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Integrating Power of a Linear Function from First Principles

Integration from First Principles

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

By: Chrysanthus Date Published: 14 May 2019

Introduction

In this part of the paper, I explain how to integrate Power of a Linear Function, 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 Function, (px+q)0

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 = (px+q)0

i.e.

f(x) = (px+q)0

=> f(x) = p0(x+q/p)0

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

Note that this is not 1/pX(px+q)1 you might have been expecting from the formula, 1/p(m+1)X(px+q)m+1

Integration of the Function, (px+q)1

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 = (px+q)1

i.e.

f(x) = (px+q)1

=> f(x) = p1(x+q/p)1

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

Rearranging:

S n = p 1 [ ( a + q p ) δ x + ( a + q p ) δ x + ( a + q p ) δ x + + ( a + q p ) n δ x S n = + 1 δ x 2 + 2 δ x 2 + 3 δ x 2 + + n δ x 2 ]
S n = p 1 [ r = 1 n ( a + q p ) r 0 δ x S n = + r = 1 n r 1 δ 2 x ]

Adjusting the multipliers and multiplicands, give

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

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

A = p 1 [ ( a + q p ) (b-a) + (b-a) 2 2 ] A = p 1 [ a b - a2 + q p b - q p a + b2 2 - b a + a2 2 ] A = p 1 [ - a2 + q p b - q p a + b2 2 + a2 2 ] A = p 1 [ b2 2 + q p b - a2 2 - q p a ]

Completing the square, by adding and subtracting, to balance up, gives

A = p 1 [ b2 2 + q p b + 1 2 ( q p ) 2 - a2 2 - q p a - 1 2 ( q p ) 2 ] A = 1 2 p 1 [ ( b + q p ) 2 - ( a + q p ) 2 ]

Multiplying Right-Hand-Side by p/p will not change the expression.

A = 1 2 p 2 p 1 [ ( b + q p ) 2 - ( a + q p ) 2 ] A = 1 2 p 1 [ ( p b + q ) 2 - ( p a + q ) 2 ] A = 1 2 p 1 ( p b + q ) 2 - 1 2 p 1 ( p a + q ) 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.

Let b be x and a be -q/p,

I = 1 2 p 1 ( p x + q ) 2 .the indefinite integral

Integration of the Function, (px+q)2

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 = (px+q)2

i.e.

f(x) = (px+q)2

=> f(x) = p2(x+q/p)2

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

Rearranging:

S n = p 2 [ ( a + q p ) 2 δ x + ( a + q p ) 2 δ x + ( a + q p ) 2 δ x + + ( a + q p ) n 2 δ x S n = + 2 ( a + q p ) 1 1 δ 2 x + 2 ( a + q p ) 1 2 δ 2 x + 2 ( a + q p ) 1 3 δ 2 x + + 2 ( a + q p ) 1 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 = p 2 [ ( a + q p ) 2 δ x r = 1 n r 0 S n = + 2 ( a + q p ) 1 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 = p 2 { ( a + q p ) 2 (b-a) n [ 1 1 n 1 ] S n = + 2 ( a + q p ) 1 (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 = p 2 [ ( a + q p ) 2 (b-a) S n = + ( a + q p ) 1 (b-a) 2 + ( a + q p ) 1 (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 = p 2 [ ( a + q p ) 2 (b-a) S n = + ( a + q p ) 1 (b-a) 2 S n = + (b-a) 3 3 ]

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

A = p 2 [ (b-a) ( a 2 + 2 q p a + ( q p ) 2 ) S n = + ( a + q p ) 1 ( b 2 - 2 b a + a 2 ) S n = + (b-a) 3 3 ] A = p 2 [ b a 2 + 2 q p b a + ( q p ) 2 b - a 3 - 2 q p a 2 - ( q p ) 2 a S n = + a b 2 - 2 b a 2 + a 3 + ( q p ) 1 b 2 - 2 ( q p ) 1 b a + ( q p ) 1 a 2 S n = + 1 3 b 3 - 3 3 b 2 a + 3 3 b a 2 - 1 3 a 3 ]
A = p 2 [ + ( q p ) 2 b - a 3 - 2 q p a 2 - ( q p ) 2 a S n = + ( q p ) 1 b 2 + ( q p ) 1 a 2 S n = + a 3 + 1 3 b 3 - 1 3 a 3 ]
A = p 2 [ 1 3 b 3 + ( q p ) 1 b 2 + ( q p ) 2 b - 1 3 a 3 - q p a 2 - ( q p ) 2 a ]

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 a2 sign-sum to q/p of a2; and terms in a3 sign-sum to 1/3 of a3.

Completing the cube (binomial expansion) by adding and subtracting, to balance up, gives

A = p 2 [ 1 3 b 3 + ( q p ) 1 b 2 + ( q p ) 2 b + 1 3 ( q p ) 3 - 1 3 a 3 - q p a 2 - ( q p ) 2 a - 1 3 ( q p ) 3 ]

Multiplying Right-Hand-Side by p/p will not change the expression, which results in:

A = p 3 p [ 1 3 ( b + q p ) 3 - 1 3 ( a + q p ) 3 ]
A = 1 3p ( p b + q ) 3 - 1 3p ( p a + q ) 3 .the definite integral

If b is x and a is -q/p,

I = 1 3p ( p x + q ) 3 .the indefinite integral

Integration of the Function, (px+q)3

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 = (px+q)3

i.e.

f(x) = (px+q)3

=> f(x) = p3(x+q/p)3

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

Rearranging:

