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Integration of the Polynomial Term from First Principles

Integration from First Principles

Foreword: In this part of the paper, I explain how to integrate the polynomial, from First Principles. (Use the latest Mozilla Firefox Browser, to see the equations properly.)

By: Chrysanthus Date Published: 1 May 2019

Introduction

In this part of the paper, I explain how to integrate the polynomial, from First Principles. You should have read the previous parts of the series before reaching here, as this is the continuation.

Integration is area under the curve.
Consider the following diagram:

y = f(x)

You might have seen the following:

a b f ( x ) d x = lim δ x 0 x = a x = b f ( x ) δ x

This statement is not perfect. It should actually be:

a b f ( x ) d x = lim δ x 0 x = a x = b f ( x n ) δ x

where f(xn) is a particular height from the curve to the x-axis.

My Discovery

The widths of the vertical strips are equal.

As 𝛿x turns to 0, the number of strips increases.

n is the total number of strips.

δ x = b - a n

As 𝛿x → 0, n → ∞

The series and the formula for its summation, for all the heights of the strips from a to b, is obtained. If you do not know the formula for the series, develop one, and that may take time. Summing is done for the heights, from 1 to n.

The summation expression above in the statement, is replaced with the summation formula.

As n → ∞ the right-hand-side of the statement becomes the definite integral; and that concludes the scheme.

Let total area be represented by A. So, my statement for the Integration from First Principles is:

A = lim n x = a x = b f ( x n ) b - a n

In the following analysis, Pascal's Triangle has been used in the binomial expansions.

Integration of the Constant, C

Let the definite integral be represented by A, and let the indefinite integral be represented by I. Let the summation expression (discrete summation) be Sn.

Here,

y = C

i.e.

f(x) = C

Sn = (C0 + C + C + C + - - - + Cn-1)𝛿x

      = (nC) 𝛿x

But δ x = b - a n

⸫ Sn = n C b - a n

The ns cancel:

=> Sn = C(b - a)

=> Sn = Cb - Ca

As n → ∞,

A = Cb - Ca .the definite integral

If a is 0 and b is x,

I = Cx ..the indefinite integral

Integration of the Term, x0

For the general term, rm, you need my general formula pattern for the power series, which is:

r = 1 n r m = 1 m+1 n m+1 + 1 2 n m + lower powers of n ...

Let the definite integral be represented by A, and let the indefinite integral be represented by I. Let the summation expression (discrete summation) be Sn.

Here,

y = x0

i.e.

f(x) = x0

S n = x = a x = b f ( x n ) δ x S n = [ ( a + 1 δ x ) 0 + ( a + 2 δ x ) 0 + ( a + 3 δ x ) 0 + + ( a + n δ x ) 0 ] δ x S n = [ 1 + 1 + 1 + + 1 n ] δ x S n = [ n ] δ x

But δ x = b - a n S n = [ n ] b - a n

The ns cancel:

=> Sn = b - a

=> Sn = b - a

As n → ∞,

A = b - a .the definite integral

If a is 0 and b is x,

I = x ..the indefinite integral

Integration of the Term, x1

For the general term, rm, you need my general formula pattern for the power series, which is:

r = 1 n r m = 1 m+1 n m+1 + 1 2 n m + lower powers of n ...

Let the definite integral be represented by A, and let the indefinite integral be represented by I. Let the summation expression (discrete summation) be Sn.

Here,

y = x1

i.e.

f(x) = x1

S n = x = a x = b f ( x n ) δ x S n = [ ( a + 1 δ x ) 1 + ( a + 2 δ x ) 1 + ( a + 3 δ x ) 1 + + ( a + n δ x ) 1 ] δ x S n = [ a + 1 δ x + a + 2 δ x + a + 3 δ x + + a + n δ x ] δ x

Rearranging gives:

S n = r = 1 n a r 0 δ x S n = + r = 1 n r 1 δ 2 x

Adjusting the multipliers and multiplicands, give

S n = a δ x r = 1 n r 0 S n = + δ 2 x r = 1 n r 1

My formulas for the power series of index 0 and 1 are:

r = 1 n r 0 = 1 1 n 1

and

r = 1 n r 1 = 1 2 n 2 + 1 2 n 1

δ x = b - a n

Substituting for δx and the summation series with the formulas, give

S n = a (b-a) n 1 1 n 1 S n = + (b-a) 2 n 2 [ 1 2 n 2 + 1 2 n 1 ]

The ns and the n2s cancel, giving:

S n = a (b-a) + (b-a) 2 2 + (b-a) 2 2 n

As n → ∞, the term with denominator 2n, turns to 0. So

A = a (b-a) + (b-a) 2 2 A = a b - a2 + b2 2 - 2ba 2 + a2 2 A = - a2 + b2 2 + a2 2 A = b2 2 - a2 2 .the definite integral

Notice that there is only one term in b2 in the evaluation. Also, the combined terms in a and b subtract out, while the terms in a2 sign-sum to 1/2 of a2.

