Integration

Snoopyies SG

How to integrate v^(3v^2 - 1) dv ?

Thanks.

Fortmax777 SG

How to integrate v^(3v^2 - 1) dv ?

Thanks.

Cannot integrate analytically. Use numerical methods such as Simpson's rule.

Snoopyies SG

I go google for Integration Simpson's rule and the rule says that it only applies to find the definite integration ie with the limits, the rule cannot be applied to find indefinite integration.

By the way, how to tell whether an integration question can be integrated or cannot be integrated ? Are there any rules or guidelines on it ?

The substition method in integration is used to reverse the differentiation chain's rule and the integration by parts in integration is used to reverse the differentiation product rule.

Is there a method in integration that can be used to reverse the differentiation quotient rule ?

Is there also a method in integration that can be used to reverse the implicit differentation ?

Fortmax777 SG

No hard and fast rule. Basically consider the following methods.

1. Substitution

2. Parts (This is the reverse of the product rule)

3. Trigonometric Substitution

4. Partial Fractions

If all the above fails then unlikely you can integrate analytically. Again you can try chim methods like expanding the function into a Taylor Series (but you must check for convergence) and integrate term by term.

No. I am not aware of any method that reverses the quotient rule or implicit differentiation.

Do you want to integrate v^(3v^2-1) or v*(3v^2-1). The latter case is easy but I still can't think of a way to integrate the former case.

ThunderFbolt SG

Mathematica & Maple turned up blank. I don't think this is directly integrate-able.

Lee012lee SG

I go google for Integration Simpson's rule and the rule says that it only applies to find the definite integration ie with the limits, the rule cannot be applied to find indefinite integration.

By the way, how to tell whether an integration question can be integrated or cannot be integrated ? Are there any rules or guidelines on it ?

The substition method in integration is used to reverse the differentiation chain's rule and the integration by parts in integration is used to reverse the differentiation product rule.

Is there a method in integration that can be used to reverse the differentiation quotient rule ?

Is there also a method in integration that can be used to reverse the implicit differentation ?

To TS,

Case 1

Using Substitution Method in integration to reverse the differentiation chain's rule

Differentiation by chain's rule

Let y = (3x^2 - 1)^3

dy/dx = 3(3x^2 - 1)^2 . (6x)

= 18x(3x^2 - 1)^2

Integration by Substitution Method

Integrate 18x(3x^2 - 1)^2 dx

Let u = 3x^2 - 1

du/dx = 6x

dx = du/6x

Substitute u = 3x^2 -1 and dx = du/6x into

Integrate 18x(3x^2 - 1)^2 dx

= Integrate 18x u^2 . du/6x

= Integrate 3u^2 du

= 3(u^3)/3 + c

= u^3 + c

= (3x^2 - 1)^3 + c since u = 3x^2 - 1

PS :

Case 1

Differentiation by chain's rule

Let y = g(u) where u = f(x)

dy/du = g'(u) du/dx = f '(x)

dy/dx = g'(u). f '(x)

Integration by substitution method

Integrate (dy/dx) dx = Integrate g'(u).f '(x) dx

= Integrate g'(u) (du/dx) dx

= Integrate g'(u) du

y = g(u)

Lee012lee SG

Case 2

Using Integration by parts to reverse the differentiation product rule

Differentiation by Product Rule

y = uv

dy/dx = u (dv/dx) + v (du/dx)

Integration by parts (for product rule)

Integrate (u) (dv/dx) dx = uv - Integrate v (du/dx) dx

Integrate (u) (dv/dx) dx + Integrate v (du/dx) dx = uv

Integrate [ (u)(dv/dx) + v (du/dx) ] dx = uv

Lee012lee SG

Case 3

Integration by parts to reverse differentiation by quotient rule

Differentiation by Quotient Rule

Let y = u/v where u = f(x) v = g(x)

du/dx = f '(x) dv/dx = g'(x)

dy/dx = [ v.du - u.dv ] / v^2

Integration by Parts (for quotient rule)

Integrate (du/v) dx = u/v + Integrate (u/v^2.dv) dx

Integrate (du/v) dx - Integrate (u/v^2.dv) dx = u/v

Integrate (v.du) / v^2 dx - Integrate (u.dv)/v^2 dx = u/v

Integrate [ v.du - u.dv ] / v^2 dx = u/v

Lee012lee SG

Relationship between Integration by parts (for product rule) and integration by parts (for quotient rule)

y = u/v can be easily re-expressed as y = uv^-1

and so the integration by parts (for product rule) can be used instead of the integration by parts (for quotient rule)

ThunderFbolt SG

umm... so what is the answer? or are you just showing the different methods of integration?

Lee012lee SG

umm... so what is the answer? or are you just showing the different methods of integration?

Snoopyies's question

Integrate v^(3v^2 -1) dx

is unlikely to have a unique function as the integration answer.

Hence, when you entered the function into Mathematica and maple or wolfram, you do not get an answer.

When we cannot get a unique function in the integration answer, we will need to use definite integration ie to use limits and then we can use the numerical approximate methods of integration eg Simpson's rule to find it.

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I am answering to the other questions raised by Snoopyies.

"The substitution method in integration is used to reverse the differentiation chain's rule and the integration by parts in integration is used to reverse the differentiation product rule.

Is there a method in integration that can be used to reverse the differentiation quotient rule ?

Is there also a method in integration that can be used to reverse the implicit differentation ?"

The last two questions are interesting as we do not see any integration method to reverse differentiation quotient rule and implicit differentiation in maths textbooks.

I am still thinking whether there is a method in integration that can be used to reverse the implicit differentiation.