l
Practical Calculus
Chapter 1: Derivatives
§1 Definition of derivative
§2 Limits
§3 Computing a derivative from first principles
Chapter 2: Methods of Differentiation
§4 Basic rules of differentiation
§5 Composition of functions and the chain rule
§6 Inverse functions
§7 Logarithmic derivative
§8 Implicit differentiation
Chapter 3: Properties of Functions and their Derivatives
§9 Maxima and minima
§10 Taylor-Maclaurin polynomials
§11 L’Hôpital’s rule
§12 Maxima and minima again
Chapter 4: Integration
§13 Quadrature
§14 Approximating areas
§15 Some properties of definite integrals
§16 The fundamental theorem of calculus
Chapter 5: Methods of Integration
§17 Substitution
§18 Rational functions and partial fractions
§19 Integration by parts
§20 Reduction formulae
Chapter 6: Applications
§21 Lengths, surface areas and volumes
§22 Functions defined using integrals
§23 Curve sketching
§24 Polar coordinates
§25 Differential equations
1
Practical Calculus
Chapter 1: Derivatives
§1 Definition of derivative
A function f : R → R is a rule which assigns to every real number x a real number f (x). Often the
rule is described by a single formula, such as f (x) = ex − cos(x). Sometimes the rule is only defined for
some values of x, such as the logarithm ln : R+ → R (here R+ denotes the set of positive real numbers).
Sometimes the rule is best described verbally; for example, if x is the distance (in meters) along a road,
f (x) is the height (in feet) above sea level of the point given by x.
Rate of increase of f at a particular point is the derivative. Synonym: the gradient of f .
x
x
x
x
If we have two points on a graph we can draw the line through them, but there is no easy ‘geometric’ way
of defining the tangent to a graph at a particular point. The rate of increase is the slope of the tangent.
If f : I → R is a function defined for x in some (open) interval I then the derivative of f at a
particular point a ∈ I is the value of the limit
lim
h→0
f (a + h) − f (a)
.
h
When we want to think of a fixed point we often use a to denote it whereas x usually stands for an
arbitrary point. Sometimes h is written as ∆x or δx.
(a + h, f (a + h))
(a, f (a))
has slope
f (a + h) − f (a)
−→
h
x
If this limit does not exist then f is said to be not differentiable at a.
There are several notations in common use for the derivative. For example,
df df f 0 (a),
or
simply
.
dx x=a
dx a
Autumn 2010
2
df
will be used interchangeably.
dx
Computing derivatives from first principles amounts to computing such limits. If f is differentiable
at each x for which f is defined, then we obtain a new function f 0 (x), the derivative of f (x). By
construction, the value f 0 (a) of the derivative of f at a is the gradient of the graph of f at the point with
co-ordinates (a, f (a)) (see Maple Worksheet 1). The equation of the tangent line to the graph of f (x)
at the point (a, f (a) is
The notations f 0 (x) and
y − f (a) = f 0 (a)(x − a) which is y = f 0 (a)x + f (a) − af 0 (a).
§2 Limits
You have a function f (x) given by a formula and for some value x = a, it might not be obvious what
f (a) should be.
0
x−1
. Substituting x = 1 gives . However, we know that if x 6= 1
x−1
0
then f (x) = 1, so the appropriate value for f at x = 1 is 1.
x2 + x − 2
0
(2) Let f (x) = 2
. Again, substituting x = 1 gives . However, x − 1 is a factor of the
x − 3x + 2
0
numerator and denominator. Thus for x 6= 1
Simple examples (1) Let f (x) =
f (x) =
A good value at x = 1 is
is not defined at x = 2.
x+2
(x − 1)(x + 2)
=
.
(x − 1)(x − 2)
x−2
1+2
4
= −3. Note that when we substitute x = 2 we get . Hence this function
1−2
0
Informally, to understand the limit
lim f (x) = L
x→a
we let x approach a from the right and see if f (x) approaches a finite number (L+ , say) and similarly
we let x approach a from the left and see if f (x) approaches a finite number (L− , say). If both these
conditions hold and L+ = L− then we say that the limit exists and its value is L = L+ = L− .
If there is no single value for lim f (x) = L, or if it is infinite, then we say the limit does not exist.
x→a
Examples are
1
1
lim ,
lim sin
.
x→0 x
x→0
x
3
Practical Calculus
sin x
= 1,
x→0 x
lim
Rational functions are of the form
1 − cos x
= 0.
x→0
x
lim
p(x)
where p, q are polynomials (the sum of finitely many
q(x)
terms of the form ak xk ).
p(x)
:
x→a q(x)
Factor p(x) = (x − a)n p1 (x) and q(x) = (x − a)m q1 (x) , where p1 (a) 6= 0 and q1 (a) 6= 0.
p(x)
= 0,
(i) If n > m then lim
x→a q(x)
p(x)
p1 (a)
(ii) If n = m then lim
=
,
x→a q(x)
q1 (a)
p(x)
(iii) If n < m then lim
does not exist.
x→a q(x)
Procedure to evaluate lim
Limits as x → ∞ or x → −∞ of a rational function: The ratio depends only on the highest power
of x involved, because for large x:
xn is much bigger than xn−1 .
Autumn 2010
4
1
In the case of limits of the form lim it is sometimes also useful to replace x by and consider lim
x→∞
y→0
y
−1
0
x
y
y
= sin
lim sin
= lim sin
= lim sin
= 0.
x→∞
y→0
y→0
x2 + 1
y −2 + 1
1 + y2
1+0
One sided limits. Informally, to understand the limit
lim f (x) = L
x→a+
we let x approach a from the right and see if f (x) approaches a finite number L. The definition of
lim f (x) = L
x→a−
is similar.
Examples.
lim
x→0+
|x|
= 1,
x
lim e−1/x = 0,
x→0+
lim
x→0−
|x|
= −1,
x
lim e−1/x does not exist.
x→0−
Definition. Suppose that f (x) is defined at x = a and
lim f (x) = lim f (x) = f (a),
x→a+
x→a−
then f (x) is continuous as x = a.
Informally, a continuous function is one for which you can draw its graph without lifting your pen
from the paper.
sin x
If one cannot define f (x) at x = a (e.g.
at x = 0) then lim f (x) is the value (if it exists) which
x→a
x
one would give f (a) to make f (x) continuous at x = a.
§3 Computing a derivative from first principles
We can now compute (from first principles) the derivatives of familiar functions. See also Maple
Worksheet1, fourth section.
