Download SOME SOLUTION OF CALCULUS OF STEWART OF 7.4

SOME SOLUTION OF CALCULUS OF
STEWART OF 7.4
PHILO WU / WU ZHONG TANG
September 6, 2013
Abstract
The goal (maybe just a hope) of this note is helping you to understand
what the calculus(as verb) is in calculus (as word).
But the only way to achieve that is doing the calculus (as verb) line
by line, hand by hand. So follow me, and do more and more, surpass this
note.
The style of this note, is trying to keep informal and readable to beginner, as well as I’m still a freshman student.
1
Introduction of 7.4
In some sense, the title of 7.4,
make questions, foolish and good at same time as samething, such as:
What is rational function?
Why only do integration on they?
What is ”fractions”?......
All the questions can answer by multiway, some way is easy, direct but naive.
Some way is deep, fascinating and more easy(in the sense of ”naturality”, in
higher math.)
2
2.1
the calculus
EX.11
Z
0
1
1
dx
2x2 + 3x + 1
First at all, let we forget the boundary, 1 and 0.
Z
1
dx
2x2 + 3x + 1
Since we need to make simplify, in any sense. We want to know the Quadratic
equation
2x2 + 3x + 1
can break down, in math we say, can we ”decompose” it??
1
This is one of the most important ideas in math. Anything can break down,
decompose into small pieces.
√
−b ± b2 − 4ac
x=
2a
it tell us this Quadratic equation has two solutions in real numbers
x = −1and −
1
2
thus
2x2 + 3x + 1 = (x + 1)(2x + 1)
check it!
Now
2x2
1
A
B
1
=
=
+
+ 3x + 1
(x + 1)(2x + 1)
(x + 1) (2x + 1)
So the problem of evaluation of
Z
1
dx
2x2 + 3x + 1
is reduced to the problem of evaluation of
Z
A
B
+
dx
(x + 1) (2x + 1)
and the fianl answer must be ”ln(something)+C”
To finish our problem, we need and only need to find what the A, B are??
Since we have
1
1
A
B
=
=
+
,
2x2 + 3x + 1
(x + 1)(2x + 1)
(x + 1) (2x + 1)
thus we have this equation
1 = A(2x + 1) + B(x + 1)
by mult the denominator (x+1)(2x+1) on the second ”=” both sides.
Now
1 = A(2x + 1) + B(x + 1) = (2A + B)x + (A + B)1,
so we got
0 = (2A + B)x
and
1 = (A + B)1,
this is the one of most powerful methods, in whole Mathematics,
”Comparing − coef f icients”,
it reduced a lot problems to the problem solve a system of linear equations, or
in modern term: linear algebra. On this way we should reformulate they,
0 = 2A + B
2
1=A+B
in this size, 2 by 2, we can solve it quickly, by any ways.
A = −1, B = 2
Please check it, may be I’m wrong.
Now we know that
Z
Z
1
2
−1
dx
=
+
dx
2
2x + 3x + 1
(x + 1) (2x + 1)
Z
−1
2
+
dx = (−1)ln(x + 1) + 2ln(2x + 1) + ”C”
(x + 1) (2x + 1)
Turn back our original problem,
Z
0
1
2x2
1
dx
+ 3x + 1
now
Z
0
1
1
dx = (−1)ln(1 + 1) + ln(1) + 2ln(2 + 1) − 2ln(1)
2x2 + 3x + 1
since ln(1)=0,
(−1)ln(1 + 1) + ln(1) + 2ln(2 + 1) − 2ln(1) = ln(3) − ln(2)
recall: ln(xy)=ln(x)+ln(y),
3
ln(3) − ln(2) = ln( )
2
we done.
P.S.
You can check the LOG table to find what the value is !!
2.2
Comments
What is the key idea and notion here??
It may said, degree, the max power of the equation such as
2x2 + 3x + 1, x2 + 3x + 1
it is
2!!Both!!