S n = p 3 [ ( a + q p ) 3 δ x + ( a + q p ) 3 δ x + ( a + q p ) 3 δ x + + ( a + q p ) n 3 δ x S n = + 3 ( a + q p ) 2 1 δ 2 x + 3 ( a + q p ) 2 2 δ 2 x + 3 ( a + q p ) 2 3 δ 2 x + + 3 ( a + q p ) 2 n δ 2 x S n = + 3 ( a + q p ) 1 1 2 δ 3 x + 3 ( a + q p ) 1 2 2 δ 3 x + 3 ( a + q p ) 1 3 2 δ 3 x + + 3 ( a + q p ) 1 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 = p 3 [ ( a + q p ) 3 δ x r = 1 n r 0 S n = + 3 ( a + q p ) 2 δ 2 x r = 1 n r 1 S n = + 3 ( a + q p ) 1 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 = p 3 { ( a + q p ) 3 (b-a) n [ 1 1 n 1 ] S n = + 3 ( a + q p ) 2 (b-a) 2 n 2 [ 1 2 n 2 + 1 2 n 1 ] S n = + 3 ( a + q p ) 1 (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 = p 3 [ ( a + q p ) 3 (b-a) S n = + 3 2 ( a + q p ) 2 (b-a) 2 + 3 2 ( a + q p ) 2 (b-a) 2 n 1 S n = + ( a + q p ) 1 (b-a) 3 + 3 2 ( a + q p ) 1 (b-a) 3 n 1 + 1 2 ( a + q p ) 1 (b-a) 3 n 2 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 = p 3 [ ( a + q p ) 3 (b-a) S n = + 3 2 ( a + q p ) 2 (b-a) 2 S n = + ( a + q p ) 1 (b-a) 3 S n = + (b-a) 4 4 ]

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

A = p 3 [ (b-a) ( a 3 + 3 q p a 2 + 3 ( q p ) 2 a 1 + ( q p ) 3 ) S n = + 3 2 ( a + q p ) 2 ( b 2 - 2 b a + a 2 ) S n = + ( a + q p ) 1 ( b 3 - 3 b 2 a + 3 b a 2 - a 3 ) S n = + (b-a) 4 4 ] A = p 3 [ b a 3 + 3 q p b a 2 + 3 ( q p ) 2 b a + ( qp ) 3 b - a 4 - 3 q p a 3 - 3 ( q p ) 2 a 2 - ( q p ) 3 a 1 S n = + 3 2 ( a 2 + 2 q p a + ( q p ) 2 ) ( b 2 - 2 b a + a 2 ) S n = + b 3 a - 3 b 2 a 2 + 3 b a 3 - a 4 + qp b 3 - 3 qp b 2 a 1 + 3 qp b a 2 - qp a 3 S n = + 1 4 b 4 - 4 4 b 3 a + 6 4 b 2 a 2 - 4 4 b 1 a 3 + 1 4 a 4 ]
A = p 3 [ b a 3 + 3 q p b a 2 + 3 ( q p ) 2 b a + ( qp ) 3 b - a 4 - 3 q p a 3 - 3 ( q p ) 2 a 2 - ( q p ) 3 a 1 S n = + 32b2a2 - 3b1a3 + 32a4 + 3(qp)1b2a1 - 6(qp)1b1a2 + 3(qp)1a3 + 32(qp)2b2 - 3(qp)2b1a1 + 32(qp)2a2 S n = + b 3 a - 3 b 2 a 2 + 3 b a 3 - a 4 + qp b 3 - 3 qp b 2 a 1 + 3 qp b a 2 - qp a 3 S n = + 1 4 b 4 - 4 4 b 3 a + 6 4 b 2 a 2 - 4 4 b 1 a 3 + 1 4 a 4 ]
A = p 3 [ ( qp ) 3 b - a 4 - 3 q p a 3 - 3 ( q p ) 2 a 2 - ( q p ) 3 a 1 S n = + 32a4 + 3(qp)1a3 + 32(qp)2b2 + 32(qp)2a2 S n = - a 4 + qp b 3 - qp a 3 S n = + 1 4 b 4 + 1 4 a 4 ]
A = p 3 [ 1 4 b 4 + qp b 3 + 32(qp)2b2 + ( qp ) 3 b - 1 4 a 4 - q p a 3 - 3 2 ( q p ) 2 a 2 - ( q p ) 3 a 1 ]

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 a2 and a3 sign-sum to their different fractions.

Completing the binomial expansions by adding and subtracting, to balance up, gives

A = p 3 [ 1 4 b 4 + qp b 3 + 32(qp)2b2 + ( qp ) 3 b + 1 4 ( qp ) 4 - 1 4 a 4 - q p a 3 - 3 2 ( q p ) 2 a 2 - ( q p ) 3 a 1 - 1 4 ( qp ) 4 ]

Multiplying Right-Hand-Side by p/p will not change the expression; and that results in:

A = p 4 p [ 1 4 ( b + q p ) 4 - 1 4 ( a + q p ) 4 ]
A = 1 4p ( p b + q ) 4 - 1 4p ( p a + q ) 4 .the definite integral

If b is x and a is -q/p,

I = 1 4p ( p x + q ) 4 .the indefinite integral

Integration of the General Function, (px+q)m

Here,

y = (px+q)m

i.e.

f(x) = (px+q)m

=> f(x) = pm(x+q/p)m

From above:

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

If b is x and a is -q/p,

I = 1 (m+1)p ( p x + q ) (m+1) .the indefinite integral

Note that when m = 0, the definite integral is b - a and the indefinite integral is x.

That is it for this part of the series. We stop here and continue in the next part.

Chrys


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