Example Consider the function f (x) = 1/x which is defined for all x 6= 0. If we want to calculate its
derivative at x = a from first principles then we look at the limit (as h → 0) of
f (a + h) − f (a)
1
1
a − (a + h)
−h
−1
=
−
=
=
=
.
h
h(a + h) ha
ha(a + h)
ha(a + h)
a(a + h)
This limit is −
1
1
. So, f 0 (a) = − 2 .
2
a
a
5
Practical Calculus
Autumn 2010
6
Summary
In this chapter, you should have learnt
• The definition of the derivative
• How the derivative relates to the gradient of the graph of a function
• The definition of limit
• The notion of continuous function
• The definition of rational function
• Some special limits
• How to start Maple and define functions in Maple ( x->x3 +sin(x); )
• How to plot functions with Maple ( plot(f(x),x=1..3,y=-2..2); )
• How to take limits using Maple ( limit(sin(x)/x,x=0); )
• How to simplify algebraic formulae using Maple ( simplify((x+3)*(x-5)2 -x); )
• How to add, multiply, divide and exponentiate algebraic expressions in Maple ( a+b; a*b; a/b;
• How to evaluate functions in Maple ( f(a); or evalf(f(a)); )
etc.)
You should be able to
• Write down the equation of tangent lines to curves given by graphs of functions.
• Take limits of f (x) at points where f is continuous ( lim f (x) = f (a))
x→a
•
•
•
•
Take limits of f (x) as x → ±∞.
Take limits of rational functions
Take limits of trigonometric functions
Take limits arising in the definition of the derivative (to compute the derivative of standard functions
from first principles.)
Chapter 2: Methods of Differentiation
§4 Basic rules of differentiation
Often it is not really practical to use the definition (first principles) to calculate derivatives but once
a few important ones have been studied and some general rules established, one can differentiate many
functions. It is helpful to have various ‘Rules of Differentiation’.
The basic rules are :
d
d
d f (x) + g(x) =
f (x) +
g(x)
= f 0 (x) + g 0 (x)
dx
dx
dx
d d
d
f (x)g(x) = g(x) f (x) + f (x) g(x) = f 0 (x)g(x) + f (x)g 0 (x)
dx
dx
dx
d f (x)
g(x)f 0 (x) − f (x)g 0 (x)
=
dx g(x)
g(x)2
They are established using the ‘first principles’ definition.
product rule
quotient rule
7
Practical Calculus
§5 Composition of functions and the chain rule
Notation for functions
f : A → B denotes a function whose domain is the set A and whose range is the set B. In other
words, given x ∈ A we have a rule that gives a single element f (x) ∈ B.
If f : A → B and g : B → C then g ◦ f : A → C denotes the composite function x → g(f (x)) (as
above). In Maple, if f, g are defined to be functions using -> , the composite is (g @ f)
Autumn 2010
8
The Chain Rule gives the derivative of the composite of two functions:
d
g(f (x)) = g 0 (f (x))f 0 (x)
dx
Informally we could prove the chain rule via
g(f (a + h)) − g(f (a))
g(f (a + h)) − g(f (a)) f (a + h) − f (a)
=
·
h
f (a + h) − f (a)
h
(∗)
Let f (a + h) = f (a) + k. Since f (a + h) is continuous when h → 0 we also have k → 0 and can write (∗)
as
g(f (a + h)) − g(f (a))
g(f (a) + k) − g(f (a)) f (a + h) − f (a)
=
·
→ g 0 (f (a)) · f 0 (a)
h
k
h
as h, k → 0.
Note: this argument is not correct if f (a + h) − f (a) = 0.
§6 Inverse functions
If f : A → B and g : B → A are such that g ◦ f : A → A and f ◦ g : B → B satisfy
(i) g ◦ f (a) = a for all a ∈ A
and
(ii) f ◦ g(b) = b for all b ∈ B,
they are called a pair of inverse functions. We will mainly discuss cases where both A, B are intervals.
I will usually consider open intervals but the discussion can easily be modified to include the case of
closed (or even half-closed) intervals. We often denote the inverse of f by f −1 .
If f : (a, b) → (c, d) maps onto the whole interval (c, d) then its inverse function f −1 : (c, d) → (a, b)
has the properties
f −1 (f (x)) = x for all x ∈ (a, b)
and
f (f −1 (y)) = y for all y ∈ (c, d).
Familiar examples of inverse functions are f (x) = exp(x) and f −1 (x) = ln(x) (What are a, b, c and d in
this case ?)
9
Practical Calculus
We can use these identities to find derivatives: By definition of arcsin x we have
sin(arcsin x) = x
for x ∈ [−1, 1].
For the moment write g(x) = arcsin x. Differentiate the identity
sin(g(x)) = x
Thus
g 0 (x) =
Hence,
=⇒
cos(g(x))g 0 (x) = 1.
1
1
1
=q
=√
.
cos(g(x))
1 − x2
1 − sin2 (g(x))
d
1
arcsin x = √
.
dx
1 − x2
When does a function f have an inverse? We need that the graph of f takes any value once only, in
other words, that every line y = c (c is a constant) meets the graph precisely once.
y
x
x = f −1 (y)
c
c
b
a
y = f (x)
x
b
y
a
If f : (a, b) → (c, d) has an inverse then for every y ∈ (c, d) there is exactly one x ∈ (a, b) such that
f (x) = y.
i.e. f is both onto and 1–1.
Hence one has to consider the domain and the range of the function.
Autumn 2010
10
Definition. Function f : (a, b) → (c, d) is monotonic increasing if
f (x1 ) < f (x2 )
whenever
a < x1 < x2 < b.
The definition for monotonic decreasing is similar. When a function is either monotonic increasing or
decreasing then we say that it is monotonic.
Facts:
(1) If f 0 (x) > 0 for every x ∈ (a, b) then f is monotonic increasing.
(2) A function f : (a, b) → (c, d) has an inverse g : (c, d) → (a, b) if f is monotonic on (a, b), and
f monotonic increasing ⇐⇒
g monotonic increasing
f monotonic decreasing ⇐⇒
g monotonic decreasing
1
- π2
π
2
x
-1
π
2
sin x
-1
arcsin y
1
y
- π2
Note that inverse functions need not exist; for example x2 does not have an inverse on any open
interval that contains 0. However, such a function can usually be made to be invertible by restricting
its domain and/or its range. Normally we pick a standard largest domain on which the function is
monotonic, for example, for the sine function we use sin : [−π/2, π/2] → [−1, 1]. The inverse function is
denoted either sin−1 or arcsin.
