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Lecture Notes on Integral Calculus
UBC Math 103 Lecture Notes by Yue-Xian Li (Spring, 2004)
1 Introduction and highlights
Differential calculus you learned in the past term was about differentiation. You may feel embarrassed to find out that you have already forgotten a number of things that you learned differential calculus. However, if you still remember that differential calculus was about the rate of change, the slope of a graph, and the tangent of a curve, you are probably OK.
- The essence of differentiation is finding the ratio between the difference in the value of f (x) and the increment in x.
Remember, the derivative or the slope of a function is given by
f ′(x) =
df dx
= lim ∆x→ 0
f (x + ∆x) − f (x) ∆x
Integral calculus that we are beginning to learn now is called integral calculus. It will be mostly about adding an incremental process to arrive at a “total”. It will cover three major aspects of integral calculus:
- The meaning of integration.
- We’ll learn that integration and differentiation are inverse operations of each other. They are simply two sides of the same coin (Fundamental Theorem of Caclulus).
- The techniques for calculating integrals.
- The applications.
2 Sigma Sum
2.1 Addition re-learned: adding a sequence of numbers
In essence, integration is an advanced form of addition. We all started learning how to add two numbers since as young as we could remember. You might say “Are you kidding? Are you telling me that I have to start my university life by learning addition?”.
The answer is positive. You will find out that doing addition is often much harder than calculating an integral. Some may even find sigma sum is the most difficult thing to learn in integral calculus. Although this difficulty is by-passed by using the Fundamental Theorem of Caclulus, you should NEVER forget that you are actually doing a sigma sum when you are calculating an integral. This is one secret for correctly formulating the integral in many applied problems with ease!
Now, I use a couple of examples to show that your skills in doing addition still need improve- ment.
Example 1a: Find the total number of logs in a triangular pile of four layers (see figure).
Solution 1a: Let the total number be S 4 , where ‘S’ stands for ‘Sum’ and the subscript reminds us that we are calculating the sum for a pile of 4 layers.
S 4 = (^) ︸︷︷︸ 1 in layer 1
in layer 2
in layer 3
in layer 4
A piece of cake!
Example 1b: Now, find the total number of logs in a triangular pile of 50 layers, i.e. find S 50! (Give me the answer in a few seconds without using a calculator).
Solution 1b: Let’s start by formulating the problem correctly.
S 50 = (^) ︸︷︷︸ 1 in layer 1
in layer 2
in layer 49
in layer 50
where ‘· · · ’ had to be used to represent the numbers between 3 and 48 inclusive. This is because there isn’t enough space for writing all of them down. Even if there is enough space, it is tedious and unnecessary to write all of them down since the regularity of this sequence makes it very clear what are the numbers that are not written down.
Still a piece of cake? Not really if you had not learned Gauss’s formula. We’ll have to leave it unanswered at the moment.
Example 2: Finally, find the total number of logs in a triangular pile of k layers, i.e. find Sk (k is any positive integer, e.g. k = 8, 888 , 888 is one possible choice)!
Solution 2: This is equivalent to calculating the sum of the first k positive integers.
Sk = 1 + 2 + · · · + (k − 1) + k.
The only thing we can say now is that the answer must be a function of k which is the total number of integers we need to add. Again, we have to leave it unanswered at the moment.
2.3 The sigma notation
In order to short-hand the mathematical exression of the sum of a regular sequence, a conve- nient notation is introduced.
Definition (Σ sum): The sum of the first k terms of a sequence generated by the sequence generator f (i) can be denoted by
Sk = f (1) + f (2) + · · · + f (k) ≡
∑^ k
i=
f (i)
where the symbol Σ (the Greek equivalent of S reads “sigma”) means “take the sum of”, the general expression for the terms to be added or the sequence generator f (i) is called the summand, i is called the summation index, 1 and k are, respectively, the starting and the ending indices of the sum.
Thus, ∑k
i=
f (i)
means calculate the sum of all the terms generated by the sequence generator f (i) for all integers starting from i = 1 and ending at i = k.
Note that the value of the sum is independent of the summation index i, hence i is called a “dummy” variable serving for the sole purpose of running the summation from the starting index to the ending index. Therefore, the sum only depends on the summand and both the starting and the ending indices.
Example 5: Express the sum Sk =
∑k i=
i^2 in an expanded form.
Solution 5: The sequence generator is f (i) = i^2. Note that the starting index is not 1 but 3!. Thus, the 1st term is f (3) = 3^2. The subsequent terms can be determined accordngly. Thus,
Sk =
∑^ k
i=
i^2 = f (3) + f (4) + f (5) + · · · + f (k) = 3^2 + 4^2 + 5^2 + · · · + k^2.
