Thursday, April 27, 2017

Unizor - Definite Integrals - Area under Curve

Notes to a video lecture on

Definite Integrals -
Area under Curve

Consider the following problem.
Given a smooth function f(x) (we will always consider smooth functions in terms of continuity and sufficient differentiability), with non-negative values (that is, f(x) ≥ 0), defined on a closed segment [a, b] and represented by its graph on (X,Y) coordinate plane.

In the following we will use the word "area" in a sense of a two-dimensional part of a plane and as a quantitative measure of this part of plane. The context would clarify which one it is used in every case.

Our task is to find the "area under curve" - the measure of a part of coordinate plane bounded on the top by a graph of this function, on the bottom - the X-axis, on the left - by a line x = a and on the right - by a line x = b.
This area looks like this gray shaded part of a plane (for now, do not pay attention to what's written inside this area):

There is no ready to use formula for such an area. We do know how the area of a rectangle is defined, it is a product of its two dimensions - length multiplied by width, but not of such a complicated figure as the one we consider now.
We really have to define what the area of this figure is and then attempt to calculate it based on values of function f(x) on segment [a, b].
We did have a similar problem in Geometry with the area of a circle, and approached it as a sequence of approximations of a complex figure with simple ones. Let's do the same now.

We will use certain intuitive considerations to define the "area under curve" and will prove that this definition is mathematically valid.

The process of approximation starts with dividing segment [a, b] into N intervals by points a=x0x1x2x3...xN=b, and constructing rectangles, based on each interval [xi−1, xi], having the height equaled to a minimum value mi of function f(x) on this interval, similar to the following picture (consider only light blue rectangles)

All the light blue rectangles are completely below the curve (since their heights are minimums of the function values on each interval) and, therefore, the sum of their areas does not exceed the intuitively understood "area under curve".

Alternatively, we can consider a set of rectangles built on the same intervals but of the height equaled to a maximum value Mi of function f(x) on each interval (consider now pink extension of light blue rectangles).
New taller rectangles are higher than the curve and, therefore, the sum of their areas is not less than the intuitively understood "area under curve".

Notice that the union of all blue rectangles resembles the figure we are trying to determine the area of. So, the sum of areas of all rectangles is close to the "area under curve", while always being less than "area under curve".
This sum of areas of rectangles can be expressed by a formula
sN = Σi∈[1,N]mi·(xi−xi−1) = Σi∈[1,N]mi·Δxi−1
where Δxi−1 = xi−xi−1, which is the width of each interval.

Adding pink extension, we note that the sum of these taller rectangles also approximates the "area under curve", while being larger than it.
This sum of areas of rectangles can be expressed by a formula
SN = Σi∈[1,N]Mi·(xi−xi−1) = Σi∈[1,N]Mi·Δxi−1
where Δxi−1 = xi−xi−1, which is the width of each interval.

Both approximations seem to be better if the number of intervals N we divide our segment [a, b] is large. And the larger is N - the better our approximation will be.

This can be confirmed by the following obvious statements.
If we divide any existing interval in two parts, build two rectangles instead of one and calculate the sum of areas of these rectangles, sum sN, based on minimum values of function f(x), will increase and sum SN, based on maximum values of function f(x), will decrease.

Let's continue adding points of partitioning to infinity such that the largest interval's width converges to zero.
During this process sum sN will monotonically increase or stay the same, but not decrease, while being bounded from above by any SN; and sum SNwill monotonically decrease or stay the same, but not increase, while being bounded from below by any sN.
These properties of sequences sN and SN are sufficient to state that both have limits:
lim sN = s
lim SN = S
where s ≤ S, and the limit is understood in terms of making partitioning of segment [a, b] finer and finer with the largest interval shrinking in width to zero.

So, our next step in the process of approximation of the "area under curve" is to divide each of N intervals in two parts, getting twice as many intervals and build twice as many narrower rectangles.
Blue rectangles lying below the curve will be inscribed "tighter" to the area under the graph of function f(x), so the approximation of the "area under curve" with sN will be better.
Blue rectangles with pink extensions lying above the curve will encompass "tighter" the area under the graph of function f(x), so the approximation of the "area under curve" with SN will also be better.

As the number of intervals grows and the width of the largest interval becomes an infinitesimal variable, the difference between sN and SNbecomes smaller and smaller.
If we can prove that this difference converges to zero, as long as the width of the largest interval converges to zero, it would be a sufficient foundation to call the limits sor S (they are the same) the area under curve.


Recall that we are proving this theorem for sufficiently smooth functions. They and their derivatives are assumed to be continuous. In general, the theorem can be proven under weaker conditions (differentiability is not a necessary condition), but for the purposes of this course we are choosing the easier proof that is valid for smooth functions.

Consider the i-th interval [xi−1, xi].
Assume that function f(x) reaches its maximum Mi on this interval at point ξi∈[xi−1, xi] and it reaches its minimum mion this interval at point ηi∈[xi−1, xi].

According to Lagrange Mean Value Theorem, there is a point ζi∈[ξi, ηi]. such that
f Ii)·(ξi−ηi) = Mi−mi
from which we can derive the upper boundary for |Mi−mi|:
|Mi−mi| ≤ max[a,b]{|f I(x)|· maxi{|xi−xi−1|}

Here max[a,b]{|f I(x)|} represents the maximum of the absolute value of the first derivative of f(x) on segment [a,b], which is some constant since f(x) is a smooth function, let's call it K.
The second multiplier maxi{|xi−xi−1|} represents the maximum width of intervals in our partitioning of the original segment into N parts, let's call it WN.

So, we conclude that
|Mi−mi| ≤ K·WN
where WN (the widest interval) is assumed to be an infinitesimal variable as N→∞.

Now we can evaluate the difference SN−sN:
SN−sN = Σi∈[1,N](Mi−mi)·(xi−xi−1) ≤ K·WN·Σi∈[1,N](xi−xi−1) = K·WN·(b−a)

The last expression contains an infinitesimal variable Wmultiplied by two constants.
Therefore, we have proven that the difference between upper SN and lower sN boundaries of "area under curve" is infinitesimal variable if the widest interval of partitioning of our segment [a,b] shrinks in width to zero.

From this follows that if, instead of choosing minimum value mi or maximum value Mon each interval [xi−1,xi] as the height of a corresponding rectangle, we choose any value of a function f(x) on this interval, the limit will be the same since the corresponding sum of areas of these rectangles will always be between sN and SN.

We can conclude now that if we define the area under curve as a limit of sum of areas of rectangles based on intervals we divide the original segment into with the heights equal to values of our function in any point inside the corresponding intervals (left margin xi−1, right margin xi, maximum point ξi, minimum point ηi or any other), the limit will be the same as long as the widest interval's width converges to zero.