11
Practical Calculus
§7 Logarithmic derivative
It is sometimes easier to calculate derivatives after first taking the logarithm.
Example. Let a > 0 and take f (x) = ax . Then ln(f (x)) = ln (ax ) = x ln a. Now differentiate
f 0 (x)
= ln a
f (x)
=⇒
f 0 (x) = f (x) ln a = ax ln a.
§8 Implicit differentiation
Sometimes a problem involves two or more variables and they are related in a complicated way, but one
wants the rate of growth of one of them in terms of the other.
Example. Circle x2 + y 2 = R2 . This is a relationship between x and y, but y is not easily written as
a function of x. It is really two functions
p
p
y+ = + R 2 − x2 ,
y− = − R2 − x2 .
So the derivatives are
dy+
−x
−x
=√
=
,
dx
y+
R 2 − x2
dy−
x
−x
=√
=
.
dx
y−
R 2 − x2
Implicit differentiation avoids this complication and also applies when the relationship between x and
y cannot be solved to give y explicitly as a function of x:
In x2 + y 2 = R2 we regard y as a function of x, so
x2 + y(x)2 = R2 .
Now differentiate this equation with respect to x
2x + 2y
dy
=0
dx
=⇒
Hence, the gradient at the point x = R cos α, y = R sin α is
dy
−x
=
.
dx
y
dy
cos α
=−
= − cot α.
dx
sin α
Autumn 2010
12
4
Example. Curve x3 y + xy 3 = 2. Note that (1, 1) is a point
on the curve.
(a) Find the gradient of y as a function of x at (1, 1).
(b) Show that the curve has no vertical tangents.
(c) Calculate y 00 (x) at (1, 1).
Again we regard y as a function of x and differentiate the equation with respect to x
y 2
0
−4
−2
0
2
4
3x2 y + x3 y 0 + y 3 + 3xy 2 y 0 = 0
x
−2
y 0 (x3 + 3xy 2 ) = −(3x2 y + y 3 )
y0 = −
−4
(∗)
3x2 y + y 3
.
x3 + 3xy 2
3+1
= −1.
1+3
0
3
2
2
2
(b) If y = ∞ then x + 3xy = x(x + 3y ) = 0. Thus x = 0 or x2 + 3y 2 = 0. In the second case we
have x = 0 and y = 0. In both cases we have x = 0, but there is no point on the curve where x = 0.
(c) To calculate y 00 we differentiate (∗) again with respect to x:
(a) When we take x = 1 and y = 1 we obtain y 0 = −
y 00 (x3 + 3xy 2 ) + y 0 (3x2 + 3y 2 + 6xyy 0 ) = −(6xy + 3x2 y 0 + 3y 2 y 0 ),
and now we use that at the point we have x = 1, y = 1 and y 0 = −1. Thus
y 00 (1 + 3) − 1(3 + 3 − 6) = −(6 − 3 − 3)
=⇒
4y 00 = 0
=⇒
y 00 = 0.
Summary
In this chapter, you should have learnt
• The product, quotient and chain rules for differentiation.
• The derivatives of the basic functions (xn , sin(x), cos(x), tan(x), exp(x), ln(x), cosh(x) ).
• The definition of an inverse function.
• The formula for the derivative of an inverse function.
• The various notations for functions, operations on functions and their derivatives.
You should be able to
• Calculate derivatives using Maple ( D(f); for a function f and diff(f,x); for an expression
f ).
• Compose functions using Maple ( f @ g; or (f@g)(x); to evaluate ).
• Calculate the composition of functions.
• Calculate the inverses of functions.
• Calculate the derivative of a function made up from the basic functions.
• Calculate the derivatives of inverse trigonometric functions and hyperbolic functions.
• Calculate the gradients of graphs given implicitly.
13
Practical Calculus
Chapter 3: Properties of Functions and their Derivatives
In this chapter we will use higher derivatives. If f is a function whose derivative is denoted f 0 then
the derivative of f 0 is denoted f 00 or f (2) . Iterating this process gives the kth derivative denoted f (k) .
dk f
.
Another common notation for the kth derivative is
dxk
§9 Maxima and minima
A point x = a where f 0 (a) = 0 is called a critical or stationary point of f .
If f (2) (a) < 0 then this stationary point is a local maximum, if f (2) (a) > 0 then the point is a
local minimum. If f (2) (a) = 0 and f (3) (a) 6= 0 then x = a is a point of inflection (sometimes spelt
inflexion).
The global maximum/minimum of a function f on a closed interval is attained at either a critical
point of f or at one of the end-points of the interval. Therefore to find the global maximum/minimum
of a function f defined on a closed interval [c, d] one needs to compare the values f (c), f (d), f (a) where
a is any of the critical points a of f such that a ∈ (c, d).
Note: The value of the function is not always a reliable guide to whether a critical point is a local
max/min. An example is f (x) = x + x−1 , then f 0 (x) = 1 − x−2 . Thus the critical points are x = ±1.
The values at these points are f (1) = 2 and f (−1) = −2. However, you can check that x = 1 is a local
minimum and x = −1 is a local maximum. (Sketch the graph of f (x).)
Autumn 2010
14
§10 Taylor-Maclaurin polynomials
We suppose that f : (a, b) → R is a function which we can differentiate as often as we please at every
point of the interval (a, b). For simplicity we concentrate at x = 0.
Consider the task of finding a polynomial p of degree at most n such that p(0) = f (0) and which
has the same derivatives as f at x = 0.
For example, for n = 1 a polynomial with the same value and the same first derivative as f at x = 0
is given by the equation of the tangent line to f (x) at x = 0 : p(x) = f (0) + xf 0 (0).
More generally, if we are looking for a polynomial of degree at most n with the same first n derivatives
then we take a general polynomial of degree at most n :
p(x) = c0 + c1 x + c2 x2 + . . . + cn xn =
n
X
ck xk .
k=0
Differentiating this we get
p0 (x) = c1 + 2c2 x + 3c3 x2 + . . . + ncn xn−1 =
n
X
kck xk−1
k=1
So p0 (0) = c1 , and similarly by differentiating again and again we get
p(2) (0) = 2c2 ,
p(3) (0) = 3 · 2c3 ,
p(4) (0) = 4 · 3 · 2c4 , ... , p(k) (0) = k!ck .