An easy check for a mistake that often occurs. If you still find the “dummy” variable i in an expanded form or in the final evaluation of the sum, your answer must be WRONG.
2.4 Gauss’s formula and other formulas for simple sums
Let us return to Examples 1 and 2 about the total number of logs in a triangular pile. Let’s start with a pile of 4 layers. Imaging that you could (in a “thought-experiement”) put an
identical pile with up side down adjacent to the original pile, you obtain a pile that contains twice the number of logs that you want to calculate (see figure).
The advantage of doing this is that, in this double-sized pile, each layer contains an equal number of logs. This number is equal to number on the 1st (top) layer plus the number on the 4th (bottom) layer. In the mean time, the height of the pile remains unchanged (4). Thus, the number in this double-sized pile is 4 × (4 + 1) = 20. The sum S 4 is just half of this number which is 10.
Let’s apply this idea to finding the formula in the case of k layers. Note that
(Original) Sk = 1 + 2 + · · · + k − 1 + k. (2) (Inversed) Sk = k + k − 1 + · · · + 2 + 1. (3) (Adding the two) 2 Sk = ( ︸k + 1) + (k + 1) + · · ·︷︷ + (k + 1) + (k + 1)︸ k terms in total
= k(k + 1). (4)
Dividing both sides by 2, we obtain Gauss’s formula for the sum of the first k positive integers.
Sk =
∑^ k
i=
i = 1 + 2 + · · · + k =
k(k + 1). (5)
This actaully answered the problem in Example 2. The answer to Example 1b is even simpler.
S 50 =
∑^50
i=
i =
× 50 × 51 = 1275.
The following are two important simple sums that we shall use later. One is the sum of the first k integers squared.
Sk =
∑^ k
i=
i^2 = 1^2 + 2^2 + · · · + k^2 =
k(k + 1)(2k + 1). (6)
The other is the sum of the first k integers cubed.
Sk =
∑^ k
i=
i^3 = 1^3 + 2^3 + · · · + k^3 =
[
k(k + 1)
] 2
We shall not illustrate how to derive these formulas. You can find it in numerous calculus text books.
To prove that these formulas work for arbitrarily large integers k, we can use a method called mathematical induction. To save time, we’ll just outline the basic ideas here.
where Rules 1, 2, and 3 are used in the last two steps. Using the known formulas, we obtain
Ok =
∑^ k
1
(2i − 1) = k(k + 1) − k = k^2.
Example 8: Let us now go back to solve Example 4. It is the sum of the first k products of pairs of subsequent integers.
Solution 8:
Pk = 1 · 2 + 2 · 3 + · · · + k · (k + 1) =
∑^ k
1
i(i + 1) =
∑^ k
1
[i^2 + i] =
∑^ k
1
i^2 +
∑^ k
1
i,
where Rule 3 was used in the last step. Applying the formulas, we learned
Pk =
∑^ k
1
i(i + 1) =
k(k + 1)(2k + 1) +
k(k + 1) =
k(k + 1)(k + 2).
This is another simple sum that we can easily remember.
Example 9: Calculate the sum S =
∑k 1
(i + 2)^3.
Solution 9: This problem can be solved in two different ways. The first is to expand the summand f (i) = (i + 2)^3 which yield
S =
∑^ k
1
(i + 2)^3 =
∑^ k
1
(i^3 + 6i^2 + 12i + 8) =
∑^ k
1
i^3 + 6
∑^ k
1
i^2 + 12
∑^ k
1
i + 8k.
We can solve the resulting three sums separately using the known formulas.
But there is a better way to solve this. This involves substituting the summation index. We find it easier to see how substitution works by expanding the sum.
S =
∑^ k
i=
(i + 2)^3 = 3^3 + 4^3 + · · · + k^3 + (k + 1)^3 + (k + 2)^3.
We see that this is simply a sum of integers cubed. But the sum does not start at 1^3 and end at k^3 like in the formula for the sum of the first k integers cubed.
Thus we re-write the sum with a sigma notation with an new index called l which starts at l = 3 and ends at l = k + 2 (there is no need to change the symbol for the index, you can keep calling it i if you do not feel any confusion). Thus,
S =
∑^ k
i=
(i + 2)^3 = 3^3 + 4^3 + · · · + k^3 + (k + 1)^3 + (k + 2)^3 =
∑^ k+
l=
l^3.
We just finished doing a substitution of the summation index. It is equivalent to replacing i by l = i + 2. This relation also implies that i = 1 ⇒ l = 3 and i = k ⇒ l = k + 2. This is actually how you can determine the starting and the ending values of the new index.
