Need help beginning to solve this Integral

  • Thread starter BlackMamba
  • Start date
  • #1
187
0
Hello,

I have this integral that is stumping me. Once I get the basic start then I know I can finish the rest myself it's just getting started that's stumping me.

Here is the problem:

[itex]\int_{0}^{4} {e^\sqrt{x}} dx[/itex]

First this doesn't appear to me to be an improper integral. So I should just go ahead and solve as normal. But I'm told to find if the integral converges or diverges and evaluate it if it converges. So even though this integral is not an improper integral, I assume I can still apply the same rules.

To begin however, is where I'm getting stuck.

Substitution doesn't appear to help. if I use [itex]u = x^{1/2}[/itex]


I think I'm making this harder than it needs to be...

Any and all input would be greatly appreciated!!
 
Last edited:

Answers and Replies

  • #2
Astronuc
Staff Emeritus
Science Advisor
19,797
3,238
[itex]u^2 = x[/itex], then

[itex]d({u^2})\,=\,2u du[/itex]
 
  • #3
187
0
Thank you

I understand how you got that.

[itex]u=\sqrt{x}[/itex] then,
[itex]u^2 = x[/itex]

However, I'm getting lost as to how I'm supposed to substitute that into the original equation.

I think the part that is throwing me is the [itex]d(u^2) = 2udu[/itex]

This is as far as I'm getting:

[itex]\int_{0}^{4} {e^\sqrt{u^2}} du[/itex]

[itex]\int_{0}^{4} e^u du[/itex]

Am I just adding the [itex]2u[/itex] and then do integration by parts?
 
Last edited:
  • #4
809
0
Thank you

I understand how you got that.

[itex]u=\sqrt{x}[/itex] then,
[itex]u^2 = x[/itex]

However, I'm getting lost as to how I'm supposed to substitute that into the original equation.

I think the part that is throwing me is the [itex]d(u^2) = 2udu[/itex]

This is as far as I'm getting:

[itex]\int_{0}^{4} {e^\sqrt{u^2}} du[/itex]

[itex]\int_{0}^{4} e^u du[/itex]

Am I just adding the [itex]2u[/itex] and then do integration by parts?



Using the u-substitution that astronuc showed you, you then have:
[tex] \int e^{\sqrt{x}}dx = \int e^{\sqrt{u^2}}2u \,du[/tex]

Your idea about 'adding' 2u and applying integration by parts is correct.
 
  • #5
187
0
Thanks FrogPad and thanks again to Astronuc.

The help was greatly appreciated!
 
  • #6
dextercioby
Science Advisor
Homework Helper
Insights Author
13,131
684
The new limits of integration will be 0 and 16.

Daniel.
 

Related Threads on Need help beginning to solve this Integral

  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
6
Views
2K
Replies
1
Views
542
  • Last Post
Replies
2
Views
655
  • Last Post
Replies
9
Views
955
  • Last Post
Replies
6
Views
2K
Replies
3
Views
570
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
7
Views
2K
Top