By dividing by k! we see that if p(x) has the same first n derivatives (at x = 0) as f, that is
p(k) (0) = f (k) (0) for 0 6 k 6 n then ck = f (k) (0)/k! for these k.
In other words, the only degree n polynomial which has the same first n derivatives as f does at 0
is given by
pn (x) = f (0) + f 0 (0)x +
f (2) (0) 2 f (3) (0) 3
f (k) (0) k
f (n) (0) n
x +
x + ... +
x + ... +
x .
2!
3!
k!
n!
This is called the Taylor-Maclaurin polynomial of degree n for f at x = 0.
They are good approximations for f (x) near x = 0 (Maple Worksheet 3).
Fact: For most common functions
f (x) − pn (x)
→ 0 as x → 0.
xn
In the case n = 1 we have
f (x) − f (0) − f 0 (0)x
f (x) − f (0)
f (x) − p1 (x)
=
=
− f 0 (0).
x
x
x
f (x) − p1 (x)
→ 0 as x → 0 is in fact the definition of the derivative of f at the origin.
x
Notation.
f (x) = O (xr )
as x → 0,
Thus
means that there is a constant C > 0 such that
|f (x)| 6 C|x|r ,
for all small x.
An example is x2 = O(x) as x → 0, since x2 6 |x| for all |x| 6 1.
15
Practical Calculus
Note that
(1) xs = O (xr ) if r 6 s.
(2) If f (x) = O (xr ) and K a constant then Kf (x) = O (xr ).
(3) If f (x) = O (xr ) and g(x) = O (xr )then f (x) + g(x) = O (xr ).
(3’) If f (x) = O (xr ) and g(x) = O (xs )then f (x) + g(x) = O (xm ), where m = min(r, s).
Fact: If f (x) is a function with nth-Maclaurin polynomial pn (x) then
f (x) − pn (x) = O xn+1 ,
as x → 0.
Moreover, if q(x) is any polynomial of degree at most n and f (x) − q(x) = O xn+1 then q(x) = pn (x).
So far we have only considered expansion about the origin. The nth Taylor polynomial of f (x)
about x = a is
pn (x) = f (a) + f 0 (a)(x − a) +
f (3) (a)
f (n) (a)
f (2) (a)
(x − a)2 +
(x − a)3 + . . . +
(x − a)n .
2!
3!
n!
Autumn 2010
16
Definition. The Maclaurin series of a function f (x) is the infinite series
∞
X
f (n) (0) n
x ,
n!
n=0
and the Taylor series of f (x) about the point x = a is
∞
X
f (n) (a)
(x − a)n .
n!
n=0
Standard Examples:
• Exponential:
• Sine:
• Cosine:
• Logarithm:
1 n
x + ...
n!
1
1 3
sin(x) = x − 3!
x2n+1 + . . .
x + . . . + (−1)n
(2n + 1)!
1
cos(x) = 1 − 12 x2 + . . . + (−1)n
x2n + . . .
(2n)!
exp(x) = 1 + x + 21 x2 + . . . +
ln(1 + x) = x − 12 x2 + . . . + (−1)n−1 n1 xn + . . .
1
• Binomial: (1 + x)` = 1 + `x + 12 `(` − 1)x2 + 3!
`(` − 1)(` − 2)x3 + . . . + `(` − 1) · · · (` − n + 1)xn /n! + . . .
1
• Geometric (` = −1):
= 1 − x + x2 − · · · + (−1)n xn + · · ·
1+x
The last three converge for |x| < 1. The other three series converge for all values of x.
17
Practical Calculus
§11 L’Hôpital’s rule
We can formalise the method of using Taylor-Maclaurin polynomials to compute limits x → a of
functions of the form f (x)/g(x) when g(a) = 0 and f (a) = 0 by stating L’Hôpital’s rule :
If g(a) = 0, f (a) = 0 then
f 0 (x)
f (x)
= lim 0
lim
x→a g (x)
x→a g(x)
Sometimes one needs to apply this again (and again!).
§12 Maxima and minima again
If f 0 (a) and f 00 (a) = 0 it is useful to think about the Taylor series at x = a. Suppose that f (k) (a) 6= 0
is the first non-zero derivative at a. So
f (x) − f (a) = f (k) (a)
(x − a)k
+ O (x − a)k+1
k!
as x → a.
Case 1: k is even. Then (x − a)k > 0 if x 6= a, and so
f (x) has at x = a a local minimum if
f (k) (a) > 0,
f (x) has at x = a a local maximum if
f (k) (a) < 0.
Case 2: k is odd. Then (x − a)k > 0 for x > a and (x − a)k < 0 for x < a. Since we assume that
f (a) 6= 0 it follows that f (x) − f (a) takes positive and negative values near x = a. Hence, x = a is not
a local maximum or minimum. It is like a point of inflection.
(k)
Autumn 2010
Summary
In this chapter, you should have learnt
• How to interpret the condition f 0 (x) = 0.
• The definition of Taylor-Maclaurin polynomials.
• How to compute higher order derivatives using Maple ( (D@@n)(f); ).
• L’Hôpital’s rule.
You should be able to
• Locate the maxima, minima and critical points of functions.
• Calculate the Taylor-Maclaurin polynomials of functions.
• Use Taylor-Maclaurin polynomials to calculate some limits.
• Use L’Hôpital’s rule to compute limits.
18
19
Practical Calculus
Chapter 4: Integration
§13 Quadrature
The definite integral between x = a and x = b of a function f (x) is denoted
Z
b
f (x) dx.
a
Z
is a sum of the areas of small strips as indicated. Hence, if
a
b
f (x) is non-negative for x between a and b, with a < b then
the value of the integral is the area of the region between the
graph of f (x) and the x-axis and between the lines x = a and
x = b.
b
There are rules for extending the definition to other cases :
(i) Where the function takes negative values the area between the
graph of f (x) and the x-axis counts with a negative sign.
(ii) If a > b then the area also counts with a negative sign
(iii) If both the previous rules apply then the signs cancel.
Basically, the area is counted positively if the boundary curve
is traversed in an anti-clockwise direction.
a
Autumn 2010
20
§14 Approximating areas
Riemann integral. One way to approximate the area of the region between the graph of a function
f (x) and the x-axis and between the lines x = a and x = b is to subdivide the interval and consider lower
and upper sums. These are the sums of the areas of the grey rectangles below.
f (x)
a
a
b
lower sum
b
upper sum
As the subdivision becomes finer the value of the integral is sandwiched.