Now we can solve this sum using Rule 4 and the known formula.
S =
∑^ k
i=
(i+2)^3 =
∑^ k+
l=
l^3 =
∑^ k+
l=
l^3 −
∑^2
l=
l^3 =
[
(k + 2)(k + 3)
] 2
[
(k + 2)(k + 3)
] 2
2.6 Applications of sigma sum
The area under a curve
We know that the area of a rectangle with length l and width w is Arect = w · l.
Starting from this formula we can calculate the area of a triangle and a trapezoid. This is because a triangle and a trapezoid can be transformed into a rectangle (see Figure). Thus, for a triangle of height h and base length b
Atrig =
hb.
Similarly, for a trapezoid with base length b, top length t, and height h
Atrap =
h(t + b).
Following a very similar idea, the sum of a trapezoid-shaped pile of logs with t logs on top layer, b logs on the bottom layer, and a height of h = b − t + 1 layers (see figure) is
∑^ b
i=t
i = t + (t + 1) + · · · + (b − 1) + b =
h(t + b) =
(b − t + 1)(t + b). (8)
Now returning to the problem of calculating the area. Another important formula is for the area of a circle of radius r. Acirc = πr^2.
Now, once we learned sigma and/or integration, we can calculate the area under the curve of any function that is integrable.
Example 10: Calculate the area under the curve y = x^2 between 0 and 2 (see figure).
Solution 10: Remember always try to reduce a problem that you do not know how to solve into a problem that you know how to solve.
2.7 The sum of a geometric sequence
3 The Definite Integrals and the Fundamental Theorem
3.1 Riemann sums
Definition 1: Suppose f (x) is finite-valued and piecewise continuous on [a, b]. Let P = {x 0 = a, x 1 , x 2 ,... , xn = b} be a partition of [a, b] into n subintervals Ii = [xi− 1 , xi] of width ∆xi = xi − xi− 1 , i = 1, 2 ,... , n. (Note in a special case, we can partition it into subintervals of equal width: ∆xi = xi − xi− 1 = w = (b − a)/n for all i). Let x∗ i be a point in Ii such that xi− 1 ≤ x∗ i ≤ xi. (Here are some special ways to choose x∗ i : (i) left endpoint rule x∗ i = xi− 1 , and (ii) the right endpoint rule x∗ i = xi).
The Riemann sums of f (x) on the interval [a, b] are defined by:
Rn =
∑^ n
i=
hi∆xi =
∑^ n
i=
f (x∗ i )∆xi
which approximate the area between f (x) and the x-axis by the sum of the areas of n thin rectangles (see figure).
Example 1: Approximate the area under the curve of y = ex^ on [0, 1] by 10 rectangles of equal width using the left endpoint rule.
Solution 1: We are actually calculating the Riemann sum R 10. The width of each rectangle is ∆xi = w = 1/10 = 0.1 (for all i). The left endpoint of each subinterval is x∗ i = xi − 1 = (i − 1)/10. Thus,
R 10 =
∑^10
i=
f (x∗ i )∆xi =
∑^10
i=
e(i−1)/^10 (0.1) = 0. 1
∑^9
j=
ej/^10 = 0. 1
1 − e 1 − e^1 /^10
3.2 The definite integral
Definition 1: Suppose f (x) is finite-valued and piecewise continuous on [a, b]. Let P = {x 0 = a, x 1 , x 2 ,... , xn = b} be a partition of [a, b] with a length defined by |p| = Max 1 ≤i≤n{∆xi} (i.e. the longest of all subintervals). The definite integral of f (x) on [a, b] is
∫^ b
a
f (x)dx = lim |p|→0; n→∞
Rn = lim |p|→0; n→∞
∑^ n
i=
f (x∗ i )∆xi
where the symbol
means “to integrate”, the function f (x) to be integrated is called the integrand, x is called the integration variable which is a “dummy” variable, a and b are,
respectively, the lower (or the starting) limit and the upper (or the ending) limit of the integral. Thus, ∫b
a
f (x)dx
means integrate the function f (x) starting from x = a and ending at x = b.
Example 2: Calculate the definite integral of f (x) = x^2 on [0.2]. (This is Example 10 of Lecture 1 reformulated in the form of a definite integral).
Solution 2:
I =
∫^2
0
x^2 dx = lim n→∞
∑^ n
i=
2 i n
f (x∗ i )
n
∆xi
= lim n→∞
n^3
) (^) ∑n
i=
i^2 = lim n→∞
(n + 1)(2n + 1) n^2
Important remarks on the relation between an area and a definite integral:
- An area, defined as the physical measure of the size of a 2D domain, is always non- negative.
