Integration by substitution.

[scguy]

I have been taught to do it this way:

 

Say i wish to integrate. (2x+1)^1/2 dx

 

1) u=2x+1

2) therefore... u^1/2 dx

3) du/dx= 2

 

4) u^1/2 du/2 (because du/dx=2)

 

5) therefore...1/2 u^1/2 du.

 

At step 5 the integration is then performed, i have no problems with that. It says in the book that du/dx=2 should not be split up to make du/2=dx, but this is what is done anyway in all the examples Note, the only reason this is done in the first place is because u are not supposed to have more than 1 variable in the integrand. I cant make sense of it.

[Dapthar]

It says in the book that du/dx=2 should not be split up to make du/2=dx, but this is what is done anyway in all the examples

This is because, as you have written it, you have a [math]dx[/math] in the integral, so you must solve for something to substitute in for [math]dx[/math], namely [math]dx = \frac{1}{2} \cdot du[/math].

[Johnny5]

I have been taught to do it this way:

 

Say i wish to integrate. (2x+1)^1/2 dx

 

1) u=2x+1

2) therefore... u^1/2 dx

3) du/dx= 2

 

4) u^1/2 du/2 (because du/dx=2)

 

5) therefore...1/2 u^1/2 du.

 

At step 5 the integration is then performed' date=' i have no problems with that. It says in the book that du/dx=2 should not be split up to make du/2=dx, but this is what is done anyway in all the examples Note, the only reason this is done in the first place is because u are not supposed to have more than 1 variable in the integrand. I cant make sense of it.[/quote']

 

[math] \int (2x + 1)^{1/2} dx [/math]

 

Define U to be equal to 2x+1. That is:

[math] U \equiv 2x+1 [/math]

Differentiate both sides of the statement above, with respect to x, to obtain the following statement which has the same truth value as your definition:

 

[math] \frac{d}{dx}(U) = \frac{d}{dx}(2x+1) [/math]

 

The parentheses on the LHS are redundant, therefore:

 

[math] \frac{dU}{dx} = \frac{d}{dx}(2x+1) [/math]

 

As for the RHS, utilize a theorem you should know a-priori to this problem, which is that the derivative of a sum is the sum of the derivatives. Using that theorem of the differential calculus you get:

 

[math] \frac{dU}{dx} = \frac{d}{dx}(2x+1) = \frac{d}{dx}(2x)+\frac{d}{dx}(1)[/math]

 

1 is a numerical constant, and a numerical constant cannot have a value which changes. Therefore d(1)=0, and since 0/dx=0, the following statement is true:

 

[math] \frac{dU}{dx} = \frac{d}{dx}(2x) [/math]

 

Where I just used the following axiom of the real number system:

 

[math] \exists 0 \in \mathbb{R} \forall x \in \mathbb{R} [0+x=x] [/math]

 

Now, utilize the following theorem of the differential calculus, which you should know a-priori to solving this problem:

 

[math] \text{Theorem:} \frac{d}{dx}[ c f(x)] = c \frac{df}{dx} [/math]

 

Where c denotes an arbitrary numerical constant, and f(x) denotes an arbitrary function of the variable x.

 

So 2 denotes a specific numerical constant, and x denotes a specific function of x. So we can use the law of first order logic known as universal instantiation, together with the theorem, to obtain the following true statement:

 

[math] \frac{d}{dx}[2x] = 2 \frac{dx}{dx}=2 \cdot 1 = 2 [/math]

 

We can now use the transitive property of equality to infer that the following statement is true:

 

[math] \frac{dU}{dx} = 2 [/math]

 

Now, multiply both sides of the previous equation by dx to obtain the following statement, which must have the same truth value as the previous statement:

 

[math] dU= 2 dx[/math]

 

Now, divide both sides of the previous statement by the number two, to obtain the following statement, which must have the same truth value as the previous statement:

 

[math] \frac{dU}{2} = dx[/math]

 

Here is the integral you want to solve:

 

[math] \int (2x + 1)^{1/2} dx [/math]

 

Now, make replace 2x+1 by U, and replace dx by dU/2, to obtain the following expression:

 

[math] \int \frac{1}{2}U^{1/2} dU [/math]

 

Then finish off the integral, as you say you did.