If this does not work the function is said not to have a Riemann integral.
f (x)
Example. Monotonic increasing functions.
b−a
Split the interval into N parts: xk = a + k
.
N
Then
uppersumN =
N
X
k=1
lowersumN =
N
X
k=1
Z b
lowersumN <
f (xk )
a
x0
x2
xk
b − a
b−a = f (x1 ) + f (x2 ) + · · · + f (xN )
N
N
f (xk−1 )
b − a
b−a = f (x0 ) + f (x1 ) + · · · + f (xN −1 )
N
N
f (x) dx < uppersumN
a
In this case we have
uppersumN − lowersumN = f (xN ) − f (x0 )
b − a
b − a
= f (b) − f (a)
→0
N
N
Hence,
Z
b
f (x) dx = lim uppersumN = lim lowersumN .
a
N →∞
N →∞
as N → ∞.
xN b
21
Practical Calculus
In general this method does not work easily. However, it is often useful when used the other way:
To understand sums of series.
∞
X
1
. We take monoExample: We try to obtain lower and upper bounds for the infinite series
k2
k=1
1
tonically decreasing function f (x) = 2 and observe
x
1
x2
1
area
1 1/4 1/9
0
1
2
N −2
3
x
N
N
X
1
The sum of the areas of the first N rectangles is
. It follows from the graph that
k2
k=1
1<
Z N
N
N
X
X
1
1
1
1
=
1
+
<
1
+
dx = 2 − .
2
2
2
k
k
x
N
1
k=1
∞
X
1
< 2.
Hence, 1 <
k2
k=1
k=2
Autumn 2010
22
§15 Some properties of definite integrals
The above rules for computing areas under graphs lead immediately to the following properties
Z a
Z b
f (x) dx
f (x) dx = −
(1)
Z b
Z cb
Zab
f (x) dx
f (x) dx +
f (x) dx =
(2)
c
Zab
Z ab
(3)
f (x) dx 6
g(x) dx if f (x) 6 g(x) for all x ∈ [a, b].
a Z
Zab
b
(4)
kf (x) dx = k
f (x) dx
Zab
Za b+c
(5)
f (x + c) dx =
f (x) dx
Zab
Z kba+c
f (x)
f (kx) dx =
(6)
dx
k
a
ka
Notice that the definite integral is a number and the use of the variable x is purely symbolic. One
could equally well use any other variable name, it is sometimes useful to do so to avoid using the same
variable with different meanings.
The properties listed above allow some integrals to be computed very easily. For example,
(7) if f : [−a, a] → R is an odd function (i.e. f (−x) = −f (x) for all x ∈ [−a, a]) then its integral between
−a and a is zero.
(8) if f : [−a, a] → R is an even function (i.e. f (−x) = f (x) for all x ∈ [−a, a]) then its integral between
−a and a is twice the value of the integral between 0 and a.
(9) if a function is periodic with period p (i.e. f (x + p) = f (x) for all values of x) then, if n is an integer,
Z
a+np
a+p
Z
f (x) dx = n
a
f (x) dx
a
§16 The fundamental theorem of calculus
Z
Theorem If F (x) =
a
x
f (t) dt then F 0 (x) = f (x) and
Z
b
f (x) dx = F (b) − F (a).
a
[Often we write [F (x)]ba for F (b) − F (a).]
These equations say that integration is the inverse operation to differentiation. The Fundamental
Theorem will provide us with a large array of functions which can be integrated without having to evaluate
areas directly.
Z
The indefinite integral (or anti-derivative) of a function f (x), denoted f (x) dx is the function
F (x) given by the inverse operation to differentiation. Because the derivative of a constant is zero, the
indefinite integral is only well defined up to adding a constant:
Z
Z
0
f (x) dx = f (x) + C
f (x) dx = F (x) + C
For many problems it is VERY important NOT to omit the constant C. So it is best to get into the
habit of never omitting it.
23
Practical Calculus
R
F (x)
F 0 (x)
f (x) dx
xn+1
sin x
cos x
tan x
ex
ln x
sinh x
cosh x
f (x)
(n + 1)xn
cos x
− sin x
sec2 x
ex
1/x
cosh x
sinh x
−1/2
1 − x2
−1
1 + x2
−1/2
1 + x2
−1
1 − x2
f 0 (x) + g 0 (x)
f 0 (x)g(x) + f (x)g 0 (x)
f 0 (g(x))g 0 (x)
f 0 (x)/f (x)
1/f 0 f −1 (x)
arcsin x
arctan x
arcsinh x
arctanh x
f (x) + g(x)
f (x)g(x)
f (g(x))
ln |f (x)|
f −1 (x)
The penultimate result in the table is useful for computing integrals which can be put in the form
f 0 (x)/f (x). Using it in the case f (x) = cos x gives f 0 (x)/f (x) = − tan x and
Z
tan x dx = − ln | cos x| + C.
Autumn 2010
24
Summary
In this chapter, you should have learnt
• The definition of the definite integral via quadrature
• The definitions of odd, even and periodic functions
• The definition of an improper integral
• The Maple command int
• The Fundamental Theorem of Calculus (FTC)
• The definition of the indefinite integral (or anti-derivative).
• The integrals of some standard functions.
You should be able to
• Compare areas to evaluate and approximate (some) integrals
• Use symmetries of functions to help evaluate integrals
• Compare areas to study certain infinite sums.
• Use Maple to integrate using int(f(x), x=a..b);
• Use FTC to evaluate (some) integrals.
Chapter 5: Methods of Integration
In this chapter we shall be considering the problem of integration when it is not obvious what the “antiderivative” is. Sometimes it is enough to use algebraic manipulations to bring the integrand into a form
that one recognises how to integrate.
Examples:
2(x + 1)2 − 3(x + 1) + 1
1
2x2 + x
=
= 2(x + 1) − 3 +
,
x+1
x+1
x+1
(1)
2
with anti-derivative (x + 1) − 3x + ln |x + 1| + C.
cos(2x) = cos2 x − sin2 x = 2 cos2 x − 1 = 1 − 2 sin2 x.
(2)
Thus sin2 x =
1
2
−
1
2
cos(2x) and
Z
sin2 x dx =
Z
1
2
−
1
2
cos(2x) dx = 21 x −
1
4
sin(2x) + C.