That proves the mathematical correctness of this definition. The area under curve, as defined, exists (since the limit of the sums of rectangles exist) and unique (since the limit does not depend on how we proceed partitioning the original segment and how we choose the points where the function value for the height of rectangle is chosen).

End of proof.

Friday, April 21, 2017

Unizor - Indefinite Integrals - Problems 2

Notes to a video lecture on

Indefinite Integral -
Problems 2

All these problems require certain guessing based on known rules of integration (substitution and "by parts") and recollection of derivatives of known functions.
These problems are illustration of a notion of integration as a form of art more than a skill.

Problem 2.1:

 d/ (cos(x)−sin(x))

One way to approach this problem was offered in a previous lecture and was based on the identity
cos(x)−sin(x) =
= (cos(x)·cos(π/4) −
− sin(x)·sin(π/4)) / (√2/2) =
= √2·cos(x+π/4)

The result of the integration using this method was
(√2/4)·[ln(1+sin(x+π/4)) −
− ln(1−sin(x+π/4))
] + C

Let's consider a different approach to "rationalize" this problem.
Recall that all trigonometric functions can be expressed as rational functions of a tangent of a half-angle.
In particular,
sin(x) =
= 2·tan(x/2) / (1+tan²(x/2))

cos(x) =
= (1−tan²(x/2)) / (1+tan²(x/2))

From these identities we derive
cos(x) − sin(x) =
= (1−tan²(x/2)−2·tan(x/2)) /
/ (1+tan²(x/2))

dx / (cos(x) − sin(x)) =
= (1+tan²(x/2)) 
dx /
/ (1−tan²(x/2)−2·tan(x/2))

Derivative of tangent can also be expressed in terms of tangent:
tanI(x) = 1+tan²(x)
tan(x/2) =
= (1/2)·(1+tan²(x/2)) 

(1+tan²(x/2)) dx =
= 2· 
d tan(x/2)

Therefore, our integral can be expressed in terms of an integral of some rational function of tangent as follows:
 [2· d tan(x/2) /
/ (1−tan²(x/2)−2·tan(x/2))

Now let's substitute y=tan(x/2).

 2· dy / (1−y²−2y) =
[(y+1−√2)·(y+1+√2)] =
= (√2/2)·
dy / (y+1+√2) −
dy / (y+1−√2)
 ] =
= (√2/2)·
[ln(y+1+√2) − ln(y+1−√2)] + C

Going back through substitution, we have the result of integration:
(√2/2)·[ln(tan(x/2)+1+√2) − ln(tan(x/2)+1−√2)] + C

As we see, this result is quite different from the one we obtained in the previous lecture that we mentioned above, but it's still correct.

Let's check it by differentiation.

Dx {(√2/2)·[ln(tan(x/2)+1+√2) − ln(tan(x/2)+1−√2)] + C } =

= (√2/2)·
[1/(tan(x/2)+1+√2) −
− 1/(tan(x/2)+1−√2)
Dx tan²(x/2) } =

= (√2/2)·
[1/(tan(x/2)+1+√2) −
− 1/(tan(x/2)+1−√2)
[1+tan²(x/2)]/2 =

= −
[1+tan²(x/2)] /
] =

= −
[1+tan²(x/2)] /

The numerator equals to:
The denominator equals to:

The result of division of numerator by denominator, considering the minus sign in-front of the whole fraction, equals to:
] =
= 1/

Since differentiation of our result proves that it's correct, it should differ from the result of the previous lecture for this integral by a constant.

Whoever suggests the trigonometric proof that the formula for a result from a previous lecture
(√2/4)·[ln(1+sin(x+π/4)) −
− ln(1−sin(x+π/4))

plus some constant equals to a new result
(√2/2)·[ln(tan(x/2)+1+√2) − ln(tan(x/2)+1−√2)]
will be mentioned in the notes for this lecture.

Problem 2.2:

 d/ (cos(x)−sin(x))

Yes, still the same problem, but yet another approach to "rationalize" it.

Recall the wonderful Euler's formula that defines the exponential function of a complex argument:
eix = cos(x) + i·sin(x)
Using this formula, we can express both cos(x) and sin(x) as exponential functions of a complex argument as follows:
e−ix = ei·(−x) =
= cos(−x) + i·sin(−x) =
= cos(x) − i·sin(x)

cos(x) = (1/2)·(eix+e−ix)
sin(x) = (1/(2i))·(eix−e−ix)
Since i² = −1, we can replace 1/i with −i and the last expression for sin(x) would look like this:
sin(x) = −(i/2)·(eix−e−ix) =
= (i/2)·(e−ix−eix)

We are planning to use this representation to reduce our integral to something familiar. The problem is, of course, that we never explained what is a derivative or an integral when dealing with functions that take real arguments but take complex values, as is the case with functions like eix.
However, we will use the same techniques with these functions as we did with real functions. Strictly speaking, we have to prove that all these techniques are applicable (and they are!), but the procedures to prove all these properties are exactly the same as for real functions, So, we'll skip this and use all the available apparatus as if we have proven its applicability.

Let's convert the denominator first.
cos(x)−sin(x) =
= (1/2)·(eix+e−ix) −
− (i/2)·(e−ix−eix) =
= (1/2)·
[(1+i)·eix+(1−i)·e−ix] =
= (e−ix/2)
[(1+i)·e2ix+(1−i)] =
= (1−i)(e−ix/2)·

Notice that
(1+i)/(1−i) = i
(1−i)·i = i−i² = 1+i
Also, e2ix = (eix
and e−ix = 1 / eix

That makes our denominator look like this:
cos(x)−sin(x) =
= (1−i)·
[i·(eix)² +1] / (2·eix)

As we see, all depends now on eix.
This prompts us to use substitution y=eix.
We also have to express dx in terms of dy as follows:
x = ln(y)/i = −i·ln(y)
dx = −i·d/ y
Now our integral equals to
2·y·(−i·dy/y)/{(1−i)·[i·y² +1]} =
dy/[i·y² +1]

One more trivial substitution z=√i·y would result in the denominator within this integral to be equal to 1+z² familiar from differentiation of arctan(z).
In this case dy=dz/√i and our integral equals to
[−2√i/(1−i)]· dz/[z² +1]

The integral itself equals to arctan(z)+C.
As for a multiplier, as was shown in the chapter "Complex Numbers" (and trivially checked directly),
[(√2/2)·(i+1)]² = i
So, we replace i with
Also, as we saw above,
(1+i)/(1−i) = i
That simplifies the result of integration to
−√2·i·arctan(z) + C

Reversing the substitutions, we get the following result of integration
 d/ (cos(x)−sin(x)) =

Notice that, regardless of the presence of an imaginary i, this expression should be a real function. To prove it, we have to get deeper into functions of complex arguments, which is beyond the scope of this course.

