Monday, May 15, 2017

Unizor - Definite Integrals - Improper Integrals Examples 1





Notes to a video lecture on http://www.unizor.com

Improper Definite Integrals

Example 1.1

011/√x dx =
lima→0 a11/√x dx

Indefinite integral of f(x)=1/√x is F(x)=2√x.
Indeed, let's take a derivative of F(x)=2√x:
Dx F(x) = 2·(1/2)·x(1/2)−1 =
= x−1/2 = 1/√x


Using Newton-Leibniz formula,
a11/√x dx = F(1) − F(a) =
= 2√1 − 2√a

As a→0, this expression converges to 2

Answer2

__________

Example 1.2

0e−x dx =
limb→∞ 0be−x dx

Indefinite integral of f(x)=e−x is F(x)=−e−x.
Evaluating integral:
0be−x dx = F(b) − F(0) =
= (−e−b) − (−e−0) = 1 −e−b

As b→∞, this expression converges to 1

Answer1

__________

Example 1.3

01/(1+x²) dx =
limb→∞ 0b1/(1+x²) dx

Indefinite integral of f(x)=1/(1+x²) is F(x)=arctan(x).
Evaluating integral:
0b1/(1+x²) dx = F(b) − F(0) =
= arctan(b) − arctan(0) =
= arctan(b)

As b→∞, this expression converges to π/2

Answerπ/2

__________

Example 1.4

1(1/x²)·e1/x dx =
limb→∞ 1b(1/x²)·e1/x dx

Indefinite integral of
f(x)=(1/x²)·e1/x can be found by noticing that −1/x² is a derivative of 1/x and, therefore, a derivative of e1/x is e1/x·(−1/x²).
Therefore, indefinite integral of f(x)=(1/x²)·e1/x is F(x)=−e1/x.
Evaluating integral:
1b(1/x²)·e1/x dx = F(b)−F(1) =
= (−e1/b) − (−e1/1) = e − e1/b

As b→∞, this expression converges to e −1

Answere −1

Unizor - Definite Integrals - Improper Integrals





Notes to a video lecture on http://www.unizor.com

Improper Definite Integrals

Recall the definition of the definite integral:
abf(x) dx =
lim 
Σi∈[1,N] f(xiΔxi
where Δxi=xi−xi−1 represents partitioning of segment [a,b] into N parts, and it is assumed that the widest interval Δxi is shrinking to zero by length as N→∞.

Before we only defined and calculated these integrals for cases where the segment of definition [a,b] was finite and the integrated function f(x) was, at least, continuous on this segment and, as a consequence, was finite as well.

Consider now cases of integration over unbounded (infinite) domains and/or functions that go to infinity around some point(s) within their domains. Let's start with a statement that we never defined these types of integrals. Riemann sums are not applicable in these cases.

For example, you want to determine an area under the curve y=1/x² from left boundary x=1 to positive infinity.
This function diminishes to zero as x→∞, so, intuitively, the area under this infinite curve might or might not be finite, depending on how fast the function value goes to zero as its argument goes to infinity.

Our first order is to define definite integrals of this kind.

Definition for infinite intervals of integration

We will define an improper integral of each kind as a limit of the corresponding proper integral if and only if this limit exists.

1. af(x) dx =
limb→∞ abf(x) dx

provided this limit exists.

2. b−∞ f(x) dx =
lima→−∞ abf(x) dx

provided this limit exists.

When both margins are infinite, we can define the integral as a sum of two integrals, each with only one margin being infinite, provided both exists in a sense of corresponding limits as defined above.
3. −∞ f(x) dx =
0−∞ f(x) 
dx + 0 f(x) dx


Definition for functions going to infinity

Assume, we are integrating a function that asymptotically goes to infinity around one point within or on the border of a segment of integration.
For example,
01ln(x) dx
As we know, logarithm goes to negative infinity as we approach argument 0, which is a left boundary of integration segment.
To define this integral, we will cut off this point out of integration by stepping side-wise and take a limit of the result as the point of cut-off is getting closer and closer to a point where our function is not defined. If this special point is on the border of a segment of integration, we will have to take only one such limit. If it's in the middle, we will have to split the segment in two parts and integrate each one separately using this technique.

Assume, function f(x) is defined and continuous on interval (a,b] that is open on the left because f(x)→∞ as x→a.
Then we define
4. abf(x) dx =
limd→0 ba+d f(x) dx

For any d, however small, integral on the right exists. So, if there is its limit as d→0, that limit is the definition of the integral on the left.

Analogously, assume, function f(x) is defined and continuous on interval [a,b) that is open on the right because f(x)→∞ as x→b.
Then we define
5. abf(x) dx =
limd→0 ab−df(x) dx

For any d, however small, integral on the right exists. So, if there is its limit as d→0, that limit is the definition of the integral on the left.

Finally, assume, function f(x) is defined and continuous on intervals [a,b) and (b,c] - everywhere at segment [a,c] except point x=b because f(x)→∞ as x→b.
Then we define
6. acf(x) dx =
abf(x) 
dx + bcf(x) dx

provided both integrals on the right exist in a sense of limits defined above.

_______

Example 1

11/x² dx =
limb→∞ 1b1/x² dx

The indefinite integral (antiderivative) of f(x)=1/x² is function g(x)=−1/x. Therefore, the definite integral on the right in the above equality can be evaluated by Newton-Leibniz formula as
1b1/x² dx =
= (−1/b) − (−1/1) = 1 − 1/b

Now we can take a limit of this expression as b→∞:
limb→∞ (1 −1/b) = 1
Therefore, according to the definition of this improper integral,
11/x² dx = 1
_______

Example 2

01ln(x) dx =
limd→0 d1ln(x) dx

Indefinite integral (antiderivative) of f(x)=ln(x) is g(x)=x·ln(x)−x (refer to a lecture about indefinite integrals or just differentiate this expression to check that its derivative is indeed equal to ln(x)).
Using Newton-Leibniz formula, we can evaluate the integral on the right:
d1ln(x) dx =
= (1·ln(1)−1) − (d·ln(d)−d) =
= −1 − d·ln(d) + d

Going to a limit as d→0, we notice that
limd→0 d·ln(d) = 0 (see a note with the proof below) and, therefore, our integral, according to the definition, is
01ln(x) dx = −1
NOTE: Proof of the limit:
limd→0 d·ln(d) =
limd→0 ln(d)/(1/d)

Use L'Hopital's rule to replace the ratio of functions with ratio of their derivatives.
limd→0 ln(d)/(1/d) =
limd→0 (1/d)/(−1/d²) =
limd→0 (−d) = 0

Thursday, May 11, 2017

Unizor - Definite Integrals - Examples on Newton-Leibniz Formula





Notes to a video lecture on http://www.unizor.com

Definite Integrals -
Examples of Integration Using
Newton-Leibniz Formula


In this lecture we will consider exactly the same examples we used to demonstrate how definite integral can be calculated using its definition as a limit of sums.

Example 1

Find "area under curve" for f(x) = 10x on segment [a=0, b=4].

Solution

The indefinite integral (antiderivative) of function f(x) = 10x is 5x² (we can omit addition of a constant).
This function should be evaluated at the end points of integration resulting in the following:
0410x dx =
= 5·4² − 5·0² = 80−0 = 80

This corresponds to the answer we have obtained previously using limit of Riemann sums.
The end.
________________

Example 2

Find "area under curve" for f(x) = −x2+1 on segment [a=−1, b=1].

Solution

The indefinite integral (antiderivative) of function f(x) = −x2+1 is −x³/3+x (we can omit addition of a constant).
This function should be evaluated at the end points of integration resulting in the following:
−11(−x2+1) dx =
[−(1)3/3+(1)] − [−(−1)3/3+(−1)] =
= 2/3 − (−2/3) = 4/3

This corresponds to the answer we have obtained previously using limit of Riemann sums.
The end.
________________

Example 3

The driver slows its car down by pressing the brakes from initial speed 20 meters per second to complete stop in 10 seconds, reducing its speed by the same value each second (linear dependency of the speed on time).
Find the distance the car covered during this breaking process.

