Definite Integration using U-Substitution

[01010011]

Hi,
I'm I on the right track here?

Homework Statement

Evaluate the definite integral, if it exists

Homework Equations

\int^{2}_{0}\stackrel{dx/}{(2x-3)^2}

The Attempt at a Solution

Let u = 2x-3

du = 2dx

1/2du = dx

1/2\int^{2}_{0}\stackrel{du/}{u^2}

going back to our original u substitution: u = 2x - 3

= 2(0) - 3 = -3
= 2(2) - 3 = 1

1/2\int^{-3}_{1}\stackrel{du/}{u^2}

1/2\int^{-3}_{1}\stackrel{du/}{(u^3)/3}

(1/6) x du/(u)^3

[(1/6) x du/(1)^3] - [(1/6) x du/(-3)^3]

[(1/6) x du] - [(1/6) x du/(-27)]

 

[Mark44]

Hi,
I'm I on the right track here?

No, if I understand what you're trying to integrate. That stackrel command is not helpful. frac{}{} is what you should use for rational expressions.

Homework Statement

Evaluate the definite integral, if it exists

Homework Equations

\int^{2}_{0}\stackrel{dx/}{(2x-3)^2}

The Attempt at a Solution

Let u = 2x-3

du = 2dx

1/2du = dx

1/2\int^{2}_{0}\stackrel{du/}{u^2}

going back to our original u substitution: u = 2x - 3

= 2(0) - 3 = -3
= 2(2) - 3 = 1

1/2\int^{-3}_{1}\stackrel{du/}{u^2}

1/2\int^{-3}_{1}\stackrel{du/}{(u^3)/3}

(1/6) x du/(u)^3

[(1/6) x du/(1)^3] - [(1/6) x du/(-3)^3]

[(1/6) x du] - [(1/6) x du/(-27)]

Is this the integral?
\int_0^{2}\frac{dx}{(2x-3)^2}

It's a major problem that the integrand is undefined at x = 1.5.

 

[01010011]

Is this the integral?
\int_0^{2}\frac{dx}{(2x-3)^2}
It's a major problem that the integrand is undefined at x = 1.5.

Yes this is the correct problem I have to work out.
Ok, correct me if I am wrong but once there is a possibility of getting an undefined answer (because of the possibility of dividing by 0) then I can easily say that the difinite integral does not exist because it can't be defined if x = 1.5?

So therefore, if x is not 1.5, then the definite integral does exist. So how do I proceed with this question then? Where did I go wrong with my attempted answer?

 

[Mark44]

Just say that this definite integral doesn't exist because the integrand is undefined at x = 1.5.

 

[01010011]

Just say that this definite integral doesn't exist because the integrand is undefined at x = 1.5.

Ok, anytime I have a quotient like that, I'll check for this, thanks.

 

[01010011]

@Mark44,

On second thought, I was looking at the following question in the book:

Homework Statement

Evaluate \int\stackrel{2}{1}\frac{dx}{(3-5x)^{2}}

and they gave the answer as \frac{1}{14}

My question is, isn't this the same situation where some number would make this also undefined? Or is the question asking something different?

 

[Dick]

How is 1/(3-5x)^2 in any way undefined on the interval [1,2]? Just integrate it.

 

[01010011]

How is 1/(3-5x)^2 in any way undefined on the interval [1,2]? Just integrate it.

Ok, I think I finally get it now, thanks