Check by differentiation:
= −√2·i·(i+1)·i·eix /
/ √2·[1+(1+i)²e2ix/2] =
= (1+i)·eix / (1+i·e2ix) =
= 2·eix / 
[(1+i·e2ix)·(1−i)] =
= 2 / 
[(1−i)·e−ix+(1+i)·eix)] =
= 1/

The end.

Wednesday, April 12, 2017

Unizor - Indefinite Integrals - Problems 1

Notes to a video lecture on

Indefinite Integral -
Problems 1

All these problems require certain guessing based on known rules of integration (substitution and "by parts") and recollection of derivatives of known functions.
These problems are illustration of a notion of integration as a form of art more than a skill.

Problem 1.1:

 sin(x)/cos³(x) dx

We know that (cos(x))I = −sin(x).
Therefore, we can combine sin(x) in the numerator and dx to obtain −dcos(x).
Our integral is transformed into
This substitution allows to integrate as if we have a power function  xa dx = xa+1/(a+1) where a=−3.
Since  dx/x³ = −1/(2x²)+C,
our integral equals to

Checking the answer:
(1/(2cos²(x)))I = (1/2)(1/cos²(x))I = (1/2)(−2/cos³(x))·(−sin(x)) = sin(x)/cos³(x).
The end.

Problem 1.2:

 ln(sin(x))·cot(x) dx

To find this integral, notice that
cot(x) = cos(x)/sin(x)
Since (sin(x))I = cos(x),
we use sin(x) as an inner function:
ln(sin(x))·cot(x) = ln(sin(x))·(sin(x))I/sin(x)
Next, notice that, since
(ln(x))I = 1/x,
(ln(sin(x)))I = (1/sin(x))·(sin(x))I
ln(sin(x))·cot(x) = ln(sin(x))·(ln(sin(x)))I

Using the above substitution, we obtain the answer:
 ln(sin(x))·cot(x) dx =  ln(sin(x))·dln(sin(x)) = ln²(sin(x))/2 + C

Checking by differentiating the answer:
Dx (1/2)ln²(sin(x)) = ln(sin(x))·(1/sin(x))·cos(x) = ln(sin(x))·cot(x)
The end.

Problem 1.3:


To find this integral, break it in two integrals:
 ln(x)/(x·√ln(x)dx −3/(x·√ln(x) dx) =  √ln(x) dln(x) −3/√ln(x) dln(x) =

= (2/3)√ln³(x) − 6√ln(x) + C

Checking by differentiating the answer:
Dx ((2/3)√ln³(x) − 6√ln(x)) = (2/3)·(3/2)·√ln(x)·(1/x) − 6·(1/2)·(1/√ln(x)·(1/x) =
= (1/x)·(√ln(x) − 3/√ln(x)) = (ln(x)−3)/(x·√ln(x))

The end

Problem 1.4:

 x²·ex/2 dx

Let's use the fact that exponential function does not change much with differentiation.
Transform our integral into
 2x² dex/2
Now we can use integration by parts twice getting
2x²·ex/2 − 2 ex/2 dx² = 2x²·ex/2 − 2 ex/2·2x dx = 2x²·ex/2 − 2 2x·2 dex/2 =
= 2x²·ex/2−8ex/2·x+8 ex/2 
dx = 2ex/2(x²−4x+8) + C

Checking by differentiating the answer:
Dx (2ex/2(x²−4x+8)) = 2ex/2·(1/2)·(x²−4x+8) + 2ex/2·(2x−4) =
= ex/2·(x²−4x+8+4x−8) = ex/2·x²

The end

Problem 1.5:

 d/ √(−x²+4x+5)

Notice that
arcsinI(x) = 1/√(1-x²)
The fact that a coefficient with  is negative is very important. It prompts that we can try to transform our original function to be integrated into a form similar to a derivative of arcsin(x).
So, our task is to transform an expression under an integral into an expression that looks like the above derivative with linear transformation of variable x.
Basically, we have to express a quadratic polynomial under a square root as a full square expression.
−x²+4x+5 = 9−(x−2)² = 9·(1−(x−2)²/9) = 9·(1−((x−2)/3)²)

Let's substitute
y = (x−2)/3
Then x = 3y+2 and
dx = 3dy
Then our integral looks like this:
 3dy/√(9·(1−y²)) =  dy/√(1−y²) = arcsin(y) + C = arcsin((x−2)/3) + C

Checking by differentiating the answer:
Dx arcsin((x−2)/3) = 1/(3√(1−((x−2)/3))²)) = 1/√(9−x²+4x-4) = 1/(−x²+4x+5)
The end

Problem 1.6:

 d/ (cos(x)−sin(x))

It's easy to deal with either cos(...) or sin(...).
To accomplish this, we can use the property
and transform the denominator into the following form:
cos(x)−sin(x) = (cos(x)·cos(π/4) − sin(x)·sin(π/4)) / (√2/2) = √2·cos(x+π/4)

Substitute y = x+π/4.
Then dy = dx.
Our integral now looks like
 d/ (√2·cos(y)) = (√2/2) d/ cos(y)

To find the above integral, we can multiply the nominator and denominator by sinI(y) = cos(y) and express everything in terms of sin(y):
(√2/2) dsin(y) / (1−sin²(y))

New substitution z = sin(y) leads to the following integral:
(√2/2) d/ (1−z²)
Now we can use an obvious identity
/ (1−z²) = (1/2)·((1/(1−z) + 1/(1+z))

Our integral equals to a sum of two integrals:
(√2/4) d/ (1−z) + (√2/4) d/ (1+z) = (√2/4)[−ln(1−z)+ln(1+z)] + C

Going back through substitutions used above, we obtain
(√2/4)·[ln(1+sin(x+π/4)) − ln(1−sin(x+π/4))]

Checking by differentiating the answer:
Dx {(√2/4)·[ln(1+sin(x+π/4)) − ln(1−sin(x+π/4))]}
Notice that
Dx ln(1+sin(x+π/4)) = [1/(1+sin(x+π/4))] · Dx(1+sin(x+π/4)) = cos(x+π/4)/(1+sin(x+π/4))
Dx ln(1−sin(x+π/4)) = [1/(1−sin(x+π/4))] · Dx(1−sin(x+π/4)) = −cos(x+π/4)/(1+sin(x+π/4))
Therefore, our derivative equals to the following:
(√2/4)·cos(x+π/4)·[1/(1+sin(x+π/4)) + 1/(1−sin(x+π/4))] =
= (√2/4)·cos(x+π/4)·2/(1−sin²(x+π/4)) = (√2/4)·cos(x+π/4)·2/(cos²(x+π/4)) =
= (√2/2)/cos(x+π/4) = 1/(cos(x)−sin(x))