Solution

First of all, we have to find the formula that represents the speed as a function of time.
Since every second the car slows down by the same number of meters per second, and it took 10 seconds to reduce the speed from 20 to 0, the function describing the speed is V(t)=20−2t.
The indefinite integral (antiderivative) of function V(t)=20−2t is 20t−t² (we can omit addition of a constant).
This function should be evaluated at the end points of integration [0,10] resulting in the following:
010(20−2t) dx =
[20·10−10²] − [10·0−0²] =
= 100 − 0 = 100

The distance cover by car during breaking is 100 meters.
The end.
________________

Example 4

The Hooke's Law tells that a force needed to expand a string from its neutral position linearly depends on the length we expand it: F=K·x, where F is the force, x - the length of expansion and K - coefficient that depends on the physical properties of a spring.
Given a spring with K=0.5 (in newtons per meter).
Determine work W required to expand this spring by 0.1 meter.

Solution

Physical concept of work is defined as a product of a force by a distance this force is applied if this force is constant. In our case the force is changing with distance. To overcome this, we will approach this problem similarly to calculating an area under curve, where curve represents the force.
Partition the distance (spring's expansion) into small intervals and assume that the force is constant on each interval. If the force is a function of distance F(x) then the amount of its work from distance x=0 to distance x=d can be represented as
0dF(x) dx

Applying this to our case, we have to calculate
00.10.5·x dx
The indefinite integral of function 0.5·x is 0.25·x².
Therefore,
W = 00.10.5·x dx =
= 0.25·0.1² − 0.25·0² = 0.0025

The work equals to 0.0025(joules)
The end.

Tuesday, May 9, 2017

Unizor - Definite Integrals - Newton-Leibniz Formula





Notes to a video lecture on http://www.unizor.com

Definite Integrals -
Newton-Leibniz Formula


Recall the definition of the definite integral:
abf(x) dx =
lim 
Σi∈[1,N] f(xiΔxi
where Δxi=xi−xi−1 represents partitioning of segment [a,b] into N parts, and it is assumed that the widest interval Δxi is shrinking to zero by length as N→∞.

First Fundamental Theorem
of Calculus


Consider a smooth function f(x) on segment [a,b] and any point t∈[a,b].
The following definite integral can be considered as a function of t:
F(t) = atf(x) dx
The First Fundamental Theorem of Calculus states that the derivative of function F(t) is function f(t):
FI(t) = f(t)

Proof

Since, by definition,
FI(t) =
lim
Δt→0[F(t+Δt)−F(t)] /Δt
we have to prove that
limΔt→0 [at+Δtf(x) dx −
− atf(x) dx] /Δt = f(t)

From the properties of definite integrals we know that
atf(x) dx + tt+Δtf(x) dx =
at+Δtf(x) dx
from which follows that
at+Δtf(x) dx − atf(x) dx =
tt+Δtf(x) dx

Therefore, we have to prove that
limΔt→0 tt+Δtf(x) dx /Δt = f(t)

From properties of definite integrals we know that
Δt ≤ tt+Δtf(x) dx ≤ M·Δt
where m is minimum of function f(x) on segment [t,t+Δt] and M - its maximum on this segment.

This allows us to state that the expression
limΔt→0 tt+Δtf(x) dx /Δt
is bounded from below by m (minimum f(t) on segment [t,t+Δt]) and from above by M (maximum f(t) on segment [t,t+Δt]).

As Δt→0, our segment [t,t+Δt] shrinks to a point t. Since we assume sufficient "smoothness" of function f(t) (in this case we need just its continuity), both minimum and maximum of f(t) on segment [t,t+Δt] converge to the same value f(t).
That forces the limit above also to converge to f(t).

End of proof.

IMPORTANT:
Since the derivative of function F(t) above (definite integral of function f(x) on a segment from a to t) equals to f(t), function F(t) can serve as an indefinite integral (antiderivative) of f(t).
Since we proved the existence and uniqueness of a definite (Riemann) integral for any continuous function, the theorem above has proven that for any continuous function there exists indefinite integral (its antiderivative).

Newton-Leibniz Formula

Let's assume that we want to find a definite (Riemann) integral abf(x) dx of some continuous function f(x) on segment [a,b].
Assume further that we know one particular function G(t) which is the indefinite integral (antiderivative) of f(t) (we deliberately decided to use different argument symbol t instead of x as an argument of function G(t) to have x only as a variable of integration).
That is, GI(t) = f(t).

Recall that there are many functions, derivative of which equal to f(t), but we know that all of them differ from each other only by an addition of a constant.