In higher degree, like
x3 + 3x + 1, x5 + 3x4 + 1, etc
We don’t know too much, you and the whole math world both!!
For more concrete, a example, we ””can’t”” find the formula like
√
−b ± b2 − 4ac
x=
,
2a
3
for the equation it degree
≥ 5,
thus we can’t not break down the equation to small piece quickly, just like
2x2 + 3x + 1 = (x + 1)(2x + 1)
But thank to Gauss, one of the greatest mathematicians.
He gave a proof of the Fundamental Theorem of Algebra, may be the first
proof that can accept by the mathematicians that living.
The Fundamental Theorem of Algebra: Every polynomial with real coefficients can be factored into linear and quadratic factors.
Under this guarantee and long division, we always can by our hard work and
spending long time to find what the smallest piece of the equation look like!!
It0 s − always − is − a − product − of − linear − and − quadratic − f actors!!!!
So crazy idea and reality(Not just a tedious ”fact”).
Whatever we choose, the equation(polynomial) always can break-down, untill it is a product of linear and quadratic factors!!
Thus we can focus on how to integration
1
Ax2 + Bx + C
when the case
Ax2 + Bx + C
has no solution in real numbers.
Find the solutions and answers of yourself!!
2.3
EX.19
Z
x2 + 1
dx
(x − 3)(x − 2)2
x2 + 1
A
B
C
=
+
+
2
(x − 3)(x − 2)
(x − 3) (x − 2) (x − 2)2
x2 + 1 = A(x − 2)2 + B(x − 3)(x − 2) + C(x − 3)
can put
x = 2, x = 3, the − evaluation − of − polynomials
we get
A = 10, C = −5
then from the constant term of
x2 + 1 = A(x − 2)2 + B(x − 3)(x − 2) + C(x − 3)
observe that we have
1 = 4A + +6B − 3C
4
thus
B = −9
hence
x2 + 1
10
−9
−5
=
+
+
2
(x − 3)(x − 2)
(x − 3) (x − 2) (x − 2)2
thus
Z
x2 + 1
dx =
(x − 3)(x − 2)2
Z
10
−9
−5
dx =
+
+
(x − 3) (x − 2) (x − 2)2
10ln(x − 3) − 9ln(x − 2) +
2.4
5
+C
(x − 2)
Comments
By the same idea, when we choose
A
B
Cx + D
x2 + 1
=
+
+
(x − 3)(x − 2)2
(x − 3) (x − 2) (x − 2)2
If we very naive, as my first and ... try, from
x2 + 1 = A(x − 2)2 + B(x − 3)(x − 2) + (Cx + D)(x − 3)
directly,
1=A+B+C
0 = (−4)A + (−5)B + (−3)C + D
1 = 4A + 6B + (−3)D
we have four unknown, but only three equations.
In linear algebra term, it is try to solve :
 
 


A
1
1
1
1
0
B 
 , Y = 0
M = −4 −5 −3 1  , X = 
C 
1
4
6
0 −3
D
MX = Y
And all wrong methods that I tried, they can describe by I have trying to solve
it, but fail. In some case, it can be solve, has solutions but not unipuely.
But not this case, since the original problem of our equations are came from
x2 + 1 = A(x − 2)2 + B(x − 3)(x − 2) + (Cx + D)(x − 3)
it is ”decomposition of polynomials”, in our case, over the real numbers. It must
decompose by unique way, it can be show, but not now.
Thus by the knowlege in linear algebra,
 


 
A
1
1
1
0
1
B 
 , Y = 0
M = −4 −5 −3 1  , X = 
C 
4
6
0 −3
1
D
5
MX = Y
if it can be solve,then can not solve uniquely,Contradiction!!
Contradiction!!
Contradiction!!
Contradiction!!
That mean and only mean, we did something wrong!
In this case, this Contradiction came form we miss some message, just like
a boor.