The end

Tuesday, April 4, 2017

Unizor - Indefinite Integrals - Variable Substitution

Notes to a video lecture on

Indefinite Integral -
Variable Substitution

Assume, you know that
 f(x) dx = g(x) + C

What immediately follows from this is that the derivative of g(x) is the original function f(x):
gI(x) = f(x)
or, equivalently,
dg(x) = f(x) dx

Given that, consider a derivative of a compound function g(w(x)):
[g(w(x))]I = f(w(x))·wI(x)
or, equivalently,
dg(w(x)) = f(w(x))·wI(x) dx = f(w(x)) dw(x)

From equality of derivatives or differentials of two functions
f(w(x)) dw(x) = dg(w(x))
follows that original functions (before the differentiation or obtained from the differential by integration) differ only by a constant.
 f(w(x)) dw(x) =  dg(w(x))

Since diiferentiation and integration are inversed to each other, the right part represents a function g(w(x)) (plus constant, as usually with integration).
 f(w(x)) dw(x) = g(w(x)) + C
or, equivalently,
 f(w(x))·wI(x) dx = g(w(x)) + C

Compare this with original relationship between f(x) and g(x) above:
 f(x) dx = g(x) + C
As we see, to integrate f(w(x))·wI(x), it is sufficient to integrate f(x) and substitute w(x) instead of x in the answer.

This is a substitution rule of integration.

Example 1:

 x·e(x²) dx
To find this integral, notice that we can do the following:
f(x) = ex
 f(x) dx = ex + C
w(x) = x²
wI(x) = 2x
f(w(x)) = e(x²)
f(w(x))·wI(x) = e(x²)·2x = e(x²)·(x²)I
f(w(x)) dw(x) = e(x²) d(x²)

Using the above substitution, we obtain the answer:
 x·e(x²) dx = (1/2) e(x²) d(x²) = (1/2)e(x²) + C

Checking by differentiating the answer (Dx is the operation of differentiation):
Dx (1/2)e(x²) = (1/2)e(x²)·Dx x² = (1/2)e(x²)·2x = x·e(x²)
The end

Example 2:

 sin(x)·cos²(x) dx
To find this integral, notice that we can do the following:
f(x) = x²
 f(x) dx = x³/3 + C
w(x) = cos(x)
wI(x) = −sin(x)
f(w(x)) = cos²(x)
f(w(x))·wI(x) = −cos²(x)·sin(x) = cos²(x)·(cos(x))I
f(w(x)) dw(x) = −cos²(x) dcos(x)

Using the above substitution, we obtain the answer:
 sin(x)·cos²(x) dx = − cos²(x) dcos(x) = −cos³(x)/3 + C

Checking by differentiating the answer:
Dx −cos³(x)/3 = −3cos²(x)·(−sin(x))/3 = cos²(x)·sin(x)
The end.

Example 3:

 ln(sin(x))·cos(x) dx
To find this integral, notice that
(sin(x))I = cos(x),
we use sin(x) as an inner function:
ln(sin(x))·cos(x) = ln(sin(x))·(sin(x))I
Also recall from the previous lecture that
 ln(x) dx = x·ln(x) − x + C

Using the above substitution, we obtain the answer:
 ln(sin(x))·cos(x) dx =  ln(sin(x)) dsin(x) = sin(x)·ln(sin(x)) − sin(x) + C

Checking by differentiating the answer:
Dx (sin·ln(sin(x)) − sin(x)) = ln(sin(x))·cos(x) + sin(x)·(1/sin(x))·cos(x) − cos(x) = ln(sin(x))·cos(x)
The end.

Monday, April 3, 2017

Unizor - Indefinite Integrals - Integration "By Parts"

Notes to a video lecture on

Indefinite Integral -
Integration 'by Parts'

First, a reminder of integration 'by parts':
 [f(x) · gI(x)] dx = f(x) · g(x) −  [fI(x) · g(x)] dx
Different form of this rule:
 f(x) · dg(x) = f(x) · g(x) −  g(x) · df(x)
A short form can be written as:
 f·dg = f·g −  g·df

Example 1:
 x·ln(x) d
f(x)=ln(x) and
x²(2ln(x)−1)/4 + C

Example 2:
 ex·cos(x) d
Use integration 'by parts' twice.
ex(sin(x)+cos(x))/2 + C

Example 3:
 x√x+1 d
u=x; dv=√x+1 dx;
(2/3)·x·(x+1)3/2 − (4/15)·(x+1)5/2 + C

Example 4:
 x·ln²(x) d
Integrate 'by parts' twice.Answer:
(1/2)x²ln²(x) − (1/2)x²ln(x) + (1/4)x²+C

Example 5:
 arctan(x) d
x·arctan(x) − ln(1+x²)/2 + C

Example 6:
 x·arctan(x) d
(x²+1)·arctan(x)/2 − x/2 + C

Monday, February 13, 2017

Unizor - Indefinite Integrals - Basic Properties

Notes to a video lecture on

Indefinite Integral -
Basic Properties

1. Integral of differential
 df(x) =  fI(x)dx = f(x) + C
 dsin(x) =  sinI(x)dx =
= sin(x) + C

 cos(x)dx =  sinI(x)dx =
= sin(x) + C

2. Constant multiplier
 a·f(x)dx = a· f(x)dx
 5·x4dx = 5· x4dx =
= 5·x5/5 = x5

3. Sum of functions
 [f(x) + g(x)] dx =
dx +  g(x)dx

 (4x³+3x²+2x+1) dx =
dx +  3x²dx +
dx + dx =
= x4 + x³ + x² + x + C

4. Integration "by-parts"
4a.  [f(x) · gI(x)] dx =
= f(x) · g(x) −  
[fI(x) · g(x)] dx
 (x²·exdx =
= x²·ex −  (2x·ex
dx =
= x²·ex − 2x·ex +  (2·ex
dx =
= (x² − 2x + 2)·ex + C

4b.  f(x) · dg(x) =
= f(x) · g(x) −  g(x) · 

 x² dsin(x) =
= x²·sin(x) −  sin(x) 
d(x²) =
= x²·sin(x) −  2x·sin(x) 
dx =
= x²·sin(x) +  2x 
dcos(x) =
= x²·sin(x) + 2x·cos(x) −
− 2sin(x) + C

4c.  f(x) dx = f(x)·x −  x df(x)
 ln(x) dx =
= x·ln(x) −  x 
dln(x) =
= x·ln(x) −  x·(1/x) 
dx =
= x·ln(x) − x + C

Friday, February 10, 2017

Unizor - Indefinite Integrals - Simple Examples

Notes to a video lecture on

Indefinite Integral - Examples

1.  dx = x + C

2.  xa dx = xa+1/(a+1) + C        (a ≠ −1)

3.  ex dx = ex + C

4.  ax dx = ax/ln(a) + C

5.  1/x dx = ln(x) + C

 sin(x) dx = −cos(x) + C

7.  cos(x) dx = sin(x) + C

8.  (cos(x)−sin(x)) dx = cos(x)+sin(x) + C

9.  sh(x) dx = ch(x) + C

10.  ch(x) dx = sh(x) + C

Unizor - Indefinite Integrals - Speed

Notes to a video lecture on

Indefinite Integral - Speed and Distance

Consider a function s(t) that describes the distance covered by a moving object as a function of time.
Recall the procedure of differentiation as it is applied to this function.
Since a ratio of increment of distance Δs covered during an increment of time from momentt to moment tt is average speed during this interval of time, the limit of this ratio, as time interval Δt converges to zero, is the instantaneous speed of an object at moment t:
v(t) = limΔt→0s/Δt] = ds/dt = sI(t)

Consider a situation when we monitor only the speed as a function of time v(t). Can we find the distance we have covered?
The operation of integration answers this.

Since indefinite integral of v(t)describes a function, whose derivative is v(t), it fits exactly what distance, as a function of time s(t), is. Additional constant, participating in the result of integration, can be interpreted as distance covered before we started our observation of speed and does not change the character of dependency between speed as a function of time v(t) and distance as a function of times(t).

This can be expressed as
 v(t) dt = s(t)+C

Let's analyze the above expression of an integral.
Since v(t) is an instantaneous speed at moment t, the result of its multiplication by an infinitesimal increment of time dt represents a distance covered during this infinitesimal interval from t to t+dt.
To get the overall distance, we have summarize infinite number of infinitesimal distances, so the sign of integration can be interpreted as such sum. Incidentally, a sign reminds a letter S (from the word "sum") stretched vertically.

The point of this analysis is that the whole expression  v(t) dcan be interpreted as "sum of infinite number of infinitesimal products of v(t) by dt".

Monday, February 6, 2017

Unizor - Indefinite Integrals - Definition

Notes to a video lecture on

Indefinite Integral - Definition

Consider a set of sufficiently smooth (differentiable as many times as the context implies) functions and an operation of differentiations.
This operation allows for any such smooth function (an element of our set) to find a corresponding target - its derivative (also an element of our set). So, the differentiation is an unary operation on a set of smooth functions.

When we talk about unary operations, we always want to know the properties of these operations. We have already considered how differentiation works on a function multiplied by a constant, how it works on a sum of two functions, their product and ratio. We did not touch the inverse operation yet.
Integration is the inverse operation to differentiation, and we would like to define it more precisely.

The meaning of the inverse operation on a set is that, if applied on a result of a direct operation, we will get an original element of a set the direct operation was applied to.
In our case, the direct operation was a differentiation that resulted in a derivative of an original function. So, we expect that integration, if applied on a derivative of an original function, results in that original function.

Well, this is not exactly true. Integration is not an inverse operation to differentiation in a classical meaning of the word "inverse". The reason is simple, and it's the same as in case of an operation of square root as not exactly an inverse to raising to the power of two. As we know, 2² = (−2)² = 4. So, raising to the power of two is an operation that for any real number finds its square. But operation of square root from has two targets, 2 and −2, which makes this operation returning two different values, which does not correspond to a classical concept of operation delivering one exact target for any source.

As in a case with square root, integration of a function results in more than one function, derivative of which equals to a subject of integration.
Consider two functions, f(x)and g(x)=f(x)+C, where C is a constant.
Derivatives of these two functions are the same because (we use symbol Dx for operation of differentiation)
Dxg(x) = Dx(f(x)+C) =
= Dx f(x)+DxC = Dx f(x)

since derivative of a constant is zero.

Since our constant C can be any real number, we have infinite number of functions, derivatives of which are the same. It's quantitatively more complex than with an operation of square root, where we have only two numbers with equal squares, but the idea is the same, and we have deal with this somehow.

One of the ways we deal with square root, we use a symbol ±to indicate both results of this operation, implying that both positive and negative numbers, if squared, give the original element we applied the operation of square root.
Analogously, since all functions, that differ only by an additive constant, result in the same derivative, we just use symbolic +C with any specific function, derivative of which gives the function we integrated, to specify all the possibilities of integration.

Again, referring to square roots, where mathematicians invented a special symbol for this operation    , a special symbol for integration was invented as well, it's called "integral" symbol and it looks like this: .

So, if Dx f(x) = g(x), then
 g(x) = f(x)+C

To emphasize that differentiation was by argument x, we used an index in an operation of differentiation Dxor symbol d/dx. For similar purposes we would like to say that an argument of integration is x. To specify this, we add dat the end of operation of integration, so the complete notation is
 g(x) dx = f(x)+C,
which implies that the derivative of f(x) (as well as derivative of any other function that differs from it by a constant) is g(x).

Here are a few examples.

Dx xn = n·xn−1 ⇒
⇒  xn dx = xn+1/(n+1) + C
(for all n ≠ −1,
see integral of 1/x below)

Dx ex = ex ⇒
⇒  ex dx = ex + C

Dx sin(x) = cos(x) ⇒
⇒  cos(x) dx = sin(x) + C

Dx ln(x) = 1/x ⇒
⇒  1/x dx = ln(x) + C

We can now define an operation of integration more precisely.
A set of functions f(x)+C, where f(x) is some concrete smooth function and C is any real constant, is called an integral of function g(x), if derivative of f(x) is equal tog(x).
Sometimes any specific function f(x), whose derivative equals to g(x) is called anti-derivative of function g(x).

There is one unfinished detail in this definition.
Obviously, our goal is to find all such functions, derivative of which is equal to a function we integrate. Assuming we found one, we can add any real constant to it and get another function, whose derivative equals to the function we integrate. Does this procedure delivers all the answers? In other words, do all functions, derivative of which equals to a given function, differ only by a constant?

The answer is yes, and here is the proof.

Recall a lecture "Constant Function" among "Main Theorems" of derivatives. In this lecture we have proven that, if the function has derivative equal to zero at each value of an argument, then this function is constant.
Now assume that two functions,f1(x) and f2(x), have derivatives equaled to each other:
d/dx f1(x) = d/dx f2(x)
From this we conclude that
d/dx (f1(x) − f2(x)) = 0
Therefore, according to a theorem mentioned above,
f1(x) − f2(x) = C
(where C - some constant).
End of proof.