So, on one hand we have G(t) as one of the possible indefinite integrals (antiderivative) of f(t).
On the other hand, we have just proven that derivative of
F(t) = atf(x) dx
by t also equals to function f(t).
Therefore, we have two different functions, derivatives of both of which are the same function f(t):
GI(t) = f(t) and
FI(t) = [atf(x) dx]I = f(t)

Since two antiderivatives differ only by an addition of a constant, we conclude that
F(t) = atf(x) dx = G(t) + C
where C - some constant.

It's easy to find this constant. Since a definite integral on a null-segment [a,a] equals to zero, assign t=a in the above formula getting
aaf(x) dx = G(a) + C = 0
from which we get
C = −G(a)

The final formula for our definite integral, therefore, is
F(t) = atf(x) dx = G(t) − G(a)
In particular, for t=b, we obtain the Newton-Leibniz formula:
abf(x) dx = G(b) − G(a)
where G(t) - any function, derivative of which is f(t), in other words, an indefinite integral (antiderivative) of f(t).

CONCLUSION

To find a definite integral of some real function f(t) on segment t∈[a,b], it is sufficient to find any particular indefinite integral (antiderivative) G(t) of function f(t) and calculate the expression G(b)−G(a).
This establishes a connection between indefinite and definite integrals and justifies the usage of the same word integral for both.

Monday, May 8, 2017

Unizor - Definite Integrals - Other Examples





Notes to a video lecture on http://www.unizor.com

Definite Integrals -
Other Examples


Distance

Consider the following problem.
A car moves along a straight line with variable speed given by function v(t) that defines speed v at any moment of time t.
Our task is to find the distance covered by this car from time moment t=a to moment t=b.

If the speed is constant v(t)=V, the solution is easy:
S = V·(b−a)

For variable speed the problem is not that easy. Here is what can be suggested as an approximate solution.
Let's divide our time interval [a,b] into N equal short intervals [a=t0,t1], [t1,t2], [t2,t3]... [tN−1,tN=b] and assume that during each interval the speed is not significantly changing - a reasonable assumption if the time interval is small enough.
Then the distance covered during i-th time interval [ti−1,ti] is approximately equal to
ΔSi = v(ti)·(ti−ti−1)
Here we use v(ti) (the value on the right margin) as a constant speed during i-th time interval. We could have taken any other value during this interval - on the left margin, minimum on this interval, maximum or any in-between.

Notice that the approach is absolutely equivalent to our approach of finding the area under curve in the previous lecture.

Our next step is to summarize all ΔSi to get a total distance and go to a limit as the number of intervals we divide our total travel time increases to infinity.
So, the final formula is
S = lim Σi∈[1,N] v(ti)·(ti−ti−1) =
lim Σi∈[1,N] v(tiΔti
where limit is assumed to be taken when N→∞ and maximum width among all intervals Δti diminishes to zero.

As with a problem of area under curve, we can prove that the limit is independent of which point within each interval is used to get the speed value. This limit is also independent on how we break the total time travel into shorter intervals as long as the length of the longest among intervals is shrinking to zero as we proceed with dividing the total travel time into more and more intervals.


Draining

Consider the following problem.
There is a tub filled with water. When we open the drain, the water starts flowing out of the tub. The speed of water flow depends on different factors - some constant (the size of a drain pipe) and some variable (the pressure at the drain opening).

Our task is to determine the volume of water that is drained from the tub during some known period of time from t=a to t=b, provided we know the speed of draining v(t) (in some units, like liters per second) at any moment of time t.

The complication here, obviously, is that the speed of water flow through a drain is variable because it depends on the water pressure at the drain opening, which changes as the water flows out of the tub.

Our approach to this problem is similar to the one above.
Let's divide our time interval [a,b] into N equal short intervals [a=t0,t1], [t1,t2], [t2,t3]... [tN−1,tN=b] and assume that during each interval the speed of water flow through a drain is not significantly changing - a reasonable assumption if the time interval is small enough.
Then the volume of water going through a drain during i-th time interval [ti−1,ti] is approximately equal to
ΔWi = v(ti)·(ti−ti−1)
Here we use v(ti) (the value on the right margin) as a constant speed during i-th time interval. We could have taken any other value during this interval - on the left margin, minimum on this interval, maximum or any in-between.

