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Title: Integration
Description: Basic concepts of integration in a simplified manner.
Description: Basic concepts of integration in a simplified manner.
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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
...
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
...
Remember, the derivative or the slope of a function is given by
f (x) =
df
f (x + ∆x) − f (x)
= lim
...
It will be
mostly about adding an incremental process to arrive at a “total”
...
The meaning of integration
...
They are simply two sides of the same coin (Fundamental Theorem of Caclulus)
...
The techniques for calculating integrals
...
The applications
...
1
Sigma Sum
Addition re-learned: adding a sequence of numbers
In essence, integration is an advanced form of addition
...
You might say “Are you kidding? Are
you telling me that I have to start my university life by learning addition?”
...
You will find out that doing addition is often much harder than
calculating an integral
...
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
...
Example 1a: Find the total number of logs in a triangular pile of four layers (see figure)
...
S4 =
1
+
in layer 1
2
+
in layer 2
3
+
in layer 3
4
= 10
...
e
...
Solution 1b: Let’s start by formulating the problem correctly
...
This is
because there isn’t enough space for writing all of them down
...
Still a piece of cake? Not really if you had not learned Gauss’s formula
...
Example 2: Finally, find the total number of logs in a triangular pile of k layers, i
...
find Sk
(k is any positive integer, e
...
k = 8, 888, 888 is one possible choice)!
Solution 2: This is equivalent to calculating the sum of the first k positive integers
...
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
...
2
2
...
A sequence is regular if each term
of the sequence is uniquely determined, following a well-defined rule, by its position/order in
the sequence (often denoted by an integer i)
...
g
...
We can call this
formula the sequence generator or the general term
...
Thus, the 9th term is 9 while the 109th term is equal
to 109
...
Find the sequence generator
...
The ith
even number is simply 2i
...
To verify,
5 is the 3rd odd number, i
...
i = 3
...
Knowing the sequence generator, we can write down the sum of the first k odd numbers for
any positive integer k
...
Example 4: Find the sequence generator of the following sum of 100 products of subsequent
pairs of integers
...
Thus, f (i) = i(i + 1)
...
Pk = 1 · 2 + 2 · 3 + · · · + k(k + 1)
...
1415926
...
We cannot find the sequence generator since the sequence is irregular, we cannot express the
sum of the first k digits of π (for arbitrary k)
...
3
The sigma notation
In order to short-hand the mathematical exression of the sum of a regular sequence, a convenient notation is introduced
...
Thus,
k
f (i)
i=1
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
...
Therefore, the sum only depends on the summand and both the
starting and the ending indices
...
i=3
Solution 5: The sequence generator is f (i) = i2
...
Thus, the 1st term is f (3) = 32
...
Thus,
k
Sk =
i=3
i2 = f (3) + f (4) + f (5) + · · · + f (k) = 32 + 42 + 52 + · · · + k 2
...
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
...
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
...
Imaging that you could (in a “thought-experiement”) put an
4
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)
...
This number is equal to number on the 1st (top) layer plus the number on the
4th (bottom) layer
...
Thus, the
number in this double-sized pile is 4 × (4 + 1) = 20
...
Let’s apply this idea to finding the formula in the case of k layers
...
(Inversed) Sk =
k
+ k − 1 +···+
2
+ 1
...
(2)
(3)
(4)
k terms in total
Dividing both sides by 2, we obtain Gauss’s formula for the sum of the first k positive integers
...
2
(5)
This actaully answered the problem in Example 2
...
50
S50 =
i=
i=1
1
× 50 × 51 = 1275
...
One is the sum of the
first k integers squared
...
6
(6)
The other is the sum of the first k integers cubed
...
(7)
We shall not illustrate how to derive these formulas
...
To prove that these formulas work for arbitrarily large integers k, we can use a method called
mathematical induction
...
5
The only way we can prove things concerning arbitrarily large numbers is to guarantee that
this formula must be correct for k = N + 1 if it is correct for k = N
...
To guarantee that all the dominos in the
string/sequence fall one after the other, we need to guarantee that the falling of each domino
will necessarily cause the falling of the subsequent one
...
The last thing you need to do
is to knock down the first one and keep your fingers crossed
...
2
...
k
i=1
c = c + c + · · · + c = kc
...
Rule 2: Constant multiplication: multiplying a sum by a constant is equal to multiplying each
term of the sum by the same constant
...
k
k
k
f (i) +
i=1
[f (i) + g(i)]
...
k
n
f (i) =
i=1
k
f (i) +
i=1
f (i),
i=n+1
(1 ≤ n < k)
...
Solution 7:
k
Ok = 1 + 3 + · · · + (2k − 1) =
1
k
(2i − 1) =
6
1
k
(2i) −
k
1=2
1
1
i − k,
where Rules 1, 2, and 3 are used in the last two steps
...
Example 8: Let us now go back to solve Example 4
...
Solution 8:
k
Pk = 1 · 2 + 2 · 3 + · · · + k · (k + 1) =
k
k
k
2
i(i + 1) =
2
[i + i] =
1
1
i +
1
i,
1
where Rule 3 was used in the last step
...
6
2
3
Pk =
1
This is another simple sum that we can easily remember
...
1
Solution 9: This problem can be solved in two different ways
...
1
We can solve the resulting three sums separately using the known formulas
...
This involves substituting the summation index
...
k
S=
i=1
(i + 2)3 = 33 + 43 + · · · + k 3 + (k + 1)3 + (k + 2)3
...
But the sum does not start at 13 and end
at k 3 like in the formula for the sum of the first k integers cubed
...
Thus,
k
k+2
3
S=
i=1
3
3
3
3
3
(i + 2) = 3 + 4 + · · · + k + (k + 1) + (k + 2) =
7
l3
...
It is equivalent to replacing i
by l = i + 2
...
This is
actually how you can determine the starting and the ending values of the new index
...
k
k+2
3
S=
(i+2) =
i=1
2
...
2
2
3
l=1
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
...
This is
because a triangle and a trapezoid can be transformed into a rectangle (see Figure)
...
2
Similarly, for a trapezoid with base length b, top length t, and height h
1
Atrap = h(t + b)
...
2
2
(8)
Now returning to the problem of calculating the area
...
Acirc = πr 2
...
Example 10: Calculate the area under the curve y = x2 between 0 and 2 (see figure)
...
8
Let the area be A
...
Thus,
A ≈ w ·h1 +w ·h2 +w ·h3 =
w[x2 + x2 + x2 ] = 1 [02 + 1
0
1
2
3
3
2
1
2
2
w[x1 + x2 + x2 ] = 3 [ 1 +
3
3
2
2 2
],
3
2 2
3 2
+ 3 ],
3
+
left − end approx
...
By using these rectangles, we introduced large errors in our estimates using the heights based
on both the left and the right endpoints of the subintervals
...
Let us now divide A into n
rectangles of equal width w = 2/n
...
i
i=1
It is very important to keep a clear account of the height of each rectangle
...
Thus, the key is finding the x-coordinate of the right-end of each rectangle
...
x0 = 0 and ∆x = 2 for this example
...
Substitute into the above equation
2
A ≈ Sn =
n
n
x2
...
36
n
3
n2
Note that if we divide the area into infinitely many retangles with a width that is infinitely
small, the approximate becomes accurate
...
2
n→∞ 3
n
3
A = lim Sn = lim
n→∞
The volume of a solid revolution of a curve
We know the volume of a rectangular block of height h, width w, and length l is
Vrect = w · l · h
...
Example 11: Calculate the volume of the bowl-shaped solid obtained by rotating the curve
y = x2 on [0, 2] about the y-axis
...
7
3
3
...
Let P =
{x0 = a, x1 , x2 ,
...
, n
...
Let x∗ be a point in Ii such that
i
xi−1 ≤ x∗ ≤ xi
...
i
The Riemann sums of f (x) on the interval [a, b] are defined by:
n
n
f (x∗ )∆xi
i
hi ∆xi =
Rn =
i=1
i=1
which approximate the area between f (x) and the x-axis by the sum of the areas of n thin
rectangles (see figure)
...
Solution 1: We are actually calculating the Riemann sum R10
...
1 (for all i)
...
Thus,
10
10
f (x∗ )∆xi =
i
R10 =
i=1
3
...
1) = 0
...
1
j=0
1−e
≈ 1
...
1 − e1/10
The definite integral
Definition 1: Suppose f (x) is finite-valued and piecewise continuous on [a, b]
...
, xn = b} be a partition of [a, b] with a length defined by |p| = M ax1≤i≤n {∆xi }
(i
...
the longest of all subintervals)
...
Thus,
b
f (x)dx
a
means integrate the function f (x) starting from x = a and ending at x = b
...
2]
...
Solution 2:
2
n
2
I=
x dx = lim
n→∞
0
i=1
2
2i
n
f (x∗ )
i
2
n
= lim
n→∞
8
n3
n
4 (n + 1)(2n + 1)
8
=
...
• The value of a definite integral, sometimes also referred to as an “area”, can be both
positive and negative
...
But Riemann sums
are defined as
n
n
f (x∗ ) ∆xi
...
3
...
4
Areas between two curves
11
4
Applications of Definite Integrals: I
4
...
2
Density and total mass
4
...
4
The average value of a function
12
5
Differentials
Definition: The differential, dF , of any differentiable function F is an infinitely small increment or change in the value of F
...
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
...
Example: If t is time measured in units of s (seconds), then its differential, dt, is also in units
of s
...
Example: If A is area measured in units of m2 (square meters), then its differential, dA, also
is in units of m2
...
Example: If V is volume measured in units of m3 (cubic meters), then its differential, dV ,
also is in units of m3
...
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
...
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
...
dC is an infinitely small increment/change in the
concentration C
...
dm is an infinitely small increment/change in the mass m
...
Definition: The derivative of a function F with respect to another function x is defined as
the quotient between their differentials:
13
dF
an inf initely small rise in F
=
...
inf initely small time interval
dt
Remark: Many physical laws are correct only when expressed in terms of differentials
...
e
...
This is because in an infinitely short time
interval, the velocity can be considered a constant
...
Example: The formula (work) = (f orce) × (distance) is true either when the force
is a constant or in terms of differentials, i
...
, (an inf initely small work) = (f orce) ×
(an inf initely small distance)
...
Thus,
dW = f dx = f
dx
dt = f (t)v(t)dt,
dt
which simply implies: (1) f = dW/dx, i
...
, 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
...
, the rate of change in work W with
respect to time t is equal to the product between f (t) and v(t)
...
e
...
This is because in an infinitely small piece of
volume, the density can be considered a constant
...
e
...
6
The Chain Rule in Terms of Differentials
When we differentiate a composite function, we need to use the Chain Rule
...
This is because f is not an exponential function of x but
it is an exponential function of u = x2 which is itself a power function of x
...
Therefore, the Chain Rule can be simply interpreted as the quotient between df and dx is equal
to the quotient between df and du multiplied by the quotient between du and dx
...
However, if we simply regard x2 as a function different from x, the actual substitution u = x2
becomes unnecessary
...
2
dx
dx dx
dx
dx
Generally, if f = f (g(x)) is a differentiable function of g and g is a differentiable function of
x, then
df
df dg
=
,
dx
dg dx
where df , dg, and dx are, respectively, the differentials of the functions f , g, and x
...
Solution: f is a composite function of another composite function!
15
df
d sin(ln(x2 + ex )) d ln(x2 + ex ) d(x2 + ex )
1
=
= cos(ln(x2 + ex )) 2
(2x + ex )
...
dx
dx
dx
Multiply both sides by the differential dx, we obtain
d(uv) = vdu + udv,
which is the Product Rule in terms of differentials
...
8
Other Properties of Differentials
1
...
2
...
3
...
4
...
16
9
The Fundamental Theorem in Terms of Differentials
Fundamental Theorem of Calculus: If F (x) is one antiderivative of the function f (x), i
...
,
F (x) = f (x), then
f (x)dx =
F (x)dx =
dF = F (x) + C
...
This is simply saying that differential and integral are inverse
math operations of each other
...
The opposite also is
true
...
Example:
x5 dx =
d(
x6
x6
)=
+ C
...
Example:
cos(3x)dx =
d(
sin(3x)
sin(3x)
)=
+ C
...
Example:
cosh(3x)dx =
d(
sinh(3x)
sinh(3x)
)=
+ C
...
1
dx =
sech2 x
d(x +
cosh2 xdx
1
sinh(2x)
sinh(2x)
) = [x +
] + C
...
10
Integration by Subsitution
Substitution is a necessity when integrating a composite function since we cannot write down
the antiderivative of a composite function in a straightforward manner
...
However, there is a general rule in substitution, namely, to
change the composite function into a simple, elementary function
...
√
Solution: Note that sin( x) is not an elementary sine function but a composite function
...
Once you realize this, u = x is an obvious subsitution
...
Substitute into the integral, we obtain
2 x
Example:
√
sin( x)
√
dx =
x
sin(u)
2udu = 2
u
sin(u)du = −2
√
dcos(u) = −2cos(u)+C = −2cos( x)+C
...
Note that x = ( x)2 ,
18
√
sin( x)
√
dx =
x
√
sin( x) √ √
√
2 xd x = 2
x
√
sin( x) √ 2
√
d( x) =
x
√ √
√
sin( x)d x = −2cos( x)+C
...
If you feel that you cannot do it without the actual substitution, that is fine
...
I here simply want to teach you a way that actual subsitution is not
a necessity!
Example:
x4 cos(x5 )dx
...
Since du = u dx = 5x4 dx, x4 dx = 5 du
...
5
5
Or alternatively,
x4 cos(x5 )dx =
1
5
1
cos(x5 )dx5 = sin(x5 ) + C
...
√
Solution: Note that x2 + 1 = (x2 + 1)1/2 is a composite function
...
du = u dx = 2xdx implies xdx = 2 du
...
23
3
Or alternatively,
√
1
x x2 + 1dx =
2
√
x2 + 1d(x2 + 1) =
19
(x2 + 1)3/2
12 2
(x + 1)3/2 + C =
+ C
...
Example:
(ln x/x)dx (x > 0)
...
Nevertheless, its antiderivative is
not obvious to calculate
...
Therefore,
(ln x/x)dx =
udu =
1
u2
+ C = (ln x)2 + C
...
Thus,
(ln x/x)dx =
Example:
1
ln xd ln x = (ln x)2 + C
...
Solution: tan x is not a composite function
...
However, if we write tan x = sin x/ cos x, we can regard
1/ cos x as a composite function
...
Thus,
tanxdx =
sinx
dx = −
cosx
du
= − ln |u| + C = − ln |cosx| + C
...
cosx
Some substitutions are standard in solving specific types of integrals
...
In this case both x = asinu and x = acosu will be good
...
Let’s pick x = asinu in this example
...
This will help us eliminate
the half power in the integrand
...
Thus,
x
dx
d(asinu)
acosudu
√
= du = u + C = sin−1 + C
...
2
a
1−u
Similarly,
√
a2 − x2 dx =
=
√
a2 − a2 sin2 ud(asinu) = a2
1 + cos(2u)
du
2
cos2 udu = a2
√
sin(2u)
a2
1
x
a2
[u +
] + C = [u + sinucosu] + C = [a2 sin−1 ( ) + x a2 − x2 ] + C
...
x = asinh(u) is a good substitution since a2 + x2 = a2 + a2 sinh2 (u) = a2 (1 + sinh2 (u)) =
a2 cosh2 (u), where the hyperbolic identity 1 + sinh2 (u) = cosh2 (u) was used
...
Thus,
√
dx
=
a2 + x2
dasinh(u)
a2 + a2 sinh2 (u)
where sinh−1 (x/a) = ln |x +
√
a2 + x2 dx =
√
=
= ln |x +
acosh(u)du
=
acosh(u)
√
x
du = u + C = sinh−1 ( ) + C
a
a2 + x2 | + C,
a2 + x2 | − ln |a| was used
...
Example: Integrands of the type (x2 − a2 )±1/2
...
(x = a sec u is also good since
sec2 u − 1 = tan2 u!)
...
a
Similarly
√
x2 − a2 dx =
a2 cosh2 (u) − a2 da cosh(u) = a2
sinh2 udu =
a2
2
[cosh(2u) − 1]du
√
1 √
= [x x2 − a2 − a2 ln |x + x2 − a2 |] + C,
2
√
where cosh u = x/a, sinh u = x2 /a2 − 1, and u = ln |x + x2 − a2 | − ln a were used
...
More Exercises on Substitution:
22
1
...
2
2
(ii)
√
1
x
dx =
4
2x + 3
2x + 3 − 3
1
√
d(2x + 3) =
4
2x + 3
u−3
1
du =
1/2
u
4
[u1/2 − 3u−1/2 ]du
1 2
1 2
= [ u3/2 − 6u1/2 ] + C = [ (2x + 3)3/2 − 6(2x + 3)1/2 ] + C
...
7 Problems
...
2 Problems
...
2
...
1 + u2
(ii)
√
=−
x
e
2x2
−1
dx2
1
2
dx =
e
2x2
1
de−u
=−
2
1 − (e−u )2
1
2
(1 − e
−2x2
)
=
1
2
du
e2u (1 − e−2u )
=
1
2
du
eu
1 − (e−u )2
1
1
1
2
= cos−1 (y)+C = cos−1 (e−u )+C = cos−1 (e−x )+C
...
2 Problems (difficult ones!)
...
Special Trigonometric Substitutions
Example:
(i)
√
x3
dx =
1 − x2
=
sin3 ud sin u
2
1 − sin u
sin3 udu = −
=
(1 − cos2 u)d cos u
cos3 u
(1 − x2 )3/2 √
− cos u + C =
− 1 − x2 + C
...
6 Problems (difficult ones!)
...
The Product
Rule in terms of differentials reads,
d(uv) = vdu + udv
...
d(uv) = uv + C, the above equation can be expressed in the following form,
udv = uv −
vdu
...
In many cases, Integration by Parts is most efficient in
solving integrals of the product between a polynomial and an exponential, a logarithmic, or a
trigonometric function
...
Example:
xex dx
...