As we see, to find an integral of a function, that is to find all such functions, whose derivative equals to the original integrated function, it's sufficient to find just one function, whose derivative equals to our integrated function, and, adding any real constant to it, we will obtain all other functions, whose derivative is the same, and we are certain that there are no other solutions to our integration.

Wednesday, February 1, 2017

Unizor - Derivatives -Problems 4

Notes to a video lecture on

Derivatives - Problems 4

Problem 4.1
For an implicitly defined function
find the second derivative
y''xx(t) = y/dx².

First derivative dy/dx was discussed in the lecture on differentiation of implicitly defined function and is
y'x(t) = dy(t)/dx(t) = [dy(t)/dt]/[dx(t)/dt]
The second derivative is a derivative of the first derivative:
y''xx(t) = d/dx[y'x(t)] =
d/dx{[dy(t)/dt/ [dx(t)/dt]} =
d/dx[y't(t)/x't(t)] =
= {d/dt[y't(t)/x't(t)]}/[dx(t)/dt]=
= {d/dt[y't(t)/x't(t)]} / x't(t) =
= [y''tt(t)·x't(t)−y't(t)·x''t(t)/ {[x't(t)]²·x't(t)} =
= [y''tt(t)·x't(t)−y't(t)·x''t(t)/ [x't(t)]³ =

Problem 4.2
Using the results of Problem 4.1, find the second derivative of parametrically defined function

y''xx(t) = 2+2·t²

Problem 4.3
Using the results of Problem 4.1, find the second derivative of parametrically defined function

y''xx(t) = 0
This prompts that y as a function of x must be a straight line.
Indeed, since
it follows that
x+2y=1 and
y = (1−x)/2.
Graph of this function is below

The parametric definition of this function puts restrictions on its domain
−1 ≤ x ≤ 1
and range
0 ≤ y ≤ 1.
But within these boundaries y(x) is a linear function.

Problem 4.4
Using the results of Problem 4.1, find the second derivative of parametrically defined function

y''xx(t) = [sin(2t)−2t·cos(2t)/ [2sin³(2t)]

Saturday, January 28, 2017

Unizor - Derivatives - Exercises 3

Notes to a video lecture on

Derivatives - Exercise 3

Exercise 3.1
Using the rules of taking a derivative, find the derivative of
f(x) = sin²(x) + cos²(x)
Why the answer is, what it is?

Exercise 3.2
Using the rules of taking a derivative, find the derivative of
f(x) = ln(ex)
Why the answer is, what it is?

Exercise 3.3
Using the rules of taking a derivative, find the derivative of
f(x) = x+√x

[1+1/(2√x)] /2x+√x

Exercise 3.4
Using the rules of taking a derivative, find the derivative of
f(x) = arctan(1/x)


Exercise 3.5
Using the rules of taking a derivative, find the derivative of
f(x) = x1/x


Exercise 3.6
Using the rules of taking a derivative, find the derivative of this implicitly defined function
yx = xy
(of course, the derivative will also be implicitly defined)


Thursday, January 26, 2017

Unizor - Derivatives -Problems 3

Notes to a video lecture on

Derivatives - Problems 3

Problem 3.1
Prove that function y=xn·e−x is converging to zero as x→∞.

Use L'Hopitale's rule.

Problem 3.2
Prove that function y=ln(x)·x−nis converging to zero as x→∞.

Use L'Hopitale's rule.

Problem 3.3
Taking into consideration that ln(10)=2.302585 and using differential to approximate increment, calculate approximate value of ln(10.1).

(more precise calculation gives value 2.312535)

Problem 3.4
As a continuation of the previous problem, using Taylor series up to second derivative, calculate approximate value of ln(10.1).


Problem 3.5
A particle is moving along a circle of radius R with a center at the origin of coordinates.
The angle from the positive direction of X-axis to a radius to a particle's position is a function φ(t) of time t.
Determine the particle's components of velocity vector {Vx(t), Vy(t)} and absolute speed V(t).

Position of a particle in Cartesian coordinates is as follows:
X(t) = R·cos(φ(t))
Y(t) = R·sin(φ(t))
The components of the velocity vector are
Vx = X'(t) = −R·sin(φ(t))·φ'(t)
Vy = Y'(t) = R·cos(φ(t))·φ'(t)
From this we can calculate the differential along the curve
ds = √[X'(t)]²+[Y'(t)]² ·dt =
= R|φ'(t)|·dt

Therefore, absolute speed is
V = ds/dt = R|φ'(t)|
This corresponds to another way to calculate the absolute speed, using our knowledge from geometry that the length of an arc s, that corresponds to angle φ, equals to R·φ, from which follows that absolute speed as a function of time is R·|φ'(t)|.
F unction φ'(t) is called angular speed of a particle.
If angle φ(t) is changing proportionally to time, that is ifφ(t)=C·t, where C is constant (positive for counterclockwise movement, negative for clockwise), angular speedφ'(t)=C is constant too and, therefore, absolute speed of a particle moving along the circle is constant R·|C| as well.

Problem 3.6
A projectile is launched at angleφ to horizon with initial absolute speed v.
Determine the components of its velocity vector {Vx(t), Vy(t)} and absolute speed V(t) during the time it's rising on its trajectory.

Horizontal component of the velocity Vx=v·cos(φ) remains the same since there are no forces acting in that direction. During the rising part of trajectory vertical component of the speed Vy(t) would decrease from initial value v·sin(φ) by g=9.8m/sec² every second, so its value at time t isVy(t)=v·sin(φ)−g·t.
Absolute speed is
V = √[Vx(t)]²+[Vy(t) =
= √[v·cos(φ)]²+[v·sin(φ)−g·t =
= √v²−2v·sin(φ)·g·t+g²·t²

Let's analyze this formula.
First, let's determine the time our projectile will rise. This is the time of diminishing its vertical component of the velocity Vy(t) from its initial value v·sin(φ) to zero if it's diminishing by g every unit of time.
This time, obviously, is v·sin(φ)/g.
At the end of this period vertical component of the velocity is zero and absolute speed should be equal to horizontal component, which retains its initial value v·cos(φ).
Substituting t=v·sin(φ)/g into a formula for absolute speed, we get:
v²−2v²·sin²(φ)+v²·sin²(φ) =
= v·cos(φ) = Vx
as expected.

Tuesday, January 24, 2017

Unizor - Derivatives - Differential Along a Curve

Notes to a video lecture on

Differential Along the Curve

We will discuss curves on a plane defined parametrically as a pair of functions {x(t), y(t)}, where t is a parameter taking real values in some finite interval [a,b], while x(t) and y(t) are coordinates of points on a curve.
We assume that both coordinate functions are smooth functions of parameter t.