§17 Substitution
The method of substitution is used very widely. It is the inverse of the chain rule for differentiating: if
f (x) = g(h(x)) then f 0 (x) = g 0 (h(x))h0 (x). The variable of integration is changed to something more
suitable, that is, we try to Rspot a good
choice for h(x) (not always obvious).
Example: In the case x cos x2 dx let u = x2 , then du = 2x dx. Thus x dx = 12 du
Z
2
x cos x
Z
dx =
1
2
cos u du =
1
2
sin u + C =
1
2
sin x2 + C.
25
Practical Calculus
R√
a2 − x2 dx take x = a sin θ.
Often one has to use√trigonometric substitutions. Let a > 0. For
2
2
Then dx = a cos θ dθ and a − x = a cos θ
Z p
Z
Z
Z
a2 − x2 dx = a cos θ a cos θ dθ = a2 cos2 θ dθ = 12 a2 (1 + cos 2θ) dθ
= 12 a2 θ + 14 a2 sin 2θ + C
= 21 a2 θ + 12 a2 sin θ cos θ + C
= 21 a2 arcsin(x/a) + 12 x
p
a2 − x2 + C.
§18 Rational functions and partial fractions
p1 (x)
p2 (x)
If the degree of the polynomial p1 (x) is equal to or greater than the degree of p2 (x), then we can
simplify this using long division of polynomials:
p1 (x)
r(x)
= q(x) +
(∗)
p2 (x)
p2 (x)
where the degree of the polynomial r is strictly less than that of p2 .
Since any polynomial q(x) is easy to integrate, this shows that it is enough to deal with cases where
the degree of the denominator is greater than that of the numerator.
We deal with these by using an algebraic manipulation called the method of partial fractions. It
is based on the fact that any polynomial with real coefficients can be factored into a product of linear
and quadratic polynomials. The method rewrites any rational function as a sum of rational function
with only one polynomial (either linear or quadratic) in the denominator. There are then several cases
depending on the power to which these polynomials are raised.
We need from the table the following four cases
Z
Z
Z 0
Z
dx
dx
xr+1
f (x)
r
= arctan x+C,
= ln |x|+C,
x dx =
+C,
dx = ln |f (x)|+C.
x2 + 1
x
r+1
f (x)
p1 (x)
are:
Hence, the three steps to integrate rational functions
p2 (x)
(1) Rewrite the rational function in the form (∗).
(2) Factorise p2 (x).
r(x)
s(x)
(3) Use the method of partial fractions to write
as a sum of fractions of the form
or
p2 (x)
(x + a)m
t(x)
.
(x2 + ax + b)m
In this section we consider integrands which are rational functions:
Autumn 2010
Example (1): In the case of
2x3
=step
x2 − 1
and now
26
2x3
dx we start with
x2 − 1
Z
2x
=step
x2 − 1
2
2x(x2 − 1) + 2x
2x
= 2x + 2
,
2
x −1
x −1
1
2x
=step
(x − 1)(x + 1)
A
B
+
.
x−1 x+1
3
To determine constants A and B we multiply the final identity times (x − 1)(x + 1)
2x = A(x + 1) + B(x − 1).
This identity must hold for all x. Taking x = 1 gives us 2 = 2A, thus A = 1, and when we take x = −1
we obtain −2 = −2B, thus B = 1. It follows that
1
1
2x3
= 2x +
+
2
x −1
x−1 x+1
Z
Example (2): For the integral
1
(x2
− 1)
2
=
2x3
dx = x2 + ln |x − 1| + ln |x + 1| + C.
x2 − 1
Z
=⇒
1
(x2
− 1)
2
dx it is tempting to write
1
2
(x − 1) (x + 1)
2
=
A
(x − 1)
2
+
B
(x + 1)
2.
However, the final identity is false. Correct is
1
2
(x − 1) (x + 1)
=
2
A
(x − 1)
2
+
B
C
D
+
+
.
2
x − 1 (x + 1)
x+1
Multiply times (x − 1)2 (x + 1)2 .
1 = A(x + 1)2 + B(x + 1)2 (x − 1) + C(x − 1)2 + D(x − 1)2 (x + 1).
Taking x = 1 and x = −1 shows us that A = C = 41 . Now we take to different values for x.
x=0
=⇒
x=2
1=
=⇒
1=
9
4
1
4
1
4 +D
1
4 + 3D
−B+
+ 9B +
1
2
=⇒
−B + D =
=⇒
9B + 3D = − 23
From the final two equations we obtain that B = − 41 and D = 14 . Thus
1
(x2 − 1)
and
Z
1
(x2
− 1)
2
2
=
dx =
1
4 (x − 1)
2
−
1
1
1
+
+
4(x − 1) 4 (x + 1)2
4(x + 1)
−1
1
1
1
− ln |x − 1| −
+ ln |x + 1| + K.
4 (x − 1) 4
4 (x + 1) 4
27
Practical Calculus
Example (3):
3x + 4
3x + 4
=
2
x4 − 4x3 + 5x2 − 4x + 4
(x − 2) (x2 + 1)
=
A
B
Cx + D
+
+ 2
.
x − 2 (x − 2)2
x +1
2
Multiply times (x − 2) (x2 + 1)
3x + 4 = A(x − 2)(x2 + 1) + B(x2 + 1) + (Cx + D)(x − 2)2 .
Taking x = 2 we obtain 10 = 5B. Thus B = 2. To obtain A, C and D we substitute x = 0, x = 1,
x = −1 and obtain the equations:
4 = −2A + 2 + 4D,
7 = −2A + 4 + C + D,
=⇒
A = −1, C = 1, D = 0.
1 = −6A + 4 − 9C + 9D,
Thus
and
3x + 4
−1
2
x
=
+
,
+ 2
x4 − 4x3 + 5x2 − 4x + 4
x − 2 (x − 2)2
x +1
Z
x4
−
4x3
3x + 4
2
dx = − ln |x − 2| −
+
2
+ 5x − 4x + 4
x−2
1
2
ln |x2 + 1| + C.
