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

Integrating the area under a curve

  • Thread starter Luke77
  • Start date
  • #1
42
0
Hey everyone, I had a recent post similar to this one, and everyone may have not understood it because I didn't use LaTeX, so here it is.

Homework Statement


Integrate the area under [itex]\frac{x}{3x}[/itex] and above [itex]\frac{x}{3x^.5}[/itex] between x=1 and x=4.
Same as: [itex]\int[/itex] [itex]\frac{x}{3x}[/itex] - [itex]\frac{x}{3x^.5}[/itex] dx between x=1 and x=4.

Homework Equations


See above

The Attempt at a Solution


I would integrate each individual fraction by raising each by 1 power, and dividing the coefficient by that new exponent. Each time that I try this, no matter how I simplify it, the answer turns out to be way too small.

What should I try next?
 

Answers and Replies

  • #2
33,106
4,802
Hey everyone, I had a recent post similar to this one, and everyone may have not understood it because I didn't use LaTeX, so here it is.

Homework Statement


Integrate the area under [itex]\frac{x}{3x}[/itex] and above [itex]\frac{x}{3x^.5}[/itex] between x=1 and x=4.
Same as: [itex]\int[/itex] [itex]\frac{x}{3x}[/itex] - [itex]\frac{x}{3x^.5}[/itex] dx between x=1 and x=4.

Homework Equations


See above

The Attempt at a Solution


I would integrate each individual fraction by raising each by 1 power, and dividing the coefficient by that new exponent. Each time that I try this, no matter how I simplify it, the answer turns out to be way too small.

What should I try next?
Before integrating, you really should simplify the fractions. Both fractions have a factor of 1/3, so that should be factored out and brought out front of your integral.

After simplification, what does your integral look like?

Tip: itex is used for inline expressions, but it makes fractions and integrals very small. For those kinds of things use tex instead of itex.
 
  • #3
42
0
Before integrating, you really should simplify the fractions. Both fractions have a factor of 1/3, so that should be factored out and brought out front of your integral.

After simplification, what does your integral look like?

Tip: itex is used for inline expressions, but it makes fractions and integrals very small. For those kinds of things use tex instead of itex.
Factoring's not my strong topic:

[itex]\frac{1}{3}[/itex][itex]\int[/itex] 1 - [itex]\frac{x}{x^.5}[/itex] dx between 1 and 4
 
  • #4
33,106
4,802
Factoring's not my strong topic:

[itex]\frac{1}{3}[/itex][itex]\int[/itex] 1 - