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Definite Integration using U-Substitution

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  • #1
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Hi,
I'm I on the right track here?

Homework Statement


Evaluate the definite integral, if it exists


Homework Equations


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


The Attempt at a Solution


Let u = 2x-3

du = 2dx

1/2du = dx

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

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

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

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

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

(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)]
 

Answers and Replies

  • #2
33,501
5,190
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


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


The Attempt at a Solution


Let u = 2x-3

du = 2dx

1/2du = dx

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

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

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

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

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

(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?
[tex]\int_0^{2}\frac{dx}{(2x-3)^2}[/tex]

It's a major problem that the integrand is undefined at x = 1.5.
 
  • #3
48
0
Is this the integral?
[tex]\int_0^{2}\frac{dx}{(2x-3)^2}[/tex]
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?
 
Last edited:
  • #4
33,501
5,190
Just say that this definite integral doesn't exist because the integrand is undefined at x = 1.5.
 
  • #5
48
0
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.
 
  • #6
48
0
@Mark44,

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

Homework Statement


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

and they gave the answer as [tex]\frac{1}{14}[/tex]

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
Dick
Science Advisor
Homework Helper
26,258
618
How is 1/(3-5x)^2 in any way undefined on the interval [1,2]? Just integrate it.
 
Last edited:
  • #8
48
0
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
 

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