Thus,
xex dx =
Example:
xdex = xex −
ex dx = xex − ex + C
...
Solution: In order to eliminate the power function x2 , we note that (x2 ) = 2
...
25
x2 cosxdx =
= x2 sinx + 2
Example:
x2 dsinx = x2 sinx −
xdcosx = x2 sinx + 2xcosx − 2
sinxdx2 = x2 sinx −
2xd(−cosx)
cosxdx = x2 sinx + 2xcosx − 2sinx + C
...
Solution: In order to eliminate the power function x2 , we note that (x2 ) = 2
...
However, the number (−2) can prove extremely annoying
and easily cause errors
...
x2 e−2x dx =
=
−1
23
(−2x)2 e−2x d(−2x) =
−1
8
u2 eu du =
−1 2 u
[u e − 2
8
ueu du]
−1 2 u
−eu 2
−e−2x 2
[u e − 2(ueu − eu )] + C =
[u − 2u + 2] + C =
[4x + 4x + 2] + C
...
This is because (ln x) = 1/x
...
Solution:
ln xdx = x ln x −
Example:
xd ln x = x ln x −
x
dx
= x ln x − x + C
...
Solution:
x(ln x)2 dx =
1
2
1
(ln x)2 dx2 = [x2 (ln x)2 −
2
26
x2 (2 ln x)
dx
1
] = [x2 (ln x)2 −
x
2
ln xdx2 ]
x2
x2
1
1
= [x2 (ln x)2 − x2 ln x + ] + C = [(ln x)2 − ln x + ] + C
...
(ii)
tan−1 xdx = xtan−1 x −
= xtan−1 x −
1
2
xdtan−1 x = xtan−1 x −
x
dx
1 + x2
√
d(1 + x2 )
= xtan−1 x − ln 1 + x2 + C
...
3 Problems
...
The method of partial
fractions is an algebraic technique that decomposes R(x) into a sum of terms:
27
R(x) =
Pn (x)
= p(x) + F1 (x) + F2 (x) + · · · Fk (x),
Qm (x)
where p(x) is a polynomial and Fi (x), (i = 1, 2, · · · , k) are fractions that can be integrated
easily
...
It is
related to algebraic techniques that many students have not been trained to use
...
It is not our goal in this course to cover
this topics in great details (Read Edwards/Penney for more details)
...
Case I: Qm (x) is a power function, i
...
, Qm (x) = (x − a)m (Qm (x) = xm if a = 0!)
...
x3
This integral involves the simplest partial fractions:
A+B+C
A B
C
=
+ +
...
Thus,
2x2 − x + 3
dx =
x3
[
2x2
x
3
− 3 + 3 ]dx =
3
x
x
x
1
3
2
3
1
[ − 2 + 3 ]dx = ln x2 + − 2 + C
...
2
x − 1 (x − 1)
x−1
28
Case II: Pn (x) = A is a constant and Qm (x) can be factorized into the form Q2 (x) =
(x − a)(x − b), Q3 (x) = (x − a)(x − b)(x − c), or Qm (x) = (x − a1 )(x − a2 ) · · · (x − am )
...
x−3 x+1
A
B
1
Since x−3 + x+1 = (x−3)(x+1) implies that A(x + 1) + B(x − 3) = 1
...
Setting x = −1, we obtain B = −1/4
...
x−3 x+1
2
x+1
Example:
dx
=
(x − 1)(x − 2)(x − 3)
I=
[
A
B
C
+
+
]dx
...
Setting x = 1 in this equation, we obtain A = 1/2
...
Setting
x = 3, we get C = 1/2
...
x−1 x−2 x−3
2
(x − 2)2
More Exercises on Integration by Partial Fractions:
Examples:
(i)
x2
dx
=
− 3x
dx
1
=
x(x − 3)
3
[
1
1
x−3
1
− ]dx = ln |
| + C
...
x−2 x+3
5
x+3
(iii)
x−1
dx =
x+1
x+1−2
dx =
x+1
[1 −
2
]dx = x − ln(x + 1)2 + C
...
5 Problems
...
Example: I =
x−3 e1/x dx
...
The substitution to change it into a simple,
elementary function is u = 1/x, du = u dx = −x−2 dx or x−2 dx = −du = d(−1/x)
...
30
√
cos( x)dx
...
The substitution to change it into a
√
√
1
simple, elementary function is u = x, du = u dx = 2√x dx or dx = 2 xdu = 2udu
...
=2
Example: I =
sec(x)dx = (1/cos(x))dx
...
Note that 1/cos(x) is a composite fucntion
...
However, x = cos−1 (u) and
−du
dx = x du = √1−u2
...
