How Do You Solve These Integral Problems Using U-Substitution?

  • Thread starter Thread starter EasyStyle4747
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary

Homework Help Overview

The discussion revolves around solving integral problems using u-substitution, specifically focusing on integrals involving exponential functions and square roots.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different substitutions for integrals, questioning the correct interpretation of expressions involving square roots and exponential functions. There is a discussion about setting u as either the expression inside the square root or the exponent itself.

Discussion Status

Some participants have provided guidance on potential substitutions and clarified misunderstandings about the expressions involved. There is an acknowledgment of different approaches to the problems, but no consensus has been reached on the first integral.

Contextual Notes

Participants are navigating issues related to the clarity of the problem statements, particularly with the notation of the integrals and the definitions of variables used in substitutions.

EasyStyle4747
Messages
24
Reaction score
0
Ok, ill try to explain what i did:

1.
integral of [square root of( e^t-3)dt]

Sorry i didn't know how to do the square roots symbol and the integral symbol. Anyways, I tried to set e^t-3 as u and then got dt=du/e^t. Then I plugged in dt and couldn't go on after that.

2. integral of x^2[e^(x^3)]dx

so uh... u substitution? but then what?

Ok basically, i think i don't know what your supposed do when u need to do a u-sub with e^something. Your help would be most appreciated.
 
Physics news on Phys.org
1. I don't have any problem with the square root sign but is that e^(t-3) or
(e^t)-3?
If it is e^(t-3) then sqrt(e^(t-3))= (e^(t-3))^(1/2)= e^((t-3)/2). Let u= (t-3)/2 so
du= (1/2)dt or dt= 2du. The integral becomes 2 integral e^u du.

2. Yes, a substitution- seeing that "complicated" x^3 in the exponent and x^2 multiplying, you should immediately think of u= x^3 (NOT u= e^something- the e is not the problem!). Then du= 3 x^2 dx or (1/3)du= x^2 dx so the integral becomes
(1/3) integral e^u du.
 
Well, thinking of [tex]u = e^{x^3}[/tex] wouldn't hurt in this case regardless :smile:
 
thnx, i see how the second one works now.

For the first one, its actually square root of (e^t)-3. Sorry i didnt make that clear.
 
Then make the substitution
[tex]e^{t}=u[/tex]
,then the substitution
[tex]u=v^{2}+3[/tex]

Daniel.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
2K
Replies
11
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
30
Views
2K
Replies
14
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 69 ·
3
Replies
69
Views
7K