- The value of a definite integral, sometimes also referred to as an “area”, can be both positive and negative.
- This is because: a definite integral = the limit of Riemann sums. But Riemann sums are defined as Rn =
∑^ n
i=
(area of ith^ rectangle) =
∑^ n
i=
f (x∗ i ) ︸ ︷︷ ︸ height
∆ ︸︷︷︸xi width
- Note that both the hieght f (x∗ i ) and the width ∆xi can be negative implying that Rn can have either signs.
3.3 The fundamental theorem of calculus
3.4 Areas between two curves
5 Differentials
Definition: The differential, dF , of any differentiable function F is an infinitely small incre- ment or change in the value of F.
Remark: dF is measured in the same units as F itself.
Example: If x is the position of a moving body measured in units of m (meters), then its differential, dx, is also in units of m. dx is an infinitely small increment/change in the position x.
Example: If t is time measured in units of s (seconds), then its differential, dt, is also in units of s. dt is an infinitely small increment/change in time t.
Example: If A is area measured in units of m^2 (square meters), then its differential, dA, also is in units of m^2. dA is an infinitely small increment/change in area A.
Example: If V is volume measured in units of m^3 (cubic meters), then its differential, dV , also is in units of m^3. dV is an infinitely small increment/change in volume V.
Example: If v is velocity measured in units of m/s (meters per second), then its differential, dv, also is in units of m/s. dv is an infinitely small increment/change in velocity v.
Example: If C is the concentration of a biomolecule in our body fluid measured in units of M (1 M = 1 molar = 1 mole/litre, where 1 mole is about 6. 023 × 1023 molecules), then its differential, dC, also is in units of M. dC is an infinitely small increment/change in the concentration C.
Example: If m is the mass of a rocket measured in units of kg (kilograms), then its differen- tial, dm, also is in units of kg. dm is an infinitely small increment/change in the mass m.
Example: If F (h) is the culmulative probability of finding a man in Canada whose height is smaller than h (meters), then dF is an infinitely small increment in the probability.
Definition: The derivative of a function F with respect to another function x is defined as the quotient between their differentials:
dF dx
an inf initely small rise in F an inf initely small run in x
Example: Velocity as the rate of change in position x with respect to time t can be expressed as
v =
inf initely small change in position inf initely small time interval
dx dt
Remark: Many physical laws are correct only when expressed in terms of differentials.
Example: The formula (distance) = (velocity) × (time interval) is true either when the velocity is a constant or in terms of differentials, i.e., (an inf initely small distance) = (velocity) × (an inf initely short time interval). This is because in an infinitely short time interval, the velocity can be considered a constant. Thus,
dx =
dx dt
dt = v(t)dt,
which is nothing new but v(t) = dx/dt.
Example: The formula (work) = (f orce) × (distance) is true either when the force is a constant or in terms of differentials, i.e., (an inf initely small work) = (f orce) × (an inf initely small distance). This is because in an infinitely small distance, the force can be considered a constant. Thus,
dW = f dx = f
dx dt
dt = f (t)v(t)dt,
which simply implies: (1) f = dW/dx, i.e., force f is nothing but the rate of change in work W with respect to distance x; (2) dW/dt = f (t)v(t), i.e., the rate of change in work W with respect to time t is equal to the product between f (t) and v(t).
Example: The formula (mass) = (density) × (volume) is true either when the density is constant or in terms of differentials, i.e., (an inf initely small mass) = (density) ×
df dx
d sin(ln(x^2 + ex)) d ln(x^2 + ex)
d ln(x^2 + ex) d(x^2 + ex)
d(x^2 + ex) dx
= cos(ln(x^2 + ex))
(x^2 + ex)
(2x + ex).
7 The Product Rule in Terms of Differentials
The Product Rule says if both u = u(x) and v = v(x) are differentiable functions of x, then
d(uv) dx
= v
du dx
dv dx
Multiply both sides by the differential dx, we obtain
d(uv) = vdu + udv,
which is the Product Rule in terms of differentials.
Example: Let u = x^2 and v = esin(x^2 ), then
d(uv) = vdu + udv = esin(x
(^2) ) d(x^2 ) + x^2 d(esin(x
(^2) ) )
= 2xesin(x
(^2) ) dx + x^2 esin(x
(^2) ) cos(x^2 )(2x)dx = 2xesin(x
(^2) ) [1 + x^2 cos(x^2 )]dx.
8 Other Properties of Differentials
- For any differentiable function F (x), dF = F ′(x)dx (Recall that F ′(x) = dF/dx!).
