−1
i=0
f(ξ
i
)∆x
i
Solution 4.2
2
−1
|x|dx =
0
−1
√
−x dx +
2
0
√
x dx
=
1
0
√
x dx +
2
0
√
x dx
=
2
3
x
3/2
1
0
+
2
3
x
3/2
2
0
=
2
3
+
2
3
2
3/2
=
2
3
(1 + 2
√
2)
Solution 4.3
d
dx
x
2
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
x + 1
a = (1 + x + x
2
)
x=−1
= 1
b =
1
1!
d
dx
(1 + x + x
2
)
x=−1
= (1 + 2x)
x=−1
= −1
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)
3
dx =
1
(x + 1)
3
−
1
(x + 1)
2
+
1
x + 1
dx
= −
1
2(x + 1)
2
+
1
x + 1
+ ln |x + 1|
=
x + 1/2
(x + 1)
2
+ ln |x + 1|
Solution 4.5
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
(1)(1 − 2)(1 − 3)
=
1
2
c =
1
(2)(2 − 1)(2 − 3)
= −
1
2
d =
1
(3)(3 − 1)(3 − 2)
=
1
6
1
x(x − 1)(x − 2)(x − 3)
= −
1
6x
+
1
2(x − 1)
−
1
2(x − 2)
+
1
6(x − 3)
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.
If r(t) represents the position of a particle at time t, then the velocity and acceleration of the particle are
dr
dt
and
d
2
r
dt
2
,
154
respectively. The speed of the particle is |r
(t)|.
Differentiation Formulas. Let f(t) and g(t) be vector functions and a(t) be a scalar function. By writing out
components you can verify the differentiation formulas:
d
dt
(f · g) = f
· g + f · g
d
dt
(f × g) = f
× g + f × g
d
dt
(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
∂x
k
≡ lim
∆x→0
u(x
1
, . . . , x
k
+ ∆x, . . . , x
n
) − u(x
1
, . . . , x
k
, . . . , x
n
)
∆x
.
Partial derivatives have the same differentiation formulas as ordinary derivatives.
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0
2
4
6
0
2
4
6
-1
-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
≡
∂
2
u
∂x∂y
≡
∂
∂x
∂u
∂y
,
u
xxyz
≡
∂
4
u
∂x
2
∂y∂z
≡
∂
2
∂x
2
∂
∂y
∂u
∂z
.
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If u
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
∂u
∂x
k
≡ lim
∆x→0
u(x
1
, . . . , x
k
+ ∆x, . . . , x
n
) − u(x
1
, . . . , x
k
, . . . , x
n
)
∆x
.
Partial derivatives of vector fields have the same differentiation formulas as ordinary derivatives.
Gradient. We introduce the vector differential operator,
∇ ≡
∂
∂x
1
e
1
+ ··· +
∂
∂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
n
,
Directional Derivative. Suppose you are standing on some terrain. The slope of the ground in a particular
direction is the directional derivative of the elevation in that direction. Consider a differentiable scalar field, u(x). The
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