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

__Definite Integrals -__

Area Examples

Area Examples

*Example 1*

Find "area under curve" for

**f(x) = 10x***].*

**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

**x**_{1}=2So,

*,*

**a=x**_{0}=0*,*

**x**_{1}=2

**x**_{2}=4=bMinimum on the first interval

*]*

**0, 2***at*

**0***.*

**x=0**Minimum on the second interval

*]*

**2, 4***at*

**20***.*

**x=2**Therefore, the sum of areas of all rectangles equals to

*.*

**S**_{1}= 0·2 + 20·2 = 40Case 2.

**N=4**Two additional points of division in four equal parts are

*and*

**x**_{1}=1

**x**_{3}=3So,

*,*

**a=x**_{0}=0*,*

**x**_{1}=1*,*

**x**_{2}=2*,*

**x**_{3}=3

**x**_{4}=4=bMinimum on the first interval

*]*

**0, 1***at*

**0***.*

**x=0**Minimum on the second interval

*]*

**1, 2***at*

**10***.*

**x=1**Minimum on the third interval

*]*

**2, 3***at*

**20***.*

**x=2**Minimum on the fourth interval

*]*

**3, 4***at*

**30***.*

**x=3**Therefore, the sum of areas of all rectangles equals to

*.*

**S**

+ 20·1 + 30·1 = 60_{2}= 0·1 + 10·1 ++ 20·1 + 30·1 = 60

To proceed to a general case, assume we have divided our segment into

*equal parts, and find the limit of the area of all rectangles as*

**N***.*

**N→∞**In this case the width of each interval equals to

**Δ**

*x*_{i}= 4/NRight margin of

*-th interval is*

**i**

**x**_{i}= i·4/NThe function value at this right margin is

**f(x**_{i}) = 10·i·4/NThe area of the

**-th rectangle, constructed on the**

*i***-th interval as a base and having height calculated above, equals to**

*i*

**S**

= (160/N²)·i_{i}= 10·i·(4/N)·(4/N) == (160/N²)·i

Our task is to sum these areas for

*changing from*

**i***to*

**1***and find the limit of this sum as*

**N***.*

**N→∞**Summation by

*is simple, we did this before (see "Algebra - Sequence and Series" chapter of this course or prove the following by induction).*

**i**Recall that

**Σ**

_{[1,K]}

**i = K·(K+1)/2**Therefore, the result of summation of the areas of

*rectangles is*

**N****Σ**

_{[1,N]}

**S**

= 80·(1+(1/N))_{i}=(160/N²)·N·(N+1)/2== 80·(1+(1/N))

As

*, this value converges to*

**N→∞***, which constitutes the "area under curve".*

**80**Incidentally, our "area under curve" can be calculated as the area of a right triangle with base

**and height**

*4**, which equals to*

**10·4=40***- the same answer as we have received through rather complicated summation and taking the limit.*

**S=(1/2)·4·40=80**That confirms the validity of our answer.

The end.

________________

*Example 2*

Find "area under curve" for

**f(x) = −x**^{2}+1*].*

**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

**x**_{1}=0So,

*,*

**a=x**_{0}=−1*,*

**x**_{1}=0

**x**_{2}=1=bThe function value on the right margin of the first interval

*]*

**−1, 0***at*

**1***.*

**x=0**The function value on the right margin of the second interval

*]*

**−1, 0***at*

**0***.*

**x=1**Therefore, the sum of areas of all rectangles equals to

*.*

**S**_{1}= 1·1 + 0·1 = 1Case 2.

**N=4**Two additional points of division in four equal parts are

*and*

**x**_{1}=−0.5

**x**_{3}=0.5So,

*,*

**a=x**_{0}=−1*,*

**x**_{1}=−0.5*,*

**x**_{2}=0*,*

**x**_{3}=0.5

**x**_{4}=1=bThe function value on the right margin of the first interval

*]*

**−1, −0.5***at*

**3/4***.*

**x=−0.5**The function value on the right margin of the second interval

*]*

**−0.5, 0***at*

**1***.*

**x=0**The function value on the right margin of the third interval

*]*

**0, 0.5***at*

**3/4***.*

**x=0.5**The function value on the right margin of the fourth interval

*]*

**0.5, 1***at*

**0***.*

**x=1**Therefore, the sum of areas of all rectangles equals to

*.*

**S**

+ (3/4)·0.5 + 0·0.5 = 5/4_{2}= (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

*equal parts, and find the limit of the area of all rectangles as*

**N***.*

**N→∞**In this case the width of each interval equals to

**Δ**

*x*_{i}= 2/NRight margin of

*-th interval is*

**i**

**x**_{i}= −1+i·2/NThe function value at this right margin is

**f(x**

= (4/N)·i−(4/N²)·i²_{i}) = 1−(−1+i·2/N)² == (4/N)·i−(4/N²)·i²

The area of the

**-th rectangle, constructed on the**

*i***-th interval as a base and having height calculated above, equals to**

*i*

**S**_{i}=**Δ**[

*x*

= (2/N)·_{i}·f(x_{i}) == (2/N)·

*]*

**(4/N)·i−(4/N²)·i²**

**=**

= (8/N²)·i−(8/N³)·i²= (8/N²)·i−(8/N³)·i²

Our task is to sum these areas for

*changing from*

**i***to*

**1***and find the limit of this sum as*

**N***.*

**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

*rectangles is*

**N****Σ**

_{[1,N]}

**S**

− (8/N³)·N·(N+1)·(2N+1)/6_{i}= (8/N²)·(N·(N+1)/2 −− (8/N³)·N·(N+1)·(2N+1)/6

As

*, this value converges to*

**N→∞***, which constitutes the "area under curve".*

**4−(16/6)=4/3**The end.

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