Here
is how we can deal with it
...
dx
=
cos(x)
I=
= ln
|
cos(x)dx
=
cos2 (x)
dsin(x)
=
1 − sin2 (x)
du
=−
1 − u2
[
1
1 du
−
]
u−1 u+1 2
u+1
u+1
sin(x) + 1
| + C = ln | √
| + C = ln |tan(x) + sec(x)| + C,
| + C = ln |
2
u−1
cos(x)
1−u
where u = sin(x) and
√
1 − u2 = cos(x) were used!
More Exercises on Integration by Multiple techniques:
1
...
2
2
31
(ii)
√
xtan−1 xdx =
1
= [u4 tan−1 u−
2
√
√ √
( x)2 tan−1 xd( x)2 =
u2 tan−1 udu2 =
1
2
tan−1 udu4
u4 − 1 + 1
1
1
du] = [u4 tan−1 u− (u2 −1+
)du]
2
1+u
2
1 + u2
1
u4 dtan−1 u] = [u4 tan−1 u−
2
√
x3/2 √
u3
1
1
+ u − tan−1 u] + C = [(x2 − 1)tan−1 x −
+ x] + C
...
3 Problems
...
Substitution followed by Integration by Partial Fractions
Example:
(i)
cos xdx
=
sin x − sin x − 6
2
d sin x
=
sin x − sin x − 6
du
1
u−3
1
sin x − 3
= ln |
|+C = ln |
|+C
(u − 3)(u + 2)
5
u+2
5
sin x + 2
2
Edwards/Penney 5th − ed 9
...
Integration by Parts followed by Partial Fractions
32
1 + ln t
dt
t(3 + 2 ln t)2
Example:
(i)
(2x + 2) tan−1 xdx =
= (x2 + 2x) tan−1 x −
tan−1 xd(x2 + 2x) = (x2 + 2x) tan−1 x −
1 + x2 + 2x − 1
dx = (x2 + 2x) tan−1 x −
1 + x2
(x2 + 2x)d tan−1 x
[dx +
d(1 + x2 )
dx
−
]
2
1+x
1 + x2
= (x2 + 2x) tan−1 x − x − ln(1 + x2 ) + tan−1 x + C = (x + 1)2 tan−1 x − x − ln(1 + x2 ) + C
...
14
...
We know the area of a rectangle is:
Area = height × width,
(see Fig
...
This formula can not be applied to the area under the curve in Fig
...
However, if we divide the
area into infinitely many rectangles with infinitely thin width dx, the infinitely small area of
the rectangle located at x is
dA = height × inf intely small width = f (x)dx
...
2
x
a
b
f (x)dx,
a
f (s)ds is the area under the curve between a and x
...
We know the volume of a cylinder is
V olume = (cross − section area) × length,
(see Fig
...
This formula does not apply to the volume of the solid obtained by rotating
a curve around the x-axis (see Fig
...
This is because the cross-section area is NOT a
34
constant but changes along the length of this “cylinder”
...
Adding up the volume of all such thin disks, we obtain the volume of the solid
V (b)
V olume = V (b) − V (a) =
b
0
where V (x) = π
14
...
Arc length
...
3(a))
...
3(b) because the
curve is NOT a straight line and the geometric shape of the area enclosed in solid lines is NOT
w
(a)
h
a
¢ ¢¡
¡
¢¡
¢¢¡
¢¡
¡
¢¡
¢¡
¢ ¢¡
¢¡
¢¡
¡
A=hw
y=f(x)
x
dx
b
(b)
Figure 1: Area under a curve
...
a right triangle
...
Thus, the length of one such curve
segments located at x is
(inf initely small base)2 + (inf initely small height)2
dl =
=
dx2 + dy 2 =
[1 + (
dy 2 2
) ]dx = [
dx
1 + (y )2 ]dx
...
b
l(b)
a
0
where l(x) =
x
[
a
[ 1 + (y )2 ]dx,
dl =
L=
1 + [y (s)]2 ]ds is the length of the curve between a and x
...
Solution: Because of the symmetry, we only need to calculate the length on the interval [0, 2]
and multiply it by two
...
√
where F (x) = [ 1 + x2 ]dx
...
Note that 1 + x2 = 1 + sinh2 (u) = cosh2 (u) and dx = dsinh(u) = cosh(u)du, we
obtain by subsitution
F (x) =
√
[ 1 + x2 ]dx =
cosh(u)cosh(u)du =
cosh2 (u)du =
1
2
[1 + cosh(2u)]du
√
√
1
1
1
1
= [u + sinh(2u)] + C = [u + sinh(u)cosh(u)] + C = [ln |x + 1 + x2 | + x 1 + x2 ] + C,
2
2
2
2
where the inverse substitution sinh(u) = x, cosh(u) =
√
1 + x2 | were used in the last step
...