Notice that the approach is absolutely equivalent to our approach of finding the area under curve in the previous lecture.

Our next step is to summarize all ΔWi to get a total volume of drained water and go to a limit as the number of intervals we divide our total drainage time increases to infinity.
So, the final formula is
W = lim Σi∈[1,N] v(ti)·(ti−ti−1) =
lim Σi∈[1,N] v(tiΔti
where limit is assumed to be taken when N→∞ and maximum width among all intervals Δti diminishes to zero.

As with a problem of area under curve, we can prove that this limit is independent of which point within each interval is used to get the speed value. This limit is also independent on how we break the total time of drainage into shorter intervals as long as the length of the longest among intervals is shrinking to zero as we proceed with dividing the total time into more and more intervals.


Volume of Solids of Revolution

Consider the following problem.
There is a solid obtained by a revolution of some curve on a plane around an axis which also lies on this plane.
Let's assume that the curve is defined as a graph of a smooth function y=f(x), where argument x varies from a to b, and the axis of rotation is the X-axis.

Our task is to determine the volume of this solid.

If our function is constant, that is f(x)=c, our solid is a cylinder of height H=b−a and radius equal to that constant c.
If function f(x) is linear (so, its graph is a straight line), our solid is a truncated cone of height H=b−a and radiuses of its two bases equal to Ra=f(a) and Rb=f(b).
In either of the above cases we know classic geometric formulas to calculate the volume of a solid.
The complexity of our problem, obviously, is that the shape of our solid is not one of those well known types.

Let's approach our problem analogously to determining the area under curve.
Let's divide our argument interval [a,b] into N equal short intervals [a=x0,x1], [x1,x2], [x2,x3]... [xN−1,xN=b] and assume that on each interval the value of function f(x) is not significantly changing - a reasonable assumption if the interval is small enough.

Replacing the function values on each interval [xi−1,xi] with a constant value at its right margin f(xi) and rotating the obtained step-function around the X-axis, we obtain a different solid, but the one that approximates the original one to certain degree.
The approximation will be better if the number of intervals we divide segment [a,b] is increasing and the size of the largest interval diminishes to zero.

The result of rotation of a step-function on each interval will be a cylinder with height hi=xi−xi−1 and radius f(xi), its volume will be equal to
ΔVi = π·f 2(xi)·(xi−xi−1) =
= π·f 2(xi
Δxi


Our next step is to summarize all ΔVi to get a total volume of a solid and go to a limit as the number of intervals we divide our segment [a,b].
So, the final formula is
V = lim Σi∈[1,N] π·f 2(xiΔxi
where limit is assumed to be taken when N→∞ and maximum width among all intervals Δxi diminishes to zero.

As with a problem of area under curve, we can prove that this limit is independent of which point within each interval is used to get the radius of a cylinder. This limit is also independent on how we break the total range of arguments into shorter intervals as long as the length of the longest among intervals is shrinking to zero as we proceed with dividing the total range into more and more intervals.


CONCLUSION

In all the cases we considered so far we came up with an expression
lim Σi∈[1,N] f(xiΔxi
where f(x) is some smooth function defined on a segment [a,b],
{xi} is partitioning of segment [a,b] into N parts,
and we assume that N→∞ and the maximum width of intervals Δxi=xi−xi−1 converges to zero.

We have also proven that this limit exists for any smooth function f(x), that it does not depend on how we partition our segment [a,b] (as long as the widest interval of division shrinks to zero length) and it does not depend on which point within each interval of division we use to determine the function value on this interval.

Friday, April 28, 2017

Unizor - Definite Integrals - Area Examples





Notes to a video lecture on http://www.unizor.com

Definite Integrals -
Area Examples


Example 1

Find "area under curve" for f(x) = 10x on segment [a=0, b=4].

Solution

First, let's experiment with a couple of simple cases.

Case 1. N=2
Point of division in two equal parts is x1=2
So, a=x0=0x1=2x2=4=b
Minimum on the first interval [0, 2] is 0 at x=0.
Minimum on the second interval [2, 4] is 20 at x=2.
Therefore, the sum of areas of all rectangles equals to
S1 = 0·2 + 20·2 = 40.