Application: One can use the method of partial fractions to evaluate integrals of functions
of the form
p1 (cos θ, sin θ)
θ
where p1 , p2 are polynomials in cos θ and sin θ. The substitution t = tan
changes the
p2 (cos θ, sin θ)
2
function to a rational function of a variable t. Using this substitution we get
sin θ =
2t
,
1 + t2
and these are substituted into the integral.
cos θ =
1 − t2
,
1 + t2
dθ =
2
dt
1 + t2
Autumn 2010
28
§19 Integration by parts
d
dv
du
Integration by parts is the inverse of the product rule for differentiation
(uv) = u
+
v and is
dx
dx
dx
usually written as
Z
Z
dv
du
u dx = uv −
v dx.
dx
dx
Z
To apply this method to f (x) dx, choose functions u and v so that
f =u
dv
dx
and
du
v can be integrated.
dx
§20 Reduction formulae
Sometimes one has a sequence of similar integrals one for each natural number n. A reduction formula
relates the one for n with ones corresponding to smaller values of n and then an inductive procedure can
be used to evaluate all of them. It is usual to write the integral as In and find a formula relating it with
In−1 (but possibly others such as In−2 ).
Z ∞
Example Evaluate the integrals In =
xn e−x dx.
Solution: Integrating this by parts gives In =
0
Z ∞
n
xn−1 e−x dx since the term xn e−x vanishes at 0 and at ∞. So we have the relation In = nIn−1 this
0
is the ‘reduction formula’. Repeating this (formally this uses mathematical induction) we have
In = n(n − 1)(n − 2) · · · 2 · 1I0 .
But I0 = [−e−x ]∞
0 = 1. So we have In = n!
29
Practical Calculus
Certain integrals involving powers of trigonometric functions are best calculated using reduction
formulae.
Z π
dv
Example: Evaluate In =
sinn x dx
Solution: Take u = sinn−1 x,
= sin x. Then
dx
0
h
iπ Z π
In = − cos x sinn−1 x +
cos x (n − 1) sinn−2 x cos x dx
0
0
Z π
= 0 + (n − 1)
1 − sin2 (x) sinn−2 (x) dx = (n − 1)In−2 − (n − 1)In .
0
So we have the reduction formula nIn = (n − 1)In−2 . Repeating we get
(n − 1) (n − 3)
3 1
(n − 1) (n − 3)
In =
In−4 = . . . =
...
I0 if n is even and
n
(n − 2)
n
(n − 2)
4 2
(n − 1) (n − 3)
2
=
. . . I1
if n is odd.
n
(n − 2)
3
But I0 = π and I1 = 2. So
3 1
(2k)! 1
1
2k
(2k − 1) (2k − 3)
...
π
= k
π
=
I2k =
π,
2k
(2k − 2)
4 2
2 k! 2k k!
22k k
I2k+1 =
2
(2k) (2k − 2)
... 2
2k + 1 (2k − 1)
3
=
22k+1
2k + 1
1
2k
k
.
Autumn 2010
30
Summary
In this chapter, you should have learnt
• the method of substitution
• techniques for integrating rational functions
• the method of integration by parts
• use of reduction formulae
• the Maple commands intparts, changevars, Int, convert/parfrac
You should be able to
• use substitution to evaluate integrals (definite and indefinite)
• integrate many rational functions ‘by hand’
• integrate products of certain types of functions
• use Maple to integrate using the methods of this chapter.
Chapter 6: Applications
§21 Lengths, surface areas and volumes
Consider the graph of a function f (x). We can use integration to compute the arc-length of the
curve given by this graph between two points (a, f (a)) and (b, f (b)) :
It is
Z q
b
2
1 + f 0 (x) dx.
arc-length =
a
Z
x=b
Sometimes this is written as
p
dx2 + dy 2 which shows the relationship with Pythagoras’s Theorem.
x=a
However, you should note that the integrals which arise from most examples are difficult/impossible to
integrate explicitly and are best evaluated numerically using Maple.
We can use the above formula to find the length of a curve defined implicitly. In fact once the
gradient is calculated one can put it into the formula for the integrand.
31
Practical Calculus
Surface of revolution: Consider the graph of y = f (x). Think of this in 3 dimensions. Rotate the
graph about the x-axis.
|f(a)|
ds
h
Calculate the area of the surface. We look at a ‘slice’ between x = a and x = a + h. It has
area= 2π|f (a)| ds. Hence the area of the surface between x = a and x = b is:
Z
b
b
Z
|f (x)| ds = 2π
2π
a
q
2
|f (x)| 1 + f 0 (x) dx.
a
Volume of solid of revolution: Consider the solid swept out when the area beneath a graph is rotated
about the x-axis.
Z
b
Same argument:
f 2 (x) dx.
Volume = π
a
Autumn 2010
32
§22 Functions defined using integrals
Z
x
dt
, the properties of the function follow fairly
t
1
easily using general properties of integrals. (See Problem Sheet 4, Qu 4.)
Some integrals which cannot be integrated in terms of the basic functions give rise to useful new
functions. For example, the gamma function is defined by
We can define the logarithm function by ln(x) =
Z
Γ(x) =
∞
tx−1 e−t dt.
0
This definition is for x > 0. To obtain one interesting relation for this function we integrate by parts:
dv
du
take u = tx−1 ,
= e−t , then
= (x − 1)tx−2 and v = −e−t :
dt
dt
Z ∞
h
it=∞
x−1 −t
for x > 1
Γ(x) = −t
e
+ (x − 1)
tx−2 e−t d = 0 + (x − 1)Γ(x − 1).
t=0
0
Thus for all x > 0 we have the recurrence relation
Γ(x + 1) = xΓ(x).
Note that Γ(1) = 1. Thus Γ(2) = 1 · Γ(1) = 1,
Γ(3) = 2 · Γ(2) = 2,
Γ(n + 1) = n · Γ(n) = n!,
One can also show that Γ( 12 ) =
√
π.
Γ(4) = 3 · Γ(3) = 3!. Hence,
for n = 0, 1, 2, · · · .
33
Practical Calculus
§23 Curve sketching
There is no definitive set of rules for sketching the graphs of functions but some of the following will be
useful.
(1) Does the curve have symmetries (eg about the origin or about one or both axes)?
(2) Find some particular points (x, y) that lie on the curve (eg for particular values of x or y such as
x = 0).
(3) If there are values of x which make y 2 negative, then there is no point of the curve with this value
of x. Also vice versa.
(4) Find the vertical asymptotes (the finite values a of x for which y → ±∞ as x → a).
(5) Compute lim y to give horizontal asymptotes.
x→±∞
(6) If the curve passes through the origin, find the slope(s) there.
(7) In the case of the graph of a function f , compute f 0 (x), f (2) (x) and so find and classify the stationary
points of f .
2
Example 1: y 2 = 5x
(1) Symmetric about the x-axis (y → −y).
