# Homework Help: Definite Integration using U-Substitution

1. Apr 3, 2010

### 01010011

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

1. The problem statement, all variables and given/known data
Evaluate the definite integral, if it exists

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

3. 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)]

2. Apr 3, 2010

### Staff: Mentor

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.
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.

3. Apr 3, 2010

### 01010011

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?

Last edited: Apr 3, 2010
4. Apr 3, 2010

### Staff: Mentor

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

5. Apr 3, 2010

### 01010011

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

6. Apr 3, 2010

### 01010011

@Mark44,

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

1. The problem statement, all variables and given/known data
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?

7. Apr 3, 2010

### Dick

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

Last edited: Apr 3, 2010
8. Apr 4, 2010

### 01010011

Ok, I think I finally get it now, thanks