Our main task is to discuss the length that a point covers when it moves along this curve as parameter t changes its value within its domain.

Traditional approach to this task is to approximate the curve with a series of connected segments as presented on this picture.

The idea is to increase the number of these segments, while the length of each would become smaller, and consider the sum of the lengths of these segments to approximate the length of a curve. This approximation would be better if all segments are getting smaller, so we can assume that, if there is a limit of the sum of lengths of these segments, if they are infinitesimally small, this limit is, by definition, the length of a curve.

Let's assume that we deal with a motion of a particle along this curve. It's position {x(t), y(t)} is a function of time t and the length covered by this particle during its motion along the curve would be a function of time s(t).

During the time from tn to tn+1a particle moves along the curve from point {x(tn), y(tn)} to {x(tn+1), y(t+11)}.
If all Δtn+1=tn+1−tn are sufficiently small, the length along the curve between each pair of beginning and ending points can be approximated by the length of straight segment between them. The smaller the time intervals - the better is our approximation.

So, it's reasonable to assume that if the time increment is an infinitesimal variable, the difference between exact length along the curve and the length of a straight segment is an infinitesimal of a higher order, which in subsequent summation of all infinitesimals to a total length of a curve would not affect the limit.

Let Δx(tn+1)=x(tn+1)−x(tn) and Δy(tn+1)=y(tn+1)−y(tn).
Then the length of the straight segment between two points on a curve is

As customary, with all Δtconverging to zero, we will use an infinitesimal variable dt to designate this increment of a parameter t.
With it, both Δx(tn) and Δy(tn)converge to zero and can be represented by infinitesimal variable increments dx and dy.
Analogously, increment of the length Δs(tn) is an infinitesimal variable ds.

So, the formula for infinitesimal increment of the length of a curve, called a differential along the curve, looks like this:
ds(t) = √[dx(t)]²+[dy(t)

As we know, differentials dx(t)and dy(t) can be represented as
dx(t) = x'(t)·dt and
dy(t) = y'(t)·dt
That results in the following representation of a differential along the curve:
ds(t) = √[x'(t)]²+[y'(t) ·dt

Connection to Physics

In Physics of motion we know a concept of velocity - a vector with absolute value of a ratio between distance and time (speed), directed towards the direction of motion.

If our motion is not straight forward, but goes along some curve, the direction of motion at any time is along the tangential line to this curve.
If our motion is not constant in speed, we can talk about a very small interval of time during which we can assume the speed changes just a little and calculate an average speed during this time.

Approaching this from a more mathematical viewpoint, we consider a moment of time t, a distance covered by that time as a function s(t) and infinitesimal increment of time dt.
During this infinitesimal time increment of the distance covered would be
s(t+dt) − s(t) = ds(t) =
= √[x'(t)]²+[y'(t) ·dt

Then the average speed during this infinitesimal period of time dt equals to
|V(t)| = ds(t)/dt =
= √[x'(t)]²+[y'(t)

Derivative x'(t) represents the rate of change of X-coordinate of a particle moving along the curve, derivative y'(t) is the same for Y-coordinate.
What the formula for speed along a curve shows is that velocity of an object moving along a curve, as a vector, can be represented as a sum of two vectors - X-component, a vector directed along the X-axis having value Vx(t)=x(t)=x'(t)and Y-component, a vector directed along the Y-axis having value Vy(t)=y'(t).

In vector form it looks like
V(t) = Vx(t) + Vy(t)
where overline indicates that we deal with vectors and sign + means vector addition. Here V(t) is a velocity vector, Vx(t) is its X-component - projection of velocity onto X-axis and Vy(t)is its Y-component - projection of velocity onto Y-axis.

Tuesday, January 17, 2017

Unizor - Derivatives - Properties of Differential of a Function

Notes to a video lecture on

Properties of Differential

Properties of differential of a function resemble properties of derivative and immediately follow from them.
Below are the proofs of all these properties, using Euler's notation for derivative Dx f(x).

Linear Combination
f(x) = a·g(x) + b·h(x)
df(x) = d(a·g(x) + b·h(x)) =
Dx(a·g(x) + b·h(x))·dx =
= (a·
Dxg(x) + b·Dxh(x))·dx =
= a·
Dxg(x)·dx + b·Dxh(x)·dx =
= a·
dg(x) + b·dh(x)

f(x) = g(x)·h(x)
df(x) = d(g(x)·h(x)) =
Dx(g(x)·h(x))·dx =
= (
Dxg(x)·h(x) +
+ g(x)·
Dxh(x))·dx =
Dxg(x)·dx·h(x) +
+ g(x)·
Dxh(x)·dx =

f(x) = 1/g(x)
df(x) = d(1/g(x)) =
Dx(1/g(x))·dx =
= −
[1/g²(x)]·Dxg(x)·dx =
= −

f(x) = g(h(x))
df(x) = d(g(h(x))) =
Dxg(h(x))·dx =
Dyg(y)·Dxh(x)·dx =

substituting y with h(x).
Let's illustrate this rule for compound functions.
g(x) = sin(x)
h(x) = ln(x)
f(x) = g(h(x)) = sin(ln(x))
First, let's calculate the differential directly by taking a derivative from f(x) and multiplying it by dx.
df(x) = Dx f(x)·dx =
Dxsin(ln(x))·dx =
using the chain rule of differentiation of a compound function
On the other hand, let's see what our expression for differential of a compound function gives.
df(x) = d(g(h(x))) =
Dysin(y)·dln(x) =
substituting y with ln(x).
cos(y)·(1/x)·dx =
As you see, the result is the same as with direct computing the differential.

Implicitly defined differential can be calculated using the above rule for differential of a compound function.
Assume, f(x) = g(h(x)) and we need to find dh(x).
Since df(x) = Dyg(y)·dh(x),
we conclude that
dh(x) = df(x)/Dyg(y)
substituting y with h(x).
For example, we need to find dln(x) without the knowledge that Dxln(x) = 1/x, which would immediately give us
dln(x) = (1/x)·dx
Consider an equality
x = eln(x)
If functions are equal, their differentials at any point are also equal.
dx = deln(x)
Now we can use the expression for a differential of a compound function eln(x) where g(x)=exand h(x)=ln(x).
dx = deln(x) = Dyey·dln(x)
substituting y with ln(x).
dx = ey·dln(x) =
eln(x)·dln(x) = x·dln(x)
dln(x) = (1/x)·dx
which corresponds to the result of direct calculations if we use the known expression Dxln(x)=1/x.

Unizor - Derivatives - Concept of Differential

Notes to a video lecture on

Concept of Differential

A concept of differential of a smooth function f(x) at point x=x0 was briefly introduced when we defined a derivative of a function as a linear function of an infinitesimal increment of its argument with a coefficient of proportionality equal to a derivative of this function at point x=x0.