4
√
√
√
√
1 + 22 | + 2 1 + 22 = ln(2 + 5) + 2 5 ≈ 5
...
From density to mass
...
37
The formula is no longer valid if the density is NOT a constant but is nonuniformly distributed
in the solid
...
Thus, the infinitely small mass contained in the little volume is
dm = density × (inf initely small volume) = ρdV
...
m(V (b))
m=
V (b)
dm =
0
ρdV,
V (a)
V (x)
where V (x) = 0
dV is the volume of the portion of the solid between a and x; while
m(V (x))
dm is the mass contained in the volume V (x)
...
The pipe is 100 m long with a diameter of 1 m
...
Solution: Since the density only varies as a function of the distance x, we can “cut” the pipe
into thin cylindrical disks along the axis of the pipe
...
Using the
integral form of the formula, we obtain
m(100)
m=
V (100)
dm =
m(0)
100
ρ(x)πr 2 dx =
=
0
π
4
ρ(x)dV
V (0)
100
e−x/10 dx =
0
10π
[1 − e−10 ] ≈ 7
...
4
Example: The air density h meters above the earth’s surface is ρ(h) = 1
...
000124h (kg/m3 )
...
(3
points)
Solution: Similar to the previous problem, the density only varies as a function of the altitutde
h
...
38
=
0
0
V (0)
m(0)
ρ(h)π22 dh
ρ(h)πr 2 dh =
ρdV =
dm =
25,000
25,000
V (25,000)
m(25,000)
m=
4 × 1
...
000124h 25,000 4 × 1
...
1 ] = 123, 873
...
−0
...
000124
Very often the density is given as mass per unit area (two dimensional density)
...
Similarly, if the density is given as mass per
unit length (one dimensional density), then dm = ρdx, where dx is an infinitely small length
...
What is the total number of bacteria in
the colony?
Solution: This is a problem involving a two dimensional density
...
Thus, we can divide the circular colony into infinitely many concentric ring areas
...
e
...
The area, dA, of a ring of bacteria colony with radius r and width dr is
dA = π(r + dr)2 − πr 2 = 2πrdr + πdr 2 = 2πrdr, where dr 2 is infinitely smaller than dr, thus is
ignored as dr → 0
...
e
...
Or simply, A = πr 2 , thus dA = d(πr 2 ) = πdr 2 = 2πdr
...
2
2
Example: The density of a band of protein along a one-dimensional strip of gel in an electrophoresis experiment is given by ρ(x) = 6(x − 1)(2 − x) for 1 ≤ x ≤ 2, where x is the distance
along the strip in cm and ρ(x) is the protein density (i
...
protein mass per cm) at distance x
...
Solution: This is a problem involving a one dimensional density
...
e
...
39
dm =
m(1)
14
...
Work done by a varying force
...
The formula is no longer valid if the force is NOT a constant over the distance covered but
changes
...
Thus, the
infinitely small work done on a distance dx is
dW = f orce × (inf intely small distance) = f dx
...
W (b)
W =
b
dW =
W (a)
f dx
...
This is the lower limit of energy required to launch
it into space
...
Assume that the earth is a sphere with radius Re and
that r = ∞ when the satellite is “outside” the earth’s gravitational field
...
)
Solution: We know that work done on an infinitely small distance dr when the satellite is
located at a distance r from the center of the earth is: dW = f (r)dr = (GMe m/r 2 )dr
...
r2
Integrals that involve a limit which is ∞ are called improper integrals
...
In many cases, we can just treat ∞ as a normal upper limit of
this definite integral
...
Example: For the time interval 0 ≤ t ≤ 10, write down the definite integral determining the
total work done by a force, f (t) (N ewton), experienced by a particle moving on a straight
line with a speed, v(t) > 0 (m/s)
...
Solution:
W (10)
W =
x(10)
dW =
W (0)
10
f (t)dx =
x(0)
f (t)(
0
41
dx
)dt =
dt
10
f (t)v(t)dt
...