 

You claim your book says that you cannot write dU/2=dx.

 

That follows from your own definition, and algebraic steps, as I showed.

 

Regards

 

PS: There is disagreement amongst mathematicians, as to whether or not you can 'decouple' the differential dx from the differential operator d/dx.

 

I can offer you a proof that you can in fact do this, but it will involve covering the difference calculus.

 

If you want to see a proof that you can do this, then let me know. But be prepared in advance, that others will say that you cannot decouple the dx from the differential operator. Ask them for proof of what they say too, and then decide for yourself who is right.

[scguy]

Thanks for the responses, i understand better now.

[davesbird]

This is fine for indefinite integrals, which is what I assume you were given, but don't forget that when you have a definite integral, and you make a substitution, you must also change the values you are integrating over, according to how you have done your substituition,

 

For example, if in the question given, you were integrating between 0 and 1, these need to change to 1 and 3, (ie if x = 1 originally then since u = 2x + 1, u = 3)

[Johnny5]

This is fine for indefinite integrals' date=' which is what I assume you were given, but don't forget that when you have a definite integral, and you make a substitution, you must also change the values you are integrating over, according to how you have done your substituition,

 

For example, if in the question given, you were integrating between 0 and 1, these need to change to 1 and 3, (ie if x = 1 originally then since u = 2x + 1, u = 3)[/quote']

 

This is an excellent response.

 

Suppose you are faced with the following symbolism:

 

[math] \int_a^b f(x) dx [/math]

 

Translation: The integral from a to b, of f of x, dx.

 

The notation above was an improvement over Newton's, and was first used by 1646-Gottfried Wilhelm von Leibniz-1716. As you can see, Leibniz was born just four years after Sir Isaac Newton was born.

 

1642-Sir Isaac Newton-1727

 

The two men supposedly engaged in a childish battle over who developed calculus.

 

At any rate...

 

A superior notation is:

 

[math] \int_{x=a}^{x=b} f(x) dx [/math]

 

Translation: The integral from x equals a, to x equals b, of f of x, dx.

 

Suppose that the integrand is:

 

[math] 7x^3 -3x^2+4x [/math]

 

And suppose the region that x is changing over, is the interval [a,b]. So this is an indefinite integral you are trying to evaluate. Symbolically you should write this:

 

[math] \int_{x=a}^{x=b} (7x^3 -3x^2+4x ) dx [/math]

 

 

Now, in an integral like this, there is no reason to make a substitution, however, in the following integral a substitution simplifies life a bit:

 

[math] \int_{x=a}^{x=b} (7x^3 -3x^2+4x )(63x^2-18x+12)dx [/math]

 

In order to evaluate the integral above, we could first multiply the two polynomials together, to get a single polynomial of degree five, but that is a dumb way to do it. The right way to do it, would be to make the following substitution:

 

Define the function U(x) as follows:

 

[math] U \equiv 7x^3 -3x^2+4x [/math]

 

Differentiate both sides of the equation above, to obtain:

 

[math] \frac{dU}{dx} = 21x^2 -6x+4 [/math]

 

Now, multiply both sides of the equation above, by dx, to obtain:

 

[math] dU = (21x^2 -6x+4)dx [/math]

 

Now, multiply both sides of the equation above by the number 3, to obtain:

 

[math] 3dU = (63x^2 -18x+12)dx [/math]

 

Now, multiply both sides of the equation above, by U.