- For any constant C, dC = 0.
- for any constant C and differentiable function F (x), d(CF ) = CdF = CF ′(x)dx.
- For any differentiable functions u and v, d(u ± v) = du ± dv.
9 The Fundamental Theorem in Terms of Differentials
Fundamental Theorem of Calculus: If F (x) is one antiderivative of the function f (x), i.e., F ′(x) = f (x), then
∫ f (x)dx =
F ′(x)dx =
dF = F (x) + C.
Thus, the integral of the differential of a function F is equal to the function itself plus an arbitrary constant. This is simply saying that differential and integral are inverse math operations of each other. If we first differentiate a function F (x) and then integrate the derivative F ′(x) = f (x), we obtain F (x) itself plus an arbitrary constant. The opposite also is true. If we first integrate a function f (x) and then differentiate the resulting integral F (x)+C, we obtain F ′(x) = f (x) itself.
Example: (^) ∫
x^5 dx =
d(
x^6 6
x^6 6
+ C.
Example: (^) ∫
e−xdx =
d(−e−x) = −e−x^ + C.
Example: (^) ∫
cos(3x)dx =
d(
sin(3x) 3
sin(3x) 3
+ C.
Example: (^) ∫
sec^2 xdx =
dtanx = tanx + C.
Example: (^) ∫
cosh(3x)dx =
d(
sinh(3x) 3
sinh(3x) 3
+ C.
sin(
x) √ x
dx =
sin(
x) √ x
d(
x)^2 =
sin(
x) √ x
xd
x = 2
sin(
x)d
x = − 2 cos(
x)+C.
Thus, as soon as you realize that
x is the substitution, your goal is to change the differential in the integral dx into the differential of
x which is d
x.
If you feel that you cannot do it without the actual substitution, that is fine. You can always do the actual substitution. I here simply want to teach you a way that actual subsitution is not a necessity!
Example:
x^4 cos(x^5 )dx.
Solution: Note that cos(x^5 ) is a composite function that becomes a simple cosine function only if the subsitution u = x^5 is made. Since du = u′dx = 5x^4 dx, x^4 dx = 15 du. Thus,
∫ x^4 cos(x^5 )dx =
cos(u)du =
sin(u) 5
+ C =
sin(x^5 ) 5
+ C.
Or alternatively,
∫ x^4 cos(x^5 )dx =
cos(x^5 )dx^5 =
sin(x^5 ) + C.
Example:
x
x^2 + 1dx.
Solution: Note that
x^2 + 1 = (x^2 + 1)^1 /^2 is a composite function. We realize that u = x^2 + 1 is a substitution. du = u′dx = 2xdx implies xdx = 12 du. Thus,
∫ x
x^2 + 1dx =
udu =
u^3 /^2 + C =
(x^2 + 1)^3 /^2 3
+ C.
Or alternatively,
∫ x
x^2 + 1dx =
x^2 + 1d(x^2 + 1) =
(x^2 + 1)^3 /^2 + C =
(x^2 + 1)^3 /^2 3
+ C.
In many cases, substitution is required even no obvious composite function is involved.
Example:
(ln x/x)dx (x > 0).
Solution: The integrand ln x/x is not a composite function. Nevertheless, its antiderivative is not obvious to calculate. We need to figure out that (1/x)dx = d ln x, thus by introducing the substitution u = ln x, we obtained a differential of the function ln x which also appears in the integrand. Therefore,
∫ (ln x/x)dx =
udu =
u^2 2
+ C =
(ln x)^2 + C.
It is more natural to consider this substitution is an attempt to change the differential dx into something that is identical to a function that appears in the integrand, namely d ln x. Thus,
∫ (ln x/x)dx =
ln xd ln x =
(ln x)^2 + C.
Example:
tan xdx.
Solution: tan x is not a composite function. Nevertheless, it is not obvious to figure out tan x is the derivative of what function. However, if we write tan x = sin x/ cos x, we can regard 1 / cos x as a composite function. We see that u = cosx is a candidate for substitution and du = u′dx = −sin(x)dx. Thus,
∫ tanxdx =
sinx cosx
dx = −
du u
= − ln |u| + C = − ln |cosx| + C.
Or alternatively,
∫ tanxdx =
sinx cosx
dx = −
dcosx cosx
= − ln |cosx| + C.
Some substitutions are standard in solving specific types of integrals.
Example: Integrands of the type (a^2 − x^2 )±^1 /^2.
In this case both x = asinu and x = acosu will be good. x = a tanh u also works (1−tanh^2 u = sech^2 u). Let’s pick x = asinu in this example. If you ask how can we find out that x = asinu