If a is 0 and b is x,

I = x2 2 ..the indefinite integral

Integration of the Term, x2

For the general term, rm, you need my general formula pattern for the power series, which is:

r = 1 n r m = 1 m+1 n m+1 + 1 2 n m + lower powers of n ...

Let the definite integral be represented by A, and let the indefinite integral be represented by I. Let the summation expression (discrete summation) be Sn.

Here,

y = x2

i.e.

f(x) = x2

S n = x = a x = b f ( x n ) δ x S n = [ ( a + 1 δ x ) 2 + ( a + 2 δ x ) 2 + ( a + 3 δ x ) 2 + + ( a + n δ x ) 2 ] δ x
S n = [ a 2 + 2 a 1 δ x + 1 2 δ 2 x S n = + a 2 + 2 a 2 δ x + 2 2 δ 2 x S n = + a 2 + 2 a 3 δ x + 3 2 δ 2 x S n = + S n = + a 2 + 2 a n δ x + n 2 δ 2 x ] δ x

Rearranging gives:

S n = a 2 δ x + a 2 δ x + a 2 δ x + + a n 2 δ x S n = + 2 a 1 δ 2 x + 2 a 2 δ 2 x + 2 a 3 δ 2 x + + 2 a n δ 2 x S n = + 1 2 δ 3 x + 2 2 δ 3 x + 3 2 δ 3 x + + n 2 δ 3 x S n = a 2 δ x r = 1 n r 0 S n = + 2 a δ 2 x r = 1 n r 1 S n = + δ 3 x r = 1 n r 2

My formulas for the power series of index 0, 1 and 2 are:

r = 1 n r 0 = 1 1 n 1

and

r = 1 n r 1 = 1 2 n 2 + 1 2 n 1

and

r = 1 n r 2 = 1 3 n 3 + 1 2 n 2 + 1 6 n 1

δ x = b - a n

Substituting for δx and the summation series with the formulas, give

S n = a 2 (b-a) n [ 1 1 n 1 ] S n = + 2 a (b-a) 2 n 2 [ 1 2 n 2 + 1 2 n 1 ] S n = + (b-a) 3 n 3 [ 1 3 n 3 + 1 2 n 2 + 1 6 n 1 ]

The ns, n2s and n3s cancel, giving:

S n = a 2 (b-a) S n = + a (b-a) 2 + a (b-a) 2 n 1 S n = + (b-a) 3 3 + (b-a) 3 2n1 + (b-a) 3 6n2

As n → ∞, the terms with denominators in n, turn to 0. So

A = a 2 (b-a) S n = + a (b-a) 2 S n = + (b-a) 3 3

Notice that only the first term of the resulting expressions-per-line have been taken (did not turn to 0).

A = b a 2 - a 3 + b 2 a - 2 b a 2 + a 3 + 1 3 b 3 - 3 3 b 2 a + 3 3 b a 2 - 1 3 a 3 A = - a 3 + a 3 + 1 3 b 3 - 1 3 a 3
A = 1 3 b 3 - 1 3 a 3 .the definite integral

Notice that there is only one term in b3 in the evaluation. Also, the combined terms in a and b subtract out, while the terms in a3 sign-sum to 1/3 of a3.

If a is 0 and b is x,

I = 1 3 x 3 .the indefinite integral

Integration of the Term, x3

For the general term, rm, you need my general formula pattern for the power series, which is:

r = 1 n r m = 1 m+1 n m+1 + 1 2 n m + lower powers of n ...

Let the definite integral be represented by A, and let the indefinite integral be represented by I. Let the summation expression (discrete summation) be Sn.

Here,

y = x3

i.e.

f(x) = x3

S n = x = a x = b f ( x n ) δ x S n = [ ( a + 1 δ x ) 3 + ( a + 2 δ x ) 3 + ( a + 3 δ x ) 3 + + ( a + n δ x ) 3 ] δ x
S n = [ a 3 + 3 a 2 1 δ x + 3 a 1 2 δ 2 x + 1 3 δ 3 x S n = + a 3 + 3 a 2 2 δ x + 3 a 2 2 δ 2 x + 2 3 δ 3 x S n = + a 3 + 3 a 2 3 δ x + 3 a 3 2 δ 2 x + 3 3 δ 3 x S n = + S n = + a 3 + 3 a 2 n δ x + 3 a n 2 δ 2 x + n 3 δ 3 x ] δ x

Rearranging gives:

S n = a 3 δ x + a 3 δ x + a 3 δ x + + a n 3 δ x S n = + 3 a 2 1 δ 2 x + 3 a 2 2 δ 2 x + 3 a 2 3 δ 2 x + + 3 a 2 n δ 2 x S n = + 3 a 1 2 δ 3 x + 3 a 2 2 δ 3 x + 3 a 3 2 δ 3 x + + 3 a n 2 δ 3 x S n = + 1 3 δ 4 x + 2 3 δ 4 x + 3 3 δ 4 x + + n 3 δ 4 x S n = a 3 δ x r = 1 n r 0 S n = + 3 a 2 δ 2 x r = 1 n r 1 S n = + 3 a δ 3 x r = 1 n r 2 S n = + δ 4 x r = 1 n r 3

My formulas for the power series of index 0, 1, 2 and 3 are:

r = 1 n r 0 = 1 1 n 1

and

r = 1 n r 1 = 1 2 n 2 + 1 2 n 1

and

r = 1 n r 2 = 1 3 n 3 + 1 2 n 2 + 1 6 n 1

and

r = 1 n r 3 = 1 4 n 4 + 1 2 n 3 + 1 4 n 2 + 0 n 1

δ x = b - a n

Substituting for δx and the summation series with the formulas, give

S n = a 3 (b-a) n [ 1 1 n 1 ] S n = + 3 a 2 (b-a) 2 n 2 [ 1 2 n 2 + 1 2 n 1 ] S n = + 3 a (b-a) 3 n 3 [ 1 3 n 3 + 1 2 n 2 + 1 6 n 1 ] S n = + (b-a) 4 n 4 [ 1 4 n 4 + 1 2 n 3 + 1 4 n 2 + 0 n 1 ]

The ns, n2s, n3s and n4s cancel, giving:

S n = a 3 (b-a) S n = + 3 2 a 2 (b-a) 2 + 3 2 a 2 (b-a) 2 n 1 S n = + 3a(b-a) 3 3 + 3a(b-a) 3 2n1 + 3a(b-a) 3 6n2 S n = + (b-a) 4 4 + (b-a) 4 2n1 + (b-a) 4 4n2

As n → ∞, the terms with denominators in n, turn to 0. So

A = a 3 (b-a) S n = + 3 2 a 2 (b-a) 2 S n = + 3a(b-a) 3 3 S n = + (b-a) 4 4

Notice that only the first term of the resulting expressions-per-line have been taken (did not turn to 0).

A = b a 3 - a 4 + 3 2 b 2 a 2 - 6 2 b a 3 + 3 2 a 4 + 3 3 b 3 a - 9 3 b 2 a 2 + 9 3 b a 3 - 3 3 a 4 A = + 1 4 b 4 - 4 4 b 3 a 1 + 6 4 b 2 a 2 - 4 4 b a 3 + 1 4 a 4 A = - a 4 + 3 2 a 4 - 3 3 a 4 + 1 4 b 4 - 1 4 a 4
A = 1 4 b 4 - 1 4 a 4 .the definite integral

Notice that there is only one term in b4 in the evaluation. Also, the combined terms in a and b subtract out, while the terms in a4 sign-sum to 1/4 of a4.

If a is 0 and b is x,

I = 1 4 x 4 .the indefinite integral

Integration of the Term, xm

Here,

y = xm

i.e.

f(x) = xm

Each of the above evaluations, makes use of the general formula pattern,

r = 1 n r m = 1 m+1 n m+1 + 1 2 n m + lower powers of n ...

for the different power series, which differ by their indexes.

In each evaluation, nm+1 cancels off. The other terms in n, turn to zero as n turns to infinity. The terms in b and a subtract out. The terms in a, sign-sum to 1/(m+1) of a(m+1). There is only one term for b(m+1), which is (1/(m+1))b(m+1). So,

A = 1 (m+1) b (m+1) - 1 (m+1) a (m+1) .the definite integral

If a is 0 and b is x,

I = 1 (m+1) x (m+1) .the indefinite integral Integration of the Term, Cxm

where C is a constant.

Here,

y = Cxm

i.e.

f(x) = Cxm

A = lim n x = a x = b f ( x n ) b - a n A = lim n x = a x = b C x m b - a n A = C lim n x = a x = b x m b - a n
A = C [ 1 (m+1) b (m+1) - 1 (m+1) a (m+1) ] .the definite integral

If a is 0 and b is x,

I = C [ 1 (m+1) x (m+1) ] .the indefinite integral
Integration of the Polynomial

Integration is area under the curve. A polynomial is a summation of polynomial terms. So integration of a polynomial is the summation of the integrals of each term. Some of the individual areas, may have negative values.

This is the end of this part of the series. Continue in the next part.

Chrys


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