 

[math] U3dU = U(63x^2 -18x+12)dx [/math]

 

Now, on the RHS, replace U by its definition, to obtain:

 

[math] 3UdU = (7x^3 -3x^2+4x )(63x^2 -18x+12)dx [/math]

 

Now, integrate both sides of the equation above, from x=a to x=b.

 

[math] \int_{x=a}^{x=b} 3UdU = \int_{x=a}^{x=b} (7x^3 -3x^2+4x )(63x^2 -18x+12)dx [/math]

 

Notice that the RHS is the original integral you started with, but it is also equivalent to the LHS, which looks a whole lot simpler.

 

But, and here is the point, on the LHS, the limits of integration involve the variable x, so they need to be switched to the variable U.

 

Recall the definition of U again...

 

[math] U \equiv 7x^3 -3x^2+4x [/math]

 

The equation above, is a cubic equation, and by a theorem due to Abel, must have three roots. However, trying to find those roots is a difficult task. However, it turns out that you dont really need to change the limits of integration until after the integration has terminated, at which point you can replace U by the equivalent function of x, and then evaluate the final answer using the original limits of integration. Doing things this way leads to the following legitimate lines of work:

 

[math] \int_{x=a}^{x=b} 3UdU = [\frac{3U^2}{2}]_a^b [/math]

 

Now substitute back the original function of x, to get:

 

[math] [\frac{3(7x^3 -3x^2+4x)^2}{2}]_a^b [/math]

 

Now, evalute the expression at the endpoints, and take the difference. That is:

 

[math] [\frac{3(7x^3 -3x^2+4x)^2}{2}]_a^b= [\frac{3(7b^3 -3b^2+4b)^2}{2}]-[\frac{3(7a^3 -3a^2+4a)^2}{2}][/math]

 

Now, by multiple applications of the transitive property of equality, we obtain:

 

 

[math] \int_{x=a}^{x=b} (7x^3 -3x^2+4x )(63x^2-18x+12)dx = [/math]

 

[math][\frac{3(7b^3 -3b^2+4b)^2}{2}]-[\frac{3(7a^3 -3a^2+4a)^2}{2}][/math]

 

Regards

[Dapthar]

Now' date=' multiply both sides of the equation above by the number 3, to obtain:

 

[math'] 3dU = (63x^2 -18x+12)dx [/math]

 

Now, multiply both sides of the equation above, by U.

 

[math] UdU = U(63x^2 -18x+12)dx [/math]

Just a minor correction, but I think that from this point onwards, you substituted [math]3 dU[/math] into the right hand side, but forgot to include the factor of [math]3[/math] on the left hand side.

 

Otherwise, it looked correct to me.

[Johnny5]

Just a minor correction' date=' but I think that from this point onwards, you substituted [math']3 dU[/math] into the right hand side, but forgot to include the factor of [math]3[/math] on the left hand side.

 

Otherwise, it looked correct to me.

 

I will go back and fix it right now, thank you Dapthar.

 

Regards

[scguy]

U really know your stuff johnny5, the last example was very informative.

[Johnny5]

U really know your stuff johnny5, the last example was very informative.

 

You're welcome.

 

Regards

[scguy]

This is fine for indefinite integrals' date=' which is what I assume you were given, but don't forget that when you have a definite integral, and you make a substitution, you must also change the values you are integrating over, according to how you have done your substituition,

 

For example, if in the question given, you were integrating between 0 and 1, these need to change to 1 and 3, (ie if x = 1 originally then since u = 2x + 1, u = 3)[/quote']

 

I know but you can still substitute in the function of x for u ie if u=2x+1 then once the integration is completed the original limits could be utilised once more. What is the point in keeping u in the integrand when the function of x can be substituted back in at the end? or is it just for convenience?

[Dapthar]

What is the point in keeping u in the integrand when the function of x can be substituted back in at the end? or is it just for convenience?

If you change to '[math]u[/math]limits', you don't need to substitute the functions of [math]x[/math] back in at the end, just evaluate the integral as one normally would.