The notation df(x0)/dx=f I(x0), which was introduced when we defined a concept of derivative, reflects this definition.
Here x0 is any point in the domain of function f(x)dx is an infinitesimal increment of argument x from this point and df(x0) is differential of function f(x) introduced above.

The notation that uses d instead of Δ implies that we are not talking about just any particular increment, but about a process of decreasing this increment to zero, thus making it an infinitesimal variable.

Using this type of notation, we can write the definition of a derivative as follows:
f I(x0) =
= limdx→0
This implies that
[f(x0+dx)−f(x0)]/dx = f I(x0)+ε
where ε is another infinitesimal variable.
Next transformation:
f(x0+dx)−f(x0) =
= f I(x0

or, using the definition of "littleo" as infinitesimal of higher order,
Δf(x0) = f(x0+Δx)−f(x0) =
= f I(x0
df(x0) + o(dx)

Let me emphasize again that in the above statement dx is not just any increment of argument x, but an infinitesimal variable representing an increment converging to zero.
Similarly, df(x0) is an infinitesimal variable representing an infinitesimal function increment during the process of an increment of an argument converging to zero.

From the definition of a differential
df(x0)=f I(x0dx
we see that differential is a function of two arguments: a fixed point x0 within a domain of function f(x) and an infinitesimal increment of an argument dx.
Since x0 is any fixed point, we can talk about a function differential at any value of argument x and use the notation df(x):
df(x)=f I(x)·dx

Here is an illustration of a concept of a differential.

The blue line represents a function, red line - a tangential to it at point A.
Segment BD=AC represents an increment of the argument.
Segment DE is an increment of a function.
Segment DF is function's differential - an increment of a value along the tangential line proportional to the increment of the argument.
When point C moves closer to point A, decreasing the increment of the argument, both segments DE and DF decrease as well, while points E and Fare getting closer to each other, illustrating that increment of a function and its differential are infinitesimals if the increment of the argument is infinitesimal.
What's extremely important is that these two infinitesimals are of the same order as an increment of the argument, while difference between them, segment EF, is an infinitesimal of a higher order.

Recall the Taylor series for function f(x) with expansion center x0:
f(x)=Σn≥0[f (n)(x0)·(x−x0)n/(n!)]

Let's set
and present this series as follows:
Δf(x0) =
= f I(x0
dx+f II(x0)·(dx)²/2+...
According to our definition of the differential, this can be expressed as
Δf(x0) = df(x0) + o(dx)
which illustrates the same concept: increment of a function is of the same order as its differential and they differ by an infinitesimal of a higher order than increment of the argument.

It's appropriate to note here that the concept of a differential of a function justifies the Leibniz's notation for a derivative:
f I(x) = df(x)/dx


Differential df(x0) of a function f(x) at some fixed point x0 of its argument is an infinitesimal variable proportional to infinitesimal increment of the argument dx with a coefficient of proportionality equaled to a derivative of this function at chosen point x0:
df(x0) = f I(x0dx

Differential differs from increment in a sense that increment is a fixed difference between two values, while differential is an infinitesimal variable.
Thus, Δx is a fixed number that is equal to a difference between the incremented value of argument x=x1 and its base value x=x0:
Δx = x1 − x0
But differential is an infinitesimal variable {x1 − x0} in the process of x1x0.
As soon as we switch from a fixed Δx to a process by considering Δx→0, increment of an argument Δx becomes its differential dx.

Similarly, Δf(x) is a fixed number that is equal to a difference between the value of a function at incremented value of the argument and the value of a function at the base value of the argument:
Δf(x) = f(x1) − f(x0)
As x1x0, function increment Δf(x) is getting smaller and relative difference between its values and corresponding values of differential df(x0) are getting smaller as well in a sense that
limΔx→0f(x)]/df(x) = 1
limΔx→0f(x)]/Δx = f I(x)
df(x)/dx = f I(x)
and dx means the same as Δx→0, that is a process of infinitely decreasing increment of an argument.

Friday, January 13, 2017

Unizor - Derivatives - Problem 2

Notes to a video lecture on

Derivatives - Problems 2

Problem 2.1
Artillery officer needs to choose an angle a cannonball should be launched to reach the maximum distance, provided its linear speed at the moment of launching is fixed.
Assume ideal physical conditions (no air resistance, gravity constant is not changed with the height and whatever else can be assumed to simplify the problem).
The maximum distance is reached when a cannonball is launched at an angle of π/4=45o

Problem 2.2
How many zero points does function f(x)=x³−3x+4 have?
Analyze intervals of monotonic behavior of this function and compare the signs of the function on each interval's ends.
This function has only one zero point.

Problem 2.3
Consider pulling an object by a rope on a surface with friction. If the rope is horizontal, the force needed to move it with constant speed should be equal to the force of friction. If a rope is at certain angle to horizon, the vertical component of the force applied to it partially neutralizes the friction.
What angle of a rope to horizon is needed to minimize the force applied to it and still to move forward with constant speed?
Consider ideal conditions and a coefficient of friction equal to k.
The angle or a rope to horizon should be equal to arctan(k).
Assume, the angle of a rope with horizon is φ (obviously, its range is from 0 to π/2), the weight of an object is P and the force applied to a rope is F.
Then vertical component of the pulling force is
Fv = F·sin(φ)
and horizontal component is
Fh = F·cos(φ)
Vertical component Fv reduces the weight and, therefore, reduces the friction. Therefore, the friction equals to
T = (P−Fv)·k
To pull an object with constant speed this friction force must be equal to horizontal component of the force applied to a rope Fh:
T = Fh
The above equation gives a dependency between the force applied to a rope F and angle of a rope to horizon φ. Having this as a function, we can minimize it and find an optimal angle of minimum force.
Fh = (P−Fv)·k
F·cos(φ) = [P−F·sin(φ)]·k
F·cos(φ)+F·sin(φ)·k = P·k
F = P·k/[cos(φ)+sin(φ)·k]
To minimize this function, let's take its derivative by φ and find where it's equal to zero.
dF/dφ = {−P·k/[cos(φ)+sin(φ)·k]²}·

Equation dF/dφ = 0
results in
-sin(φ)+cos(φ)·k = 0
which can be easily solved:
sin(φ) = cos(φ)·k
sin(φ)/cos(φ) = k
tan(φ) = k
φ = arctan(k)
This solution is independent of the weight of an object and means that the greater the friction - the more vertical should be an angle we pull the object to neutralize friction and minimize the force applied to a rope.
If we are talking about practical application of this, when a person pulls something by a rope, for lower friction coefficient we should use longer rope and for higher friction - shorter.