Definite Integration

  • #1
650
1
[tex] f(x)=x^2+\int_0^x e^{-t}f(x-t)dt[/tex] ........(I)

Find f(x)

Okay what i did:

For
[tex]\int_0^x e^{-t}f(x-t)dt[/tex]

I substituted h=x-t =>dh=-dt
so [tex]\int_0^x e^{h-x}f(h)dh[/tex]

Now i differentiated (I)
so i got
f'(x)=2x+f(x) after solvin this by integrating factor method i got different results which involve ex

But i had to prove f(x)=x2+x3/3
 

Answers and Replies

  • #2
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
The fundamental theorem of calculus says

[tex]
\frac{d}{dx} \int_a^x f(h) \, dh = f(x)
[/tex]

However, when you differentiate, you don't have something in this form! You have the form

[tex]
\frac{d}{dx} \int_a^x f(\mathb{x,} h) \, dh
[/tex]

Fortunately, in this case, you don't need the full messy version; you can factor [itex]e^{-x}[/itex] out of the integrand and then differentiate normally.
 
  • #3
650
1
Thanks Hurkyl I got corrected

now i have f'(x)=x2+2x

which gives me the result

Still i want to know how would u have approached the pro
 

Related Threads on Definite Integration

Replies
11
Views
926
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
6
Views
5K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
4K
Top