• Support PF! Buy your school textbooks, materials and every day products Here!

Simple or double integration?

  • Thread starter kasse
  • Start date
  • #1
366
0
If I want to compute the area of the region bounded by the graphs y=x and y=x^2-2x, can I simply compute the integral of (x-(x^2-2x)) dx from x=0 to x=3, or do I have to use double integration?

In any case, why?
 

Answers and Replies

  • #2
529
1
For two curves, f(x) and g(x) the area of the element between x and x+dx is simply |f(x)-g(x)|dx. The magnitude sign is there only if you don't want the area to go negative when the curves cross.

So the total area = Int |f(x)-g(x)|dx
 
  • #3
2,063
2
"Ordinary" integration will do.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,805
932
I'm a bit confused by the question. Since you know about "double integration", you must have been studying calculus for some time- and should have learned to find the "area between two curves" quite a while ago!

Here, y= x is above y= x2- 2x for x between 0 and 3 so the area is given by
[tex]\int_0^3 (x- (x^2- 2x))dx= \int_0^3 (3x- x^2)dx[/tex]

Of course, you could use "double integration" with '1' as integrand:
[tex]\int_{x=0}^3 \int_{y= x^2- 2x}^x dy dx[/tex]
but after the first integration, it reduces to the single integral form.
 

Related Threads on Simple or double integration?

  • Last Post
Replies
4
Views
5K
Replies
4
Views
860
  • Last Post
Replies
3
Views
898
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
8
Views
4K
  • Last Post
Replies
21
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
9
Views
681
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
1K
Top