May be the real formula is
A
B
C
x2 + 1
=
+
+
(x − 3)(x − 2)2
(x − 3) (x − 2) (x − 2)2
x2 + 1 = A(x − 2)2 + B(x − 3)(x − 2) + C(x − 3)
or
x2 + 1
A
B
Cx
=
+
+
(x − 3)(x − 2)2
(x − 3) (x − 2) (x − 2)2
x2 + 1 = A(x − 2)2 + B(x − 3)(x − 2) + Cx(x − 3)
then why I choose the first one to solve our problem, think about that, it
will help you understand what are we doing!
LINAER ALGEBRAR...... May be you asked me: what is this? Or this is
just solve
1=
A
+B
+C
0 = −4A −5B −3C +D
1 = 4A +6B
−3D
with four unknown, three equations, not thing new!
Yes, but this also what linear algebra care and only care, solve and understand the linear equations. The Understanding of
1=
A
0 = −4A
1 = 4A
+B
−5B
+6B
+C
−3C
+D
−3D
can not be solve with uniqueness, it is came from the consequences of some basis
properties of linear equations, but when in math textbook it usually reformulate
by more abstract term, since it will be more easy to Understand(in some sense)!
This is one of the most romantic characteristic of math!!
2.5
EX.23
Z
10
dx
(x − 1)(x2 + 9)
Same method
A
Bx + C
1
(−1)x − 1
10
=
+
=
+
(x − 1)(x2 + 9)
(x − 1) (x2 + 9)
(x − 1)
(x2 + 9)
6
(−1)x − 1
1
+
dx
(x − 1)
(x2 + 9)
Z
Z
(−1)x − 1
1
(−1)x − 1
+
dx
=
ln(x
−
1)
+
dx
2
(x − 1)
(x + 9)
(x2 + 9)
Z
Z
Z
(−1)x − 1
(−1)x
−1
dx
=
dx
+
dx
(x2 + 9)
(x2 + 9)
(x2 + 9)
Z
10
dx =
(x − 1)(x2 + 9)
Here to evaluate
Z
Z
(−1)x
dx
(x2 + 9)
only need to recall the differential rule
d(x2 ) = 2xdx, d(u + c) = du, d(constant ∗ u) = constant ∗ du
then we combine they, we got
(−1)xdx = (−1/2)2xdx = (−1/2)d(x2 ) = (−1/2)d(x2 + 9)
Z
Z
(−1/2)d(x2 + 9)
(−1)xdx
=
2
(x + 9)
(x2 + 9)
Z
(−1/2)d(x2 + 9)
= (−1/2)ln(x2 + 9) + C
(x2 + 9)
For
Z
−1
dx
(x2 + 9)
we observe that
(x2 + 9) = 9[(x/3)2 + 1]
then
Z
Z
−1
dx =
(x2 + 9)
Z
−1/9
dx
[(x/3)2 + 1]
−1/9
dx = (1/9)arctan(x/3) + C
[(x/3)2 + 1]
finally
Z
10
dx =
(x − 1)(x2 + 9)
Z
1
(−1)x − 1
+
dx
(x − 1)
(x2 + 9)
Z
(−1)x − 1
dx
(x2 + 9)
= ln(x − 1) +
= ln(x − 1) + (−1/2)ln(x2 + 9) + (1/9)arctan(x/3) + C
We done!!
7
2.6
comments
The notion of Ex23 is if you know and remember, correctly, some of basis rules
and formulas, such as
d(x2 ) = 2xdx, d(u + c) = du, d(constant ∗ u) = constant ∗ du
Z
1
dx = arctan(u) + C
u2 + 1
you may get some help.
But on the other hand you may lost, so keep training, this is the only way
make you safe in many sense.
2.7
Warning!!!!
W arning!
As you saw, some time we may go fast than the begin, since a beginner will
not be a beginner forever. You will to grow up. More stronger, adaptable,
freestanding. If you feel too fast. Try again, and again.
