2 the calculus

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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 the calculus

2.1 EX.11

Z 1 0

1

2x²+ 3x + 1dx First at all, let we forget the boundary, 1 and 0.

Z 1

2x²+ 3x + 1dx

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??

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This is one of the most important ideas in math. Anything can break down, decompose into small pieces.

x = −b ±√

b²− 4ac 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

1

2x²+ 3x + 1 = 1

(x + 1)(2x + 1) = A

(x + 1)+ B (2x + 1) So the problem of evaluation of

Z 1

2x²+ 3x + 1dx is reduced to the problem of evaluation of

Z A

(x + 1)+ B (2x + 1)dx 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

2x²+ 3x + 1 = 1

(x + 1)(2x + 1) = A

(x + 1)+ B (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

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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 1

2x²+ 3x + 1dx =

Z −1

(x + 1)+ 2 (2x + 1)dx

Z −1

(x + 1)+ 2

(2x + 1)dx = (−1)ln(x + 1) + 2ln(2x + 1) + ”C”

Turn back our original problem, Z 1

0

1

2x²+ 3x + 1dx now

Z 1 0

1

2x²+ 3x + 1dx = (−1)ln(1 + 1) + ln(1) + 2ln(2 + 1) − 2ln(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),

ln(3) − ln(2) = ln(3 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 x = −b ±√

b²− 4ac

2a ,

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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 coeffi- cients 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!!

It⁰s − 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, un- till 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 (x − 3)(x − 2)²dx x²+ 1

(x − 3)(x − 2)² = A

(x − 3)+ B

(x − 2)+ C (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

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thus

B = −9 hence

x²+ 1

(x − 3)(x − 2)² = 10

(x − 3)+ −9

(x − 2)+ −5 (x − 2)² thus

Z x²+ 1

(x − 3)(x − 2)²dx =

Z 10

(x − 3) + −9

(x − 2)+ −5 (x − 2)²dx = 10ln(x − 3) − 9ln(x − 2) + 5

(x − 2)+ C

2.4 Comments

By the same idea, when we choose x²+ 1

(x − 3)(x − 2)² = A

(x − 3)+ B

(x − 2)+Cx + D (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 :

M =





1 1 1 0

−4 −5 −3 1

4 6 0 −3



, X =





 A B C D





 , Y =



 1 0 1





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,

M =





1 1 1 0

−4 −5 −3 1

4 6 0 −3



, X =





 A B C D





 , Y =



 1 0 1





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MX = Y

if it can be solve,then can not solve uniquely,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 x²+ 1

(x − 3)(x − 2)² = A

(x − 3)+ B

(x − 2)+ C (x − 2)² x²+ 1 = A(x − 2)²+ B(x − 3)(x − 2) + C(x − 3) or

x²+ 1

(x − 3)(x − 2)² = A

(x − 3)+ B

(x − 2)+ Cx (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 +B +C

0 = −4A −5B −3C +D

1 = 4A +6B −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

(x − 1)(x²+ 9)dx Same method

10

(x − 1)(x²+ 9) = A

(x − 1)+Bx + C

(x²+ 9) = 1

(x − 1)+(−1)x − 1 (x²+ 9)

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Z 10

(x − 1)(x²+ 9)dx =

Z 1

(x − 1)+(−1)x − 1 (x²+ 9) dx

Z 1

(x − 1)+(−1)x − 1

(x²+ 9) dx = ln(x − 1) +

Z (−1)x − 1 (x²+ 9) dx Z (−1)x − 1

(x²+ 9) dx =

Z (−1)x (x²+ 9)dx +

Z −1

(x²+ 9)dx Here to evaluate

Z (−1)x (x²+ 9)dx 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 (−1)xdx

(x²+ 9) =

Z (−1/2)d(x²+ 9) (x²+ 9) Z (−1/2)d(x²+ 9)

(x²+ 9) = (−1/2)ln(x²+ 9) + C

For Z −1

(x²+ 9)dx we observe that

(x²+ 9) = 9[(x/3)²+ 1]

then Z −1

(x²+ 9)dx =

Z −1/9

[(x/3)²+ 1]dx

Z −1/9

[(x/3)²+ 1]dx = (1/9)arctan(x/3) + C finally

Z 10

(x − 1)(x²+ 9)dx =

Z 1

(x − 1)+(−1)x − 1 (x²+ 9) dx

= ln(x − 1) +

Z (−1)x − 1 (x²+ 9) dx

= ln(x − 1) + (−1/2)ln(x²+ 9) + (1/9)arctan(x/3) + C We done!!