They should be memorized
...
sin x dx = − cos x + C
cos x dx = sin x + C
sec2 x dx = tan x + C
csc2 x dx = − cot x + C
1
x
1
dx = tan−1 + C
2
+x
a
a
tan x dx = − ln | cos x| + C
1
x
dx = sin−1 + C
2
a
−x
cot x dx = ln | sin x| + C
sec x dx = ln | tan x + sec x| + C
csc x dx = ln | cot x − csc x| + C
a2
√
a2
The following are elementary integrals involving hyperbolic functions (for details read the
Section on Hyperbolic Functions)
...
f (g(x))g (x) dx =
f (u) du where u = g(x) (change of variables)
f (g(x)) dx =
dx
du where u = g(x) (different form of the same change of variables)
du
f (u)
1
ecx dx = ecx + C (c = 0)
c
1 x
a + C (for a > 0, a = 1)
ax dx =
ln a
ln x dx = x ln x − x + C
1
x
1
dx = arctan + C, a = 0
2
+a
a
a
x−a
1
1
dx =
ln
+C, a = 0
x2 − a 2
2a
x+a
1
x
√
dx = arcsin + C, a > 0
2 − x2
a
a
√
1
√
dx = ln |x + x2 ± a2 | + C
x2 ± a 2
1
To compute
dx we complete the square
x2 + bx + c
x2
b2
b2
x + bx + c = x + bx + + c −
=
4
4
2
2
b
x+
2
2
+c−
b2
4
If c − b2 /4 > 0, set it equal to a2 ; if < 0 equal to −a2 ; and if = 0 forget it
...
This is to be used repetaedly until you arrive at the
c
c
case n = 0, which you can do easily
...
4
for plots of these functions)
...
10
6
1
3
−4
0
−10
sinh(x)
4
−4
−3 −2 −1 0
1
2
3
−2
0
2
4
−1
tanh(x)
cosh(x)
Figure 4: Graphs of sinh x, cosh x, and tanh x
...
However, it helps us memorize these identities if we know the relationship between
sinh x, cosh x and sin x, cos x in complex analysis
...
ex + e−x
eix + e−ix
and cos x =
, we obtain
cosh(x) = cos(ix)
...
Now, we can write down these identities:
Combining cosh x =
cosh2 x − sinh2 x = 1
1 − tanh2 x = sech2 x
coth2 x − 1 = csch2 x
cosh2 x = (1 + cosh 2x)/2
sinh2 x = cosh2 x − 1 = (−1 + cosh 2x)/2
sinh 2x = 2 sinh x cosh x
sinh(x ± y) = sinh x cosh y ± cosh x sinh y
cosh(x ± y) = cosh x cosh y ± sinh x sinh y
cos2 x + sin2 x = 1
1 + tan2 x = sec2 x
cot2 x + 1 = csc2 x
cos2 x = (1 + cos 2x)/2
sin2 x = 1 − cos2 x = (1 − cos 2x)/2
sin 2x = 2 sin x cos x
sin(x ± y) = sin x cos y ± cos x sin y
cos(x ± y) = cos x cos y sin x sin y
44
The Advantage of Hyperbolic Functions
The advantage of hyperbolic functions is that their inverse functions can be expressed in terms
of elementary functions
...
By definition, u = sinh x =
ex − e−x
...
This is a quadratic in ex , Thus,
ex = [2u ±
(2u)2 + 4]/2 = u ±
√
u2 + 1
...
√
Thus, x = ln ex = ln |u + u2 + 1| which implies
x = sinh−1 u = ln |u +
√
u2 + 1|
...
Similarly, by definition, u = cosh x =
2
We can solve this equation for the two branches of the inverse:
x = cosh−1 u = ln |u ±
√
√
u2 − 1| = ± ln |u + u2 − 1|
...
ex + e−x
We can solve the equation for the inverse:
x = tanh−1 u =
1+u
1
ln |
|
...
When we use trig
substitution, it is often a necessity for us to know the definition of other trig functions that is
related to the one we use in the substitution
...
Solution: y = sin−1 x means sin y = x
...
cos y
In order to express y in terms of x, we need to express cos y in terms of x
...
However, it
is easier to construct a right triangle with an angle y
...
Therefore,
adjacent side
=
cos y =
hypotenuse
Example: Calculate the integral
√
1 − x2 √
= 1 − x2
...
Solution: This integral requires standard trig substitution x = tan u or x = sinh u
...
Recall that 1 + sinh2 u = cosh2 u and that dx = d sinh u = cosh udu,
√
=
1
2
1 + x2 dx =
1 + sinh2 ud sinh u =
cosh2 udu
1
1
1
[1 + cosh(2u)]du = [u + sinh(2u)] + C = [u + sinh u cosh u] + C
...
However, we need to express the solution in terms of x
...
cosh u = cosh2 u =
√
1 + sinh2 u = 1 + x2
...
2
2
46
However, the following table shows how can we exploit the similarity between trig and hyperbolic functions to find out the relationship between different hyperbolic functions
Title: Integration
Description: Basic concepts of integration in a simplified manner.
Description: Basic concepts of integration in a simplified manner.