2.8
Ex.31
Z
x3
1
dx
−1
we observe that
x3 − 1 = (x − 1)(x2 + x + 1)
You can find it, by long division, or just check it by explanation the product.
Again, the old but useful method
x3
1
1
A
Bx + C
=
=
+ 2
2
−1
(x − 1)(x + x + 1)
x−1 x +x+1
then
1 = A(x2 + x + 1) + (Bx + C)(x − 1)
and
C = −2/3, B = −1/3, A = 1/3
Check it, may be I fail it. Now we knew
1
1
1
(−1)(x + 2)
=
=
+
x3 − 1
(x − 1)(x2 + x + 1)
3(x − 1) 2(x2 + x + 1)
So
then
Z
1
dx +
3(x − 1)
Z
Z
1
dx =
3
x −1
Z
(−1)(x + 2)
dx = (1/3)ln(x − 1) − (1/6)
3(x2 + x + 1)
1
(−1)(x + 2)
+
dx
3(x − 1) 3(x2 + x + 1)
8
Z
d(x2 + x + 1)
(x2 + x + 1)
Did you feel something wrong here?
since
d(x2 + x + 1) = (2x + 1)dx
Finally
Z
x3
1
dx = (1/2)ln(x − 1) − (1/4)ln(x2 + x + 1) + C
−1
The answer in here is wrong! Try to correct it. If you stuck, see EX.23!
2.9
comments
Here the final answer of this form can be better,that is with more insight
(1/2)ln(x − 1) − (1/4)ln(x2 + x + 1) = ln(x − 1)1/2 + ln
thus
Z
1
(x2 + x + 1)1/4
1
(x − 1)1/2
dx
=
ln
x3 − 1
(x2 + x + 1)1/4
The mistake still here.
Here we can take limit as
lim ln
x→∞
(x − 1)1/2
+ x + 1)1/4
(x2
Try to do some calculus(as verb), you may enjoy it!
Evening we have a mistake, the limit still give us some insight, about what
the graph of the integral.
Try to clarify what I said and what I intimate. And why they hold?
2.10
Ex.39
Z √
x+1
dx
x
Observe that
def
x + 1 = u2 → dx = 2udu
thus
Z √
Z
u(2udu)
(u − 1)(u + 1)
Z
Z
Z
u(2udu)
1
= 2[ du +
du] =
(u − 1)(u + 1)
u2 − 1
Z
1/2
−1/2
u+
+
du
u−1 u+1
x+1
dx =
x
The next step is the most important and the easy be skip.
9
If the mistake does
not exist
any more!
What
the limit
meaning?
Recall
√
def
x + 1 = u2 ↔ x + 1 = u
√
√
√
u+(1/2)ln(u−1)+(−1/2)ln(u+1)+C = 1 + x+(1/2)ln( x + 1−1)+(−1/2)ln( x + 1+1)+C
that is
Z √
2.11
√
√
√
x+1
dx = 1 + x + (1/2)ln( x + 1 − 1) + (−1/2)ln( x + 1 + 1) + C
x
comments
When you say Something equivalent to something, do not mean you said Something equal to something. This ”miss”, she also came in the badroom of mathematicians.....
As EX, find some example in your daily life.
2.12
EX.43
Z
√
3
x3
x2 + 1
dx
my method is do the easiest job, let
x = u3 → x2 = u6 and, x3 = u9
then
Z
dx = 3u2 du
Z
x
u9
√
dx
=
3u2 dx
3
u2 + 1
x2 + 1
3
by spend the time on long division, we got
u11
u
= u9 − u7 + u5 − u3 + 2
+1
u +1
u2
Check it, may be I miss something, then
Z
Z
u11
u
3
du
=
3
u9 − u7 + u5 − u3 + 2
du
u2 + 1
u +1
then everything you knew them, don’t forget”+C” and
x = u3
2.13
EX.46
Z p
my way is
1+
√
1+
x
√
x
dx
√
x = u2 , x = u2 − 1
then
x = (u2 − 1) = u4 − 2u2 + 1
dx = (4u3 − 4u)du
10
thus
Z
p
√
Z
Z 2 2
( 1 + x)dx
u2 du
u (u − 1)du
=
4
=4
x
(u2 − 1)2
u2 − 1
since
a + 0 = a, 0 = 1 − 1
u2
u2 − 1 + 1
=
−1
u2 − 1
u2
u2 − 1 + 1
u2 − 1
1
=
+
u2 − 1
u2 − 1 u2 − 1
thus
Z
u2 du
=
u2 − 1
Z
Z
du +
u2
1
du
−1
Since we knew
u2 − 1 = (u − 1)(u + 1)
then
Z
1
du =
2
u −1
Z
−1/2
du +
u+1
Z
−1/2
du
u−1
and thank to Leibniz and Fundamental Theorem of Calculus, we get
Z
du = u + c
then everything are almost done, complete the puzzle by yourself!
Don’t forget this
√
√
1 + x = u2 , x = u2 − 1!!
2.14
comments
Now we got
Z
q
q
q
√
√
√
x3
√
dx
=
4[
1
+
x−(1/2)ln(
1
+
x+1)−(1/2)ln(
1 + x−1)]+C
3
x2 + 1
Again, we can make the form better, that mean with more insight.
q
q
√
√
(−2)ln( 1 + x + 1) + (−2)ln( 1 + x − 1) =
(−2)ln(1 +
√
√
1
x − 1) = (−2)ln( x) = ln( )
x
Since
a2 − b2 = (a + b)(a − b), ln(xy) = ln(x) + ln(y)
Z
q
√
1
x3
√
dx
=
4
1 + x + ln( ) + C
3
2
x
x +1
Funny thing come in,
q
√
1
lim [4 1 + x + ln( )] =??
x→∞
x
11
What
the limit
meaning?
Hint : lim
q
x→∞
1+
√
x = lim
q
√
x→∞
x
x1/4
=∞
x→∞ ln( 1 )
x
lim
here I used” equal to Infinity”, this is a bad but useful notaion, try to use it
correctly.
2.15
EX.51
Z
1
dx
1 + ex
Observe that
ex = u, x = lnu, dx = (1/u)du
then
thus
Z
1
dx =
1 + ex
Z
du
u(1 + u)
Z
Z
du
−1
1
=
du +
du
u(1 + u)
(1 + u)
u
Z
Z
−1
1
u
du +
du = ln(u) − ln(u + 1) + C = ln(
)+C
(1 + u)
u
u+1
Z
thus
Z
substitute
it at the
first”=”,
what happen?
ex
1
dx
=
ln(
)+C
1 + ex
ex + 1
Q:
lim ln(
x→∞
ex
) =?
+1
ex
What it meaning?
2.16
EX.61
Z
1
dx
3 sin(x) − 4 cos(x)
As the hint they given,let
t = tan(x/2)
then
sin(x) =
2t
1 − t2
, cos(x) =
2
1+t
1 + t2
and
2 arctan(t) = x, dx =
thus
Z
1
dx =
3 sin(x) − 4 cos(x)
12
Z
2
dt
1 + t2
2
1+t2 dt
2t
1−t2
3 1+t
2 − 4 1+t2
u = ex
=
Z
Z
dt
dt
=
=
2t2 + 3t − 2
(t + 2)(2t − 1)
Z
Z
−1/5
2/5
dt +
dt =
t+2
2t − 1
(−1/5)ln(t + 2) + (2/5)ln(2t − 1) + C =
(−1/5)ln(tan(x/2) + 2) + (2/5)ln(2 tan(x/2) − 1) + C
2.17
comments
You can find smoething funny, in the reformulate of the answer, may be!?
13