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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

u²+ 1dx = arctan(u) + C 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 1

x³− 1dx 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 1

x³− 1 = 1

(x − 1)(x²+ x + 1) = A

x − 1+ Bx + C x²+ 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

x³− 1 = 1

(x − 1)(x²+ x + 1) = 1

3(x − 1)+ (−1)(x + 2) 2(x²+ x + 1)

So Z 1

x³− 1dx =

Z 1

3(x − 1)+ (−1)(x + 2) 3(x²+ x + 1)dx then

Z 1

3(x − 1)dx +

Z (−1)(x + 2)

3(x²+ x + 1)dx = (1/3)ln(x − 1) − (1/6)

Z d(x²+ x + 1) (x²+ x + 1)

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Did you feel something wrong here?

since

d(x²+ x + 1) = (2x + 1)dx Finally

Z 1

x³− 1dx = (1/2)ln(x − 1) − (1/4)ln(x²+ x + 1) + C 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 1

(x²+ x + 1)^1/4 thus

Z 1

x³− 1dx = ln (x − 1)^1/2 (x²+ x + 1)^1/4 The mistake still here.

Here we can take limit as If the mis-

take does not exist any more!

What the limit meaning?

If the mistake does not exist any more!

x→∞lim ln (x − 1)^1/2 (x²+ x + 1)^1/4 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

x dx Observe that

x + 1^def= u²→ dx = 2udu thus

Z √ x + 1

x dx =

Z u(2udu) (u − 1)(u + 1) Z u(2udu)

(u − 1)(u + 1) = 2[

Z du +

Z 1

u²− 1du] = u +

Z 1/2

u − 1+−1/2 u + 1du

The next step is the most important and the easy be skip.

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Recall

x + 1^def= 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 √ x + 1

x dx =√

1 + x + (1/2)ln(√

x + 1 − 1) + (−1/2)ln(√

x + 1 + 1) + C

2.11 comments

When you say Something equivalent to something, do not mean you said Some- thing 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 x³

√3

x²+ 1dx my method is do the easiest job, let

x = u³→ x²= u⁶and, x³= u⁹ then

dx = 3u²du Z x³

√3

x²+ 1dx =

Z u⁹

u²+ 13u²dx by spend the time on long division, we got

u¹¹

u²+ 1 = u⁹− u⁷+ u⁵− u³+ u u²+ 1 Check it, may be I miss something, then

3

Z u¹¹

u²+ 1du = 3 Z

u⁹− u⁷+ u⁵− u³+ u u²+ 1du then everything you knew them, don’t forget”+C” and

x = u³

2.13 EX.46

Z p

1 +√ x

x dx

my way is

1 +√

x = u²,√

x = u²− 1 then

x = (u²− 1) = u⁴− 2u²+ 1 dx = (4u³− 4u)du

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thus

Z (p 1 +√

x)dx

x = 4

Z u²(u²− 1)du (u²− 1)² = 4

Z u²du u²− 1 since

a + 0 = a, 0 = 1 − 1 u²

u²− 1 = u²− 1 + 1 u²− 1 u²− 1 + 1

u²− 1 =u²− 1 u²− 1+ 1

u²− 1 thus

Z u²du u²− 1 =

Z du +

Z 1

u²− 1du Since we knew

u²− 1 = (u − 1)(u + 1) then

Z 1

u²− 1du =

Z −1/2 u + 1du +

Z −1/2 u − 1du

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 x³

√3

x²+ 1dx = 4[

q 1 +√

x−(1/2)ln(

q 1 +√

x+1)−(1/2)ln(

q 1 +√

x−1)]+C Again, we can make the form better, that mean with more insight.

(−2)ln(

q 1 +√

x + 1) + (−2)ln(

q 1 +√

x − 1) =

(−2)ln(1 +√

x − 1) = (−2)ln(√

x) = ln(1 x) Since

a²− b²= (a + b)(a − b), ln(xy) = ln(x) + ln(y) Z x³

√3

x²+ 1dx = 4 q

1 +√

x + ln(1 x) + C

Funny thing come in, What

the limit meaning?

x→∞lim[4 q

1 +√

x + ln(1 x)] =??

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Hint : lim

x→∞

q 1 +√

x = lim

x→∞

q√ x

x→∞lim x^1/4 ln(_x¹) = ∞

here I used” equal to Infinity”, this is a bad but useful notaion, try to use it correctly.

2.15 EX.51

Z 1

1 + e^xdx Observe that

e^x= u, x = lnu, dx = (1/u)du

then Z 1

1 + e^xdx =

Z du

u(1 + u)

thus Z du

u(1 + u) =

Z −1

(1 + u)du + Z 1

udu

u = e^x substitute it at the first”=”, what hap- pen?

Z −1

(1 + u)du + Z 1

udu = ln(u) − ln(u + 1) + C = ln( u u + 1) + C

thus Z 1

1 + e^xdx = ln( e^x e^x+ 1) + C Q:

x→∞lim ln( e^x e^x+ 1) =?

What it meaning?

2.16 EX.61

Z 1

3 sin(x) − 4 cos(x)dx As the hint they given,let

t = tan(x/2) then

sin(x) = 2t

1 + t², cos(x) =1 − t² 1 + t² and

2 arctan(t) = x, dx = 2 1 + t²dt thus

Z 1

3 sin(x) − 4 cos(x)dx =

Z ²

1+t²dt 3_1+t^2t2 − 4^1−t_1+t²2

=

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Z dt 2t²+ 3t − 2=

Z dt

(t + 2)(2t − 1) = Z −1/5

t + 2dt +

Z 2/5 2t − 1dt = (−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!?