Case 2. N=4
Two additional points of division in four equal parts are x1=1 and x3=3
So, a=x0=0x1=1x2=2,
x3=3x4=4=b
Minimum on the first interval [0, 1] is 0 at x=0.
Minimum on the second interval [1, 2] is 10 at x=1.
Minimum on the third interval [2, 3] is 20 at x=2.
Minimum on the fourth interval [3, 4] is 30 at x=3.
Therefore, the sum of areas of all rectangles equals to
S2 = 0·1 + 10·1 +
+ 20·1 + 30·1 = 60
.

To proceed to a general case, assume we have divided our segment into N equal parts, and find the limit of the area of all rectangles as N→∞.
In this case the width of each interval equals to
Δxi = 4/N
Right margin of i-th interval is
xi = i·4/N
The function value at this right margin is
f(xi) = 10·i·4/N
The area of the i-th rectangle, constructed on the i-th interval as a base and having height calculated above, equals to
Si = 10·i·(4/N)·(4/N) =
= (160/N²)·i


Our task is to sum these areas for i changing from 1 to N and find the limit of this sum as N→∞.
Summation by i is simple, we did this before (see "Algebra - Sequence and Series" chapter of this course or prove the following by induction).
Recall that
Σ[1,K]i = K·(K+1)/2

Therefore, the result of summation of the areas of N rectangles is
Σ[1,N]Si=(160/N²)·N·(N+1)/2=
= 80·(1+(1/N))


As N→∞, this value converges to 80, which constitutes the "area under curve".

Incidentally, our "area under curve" can be calculated as the area of a right triangle with base 4 and height 10·4=40, which equals to S=(1/2)·4·40=80 - the same answer as we have received through rather complicated summation and taking the limit.
That confirms the validity of our answer.

The end.
________________

Example 2

Find "area under curve" for f(x) = −x2+1 on segment [a=−1, b=1].

Solution

Again, let's consider two particular cases prior to generalize the problem. We will use the values of a function on the right margin of each interval.

Case 1. N=2
Point of division in two equal parts is x1=0
So, a=x0=−1x1=0x2=1=b
The function value on the right margin of the first interval [−1, 0] is 1 at x=0.
The function value on the right margin of the second interval [−1, 0] is 0 at x=1.
Therefore, the sum of areas of all rectangles equals to
S1 = 1·1 + 0·1 = 1.

Case 2. N=4
Two additional points of division in four equal parts are x1=−0.5 and x3=0.5
So, a=x0=−1x1=−0.5x2=0,
x3=0.5x4=1=b
The function value on the right margin of the first interval [−1, −0.5] is 3/4 at x=−0.5.
The function value on the right margin of the second interval [−0.5, 0] is 1 at x=0.
The function value on the right margin of the third interval [0, 0.5] is 3/4 at x=0.5.
The function value on the right margin of the fourth interval [0.5, 1] is 0 at x=1.
Therefore, the sum of areas of all rectangles equals to
S2 = (3/4)·0.5 + 1·0.5 +
+ (3/4)·0.5 + 0·0.5 = 5/4
.

To proceed to a general case, assume we have divided our segment into N equal parts, and find the limit of the area of all rectangles as N→∞.
In this case the width of each interval equals to
Δxi = 2/N
Right margin of i-th interval is
xi = −1+i·2/N
The function value at this right margin is
f(xi) = 1−(−1+i·2/N)² =
= (4/N)·i−(4/N²)·i²

The area of the i-th rectangle, constructed on the i-th interval as a base and having height calculated above, equals to
Si = Δxi·f(xi) =
= (2/N)·
[(4/N)·i−(4/N²)·i²] =
= (8/N²)·i−(8/N³)·i²


Our task is to sum these areas for i changing from 1 to N and find the limit of this sum as N→∞.
Recall that
Σ[1,K]i = K·(K+1)/2
Σ[1,K]i² = K·(K+1)·(2K+1)/6

Therefore, the result of summation of the areas of N rectangles is
Σ[1,N]Si = (8/N²)·(N·(N+1)/2 −
− (8/N³)·N·(N+1)·(2N+1)/6


As N→∞, this value converges to 4−(16/6)=4/3, which constitutes the "area under curve".

The end.

Thursday, April 27, 2017

Unizor - Definite Integrals - Area under Curve





Notes to a video lecture on http://www.unizor.com

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.

Proof

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.