(2) There are no points on the curve for which x < 0.
(3) Curve passes through (0, 0).
(4) Plot some values.
5
= ∞ when y = 0.
(5) 2y(x)y 0 (x) = 5
=⇒
y0 =
2y
1
0
0
1
x
−1
y
−2
x2
y2
+
=1
16
9
Symmetric about both axes.
There are no points on the curve for which y 2 > 9 and x2 > 16.
(0, ±3) and (±4, 0) lie on the curve.
x 2y(x)y 0 (x)
−9x
0, when x = 0,
+
=0
=⇒
y0 =
=
∞, when y = 0.
8
9
16y
Example 2:
(1)
(2)
(3)
(4)
2
0
−4
−2
0
x
y
−2
2
4
Autumn 2010
x2
1 + x2
Symmetric about the y-axis (x → −x).
There are no points on the curve for which y < 0.
(0, 0) on the curve.
y 0 = 0 at x = 0.
y < 1 for all x.
lim y = 1. So y = 1 is an asymptote.
34
Example 5: y =
(1)
(2)
(3)
(4)
(5)
(6)
x→±∞
1
y
0
0
−4
4
x
4
x2
Example 6: y =
1 − x2
(1) Symmetric about the y-axis.
(2) (0, 0) on the curve.
2x
(3) y 0 =
thus y 0 = 0 only at x = 0.
(1 − x2 )2
(4) x = ±1 are asymptotes.
(5) If |x| < 1 then y > 0 and if |x| > 1 then y < 0
(6) lim y = −1. So y = −1 is an asymptote.
2
0
−4
0
−2
4
2
x
y −2
x→±∞
−4
§24 Polar coordinates
r
It is often convenient to use other choices of coordinates to describe a graph or a curve. For example, a
circle centred at 0 can be most easily expressed in coordinates P = (r, θ), where r measures the distance
from the origin and θ, the angle between OP and the positive x-axis. These are called polar coordinates
and are constrained by r > 0 and 0 6 θ < 2π although sometimes it is more convenient to use −π 6 θ < π.
Each point (except the origin) in the plane R2 can be uniquely represented
by such a pair (r, θ). In these coordinates a circle centred at the origin of
y
P
radius R is simply given by the equation r = r(θ) = R.
To convert from the usual Cartesian coordinates (x, y) we use the
formulae
x = r cos θ,
y = r sin θ.
These can be solved for r and θ to give
p
r = + x2 + y 2 ,
θ = arctan(y/x).
!
x
35
Practical Calculus
Suppose a curve is given by r = r(θ), that is, the r as a function of θ. For the arclength we need
x(θ) = r(θ) cos θ,
y(θ) = r(θ) sin θ,
Thus
x0 (θ) = r0 (θ) cos θ − r(θ) sin θ,
=⇒
y 0 (θ) = r0 (θ) sin θ + r(θ) cos θ.
x02 (θ) + y 02 (θ) = r02 (θ) cos2 θ − 2r0 (θ)r(θ) cos θ sin θ + r2 (θ) sin2 θ
+ r02 (θ) sin2 θ + 2r0 (θ)r(θ) sin θ cos θ + r2 (θ) cos2 θ
= r02 (θ) + r2 (θ).
Hence,
arc-length =
Z p
Z p
x02 + y 02 =
r02 (θ) + r2 (θ) dθ.
Example: A cardioid is given by r = a(1 − cos θ), θ ∈ [0, 2π]. Then
r2 (θ) + r02 (θ) = a2 1 − 2 cos θ + cos2 θ + sin2 θ
= 2a2 (1 − cos θ) = 4a2 sin2 (θ/2).
Thus
Z
arc-length =
2π
p
r02 (θ) + r2 (θ) dθ
Z
= 2a
sin(θ/2) dθ
0
0
h
i2π
= −4a cos(θ/2)
= 8a.
0
Note that there is no π this answer !
2π
Autumn 2010
Area in polar coordinates: If a curve is given by r = r(θ), θ ∈ [α, β] then
area =
1
2
Z
36
y
β
r(θ)2 dθ.
α
Example: A circle with radius r = r(θ) = R.
R 2π
Then area= 12 0 R2 dθ = πR2 .
!
"
x
§25 Differential equations
In many ‘real life’ problems one obtains equations involving the derivatives of functions. To solve these
equations one needs new methods for their integration. Here we look at one very common example, the
decay equation.
Example When an atom of radioactive substance emits radiation it becomes non-active. The rate
of emission of the radiation from a mass is proportional to the number of its atoms that have not yet
radiated. So in a fixed amount of material, let R(t) be the amount of material that has not yet radiated
at time t; this amount decreases at a rate proportional to R(t); in mathematical terms, the function R(t)
dR
satisfies the equation
= −κR where κ > 0 is a constant which depends on the particular material.
dt
This is the decay equation.
We can solve this particular equation by rewriting it as
dR
= −κdt
R
and integrating to give
ln(R) = −κt + C
which (after applying exp to both sides and writing A for exp(C)) is rewritten as
R(t) = exp(−κt + C) = exp(C) exp(−κt) = Ae−κt .
If we know the amount of material R(0) at time t = 0, then this allows us to determine the ‘arbitrary’
constant A since from the expression for the solution with t = 0 one has A = R(0). Hence the solution
of the equation becomes
R(t) = R(0)e−κt .
The constant κ depends entirely on the particular material and is determined by its ‘half life’ T
-which is the time over which the amount of active material becomes half of the initial amount, so
R(T ) =
Now
1
ln 2
R(0)
, that is, e−κT = or κT = ln(2), so T =
.
2
2
κ
R(t + T ) = R(0)e−κt e−κT = 12 R(0)e−κt = 12 R(t).
Hence, the name ‘half life’ makes sense.
37
Practical Calculus
Summary
In this chapter, you should have learnt
• the formula for the arc-length of a curve.
• the formula for the surface area of a figure of revolution.
• the formula for the volume of a figure of revolution.
• the formula that defines the Γ function.
• what vertical and horizontal asymptotes are.
• the meaning of polar coordinate systems.
• an example of a differential equation.
You should be able to
• write down the formula for the arc-length of curves given as the graphs of functions or implicitly.
• compute the surface areas and volumes of surfaces of revolution.
• calculate the arc-length of curves described as graphs in Cartesian and polar coordinates.
• convert between Cartesian and polar coordinates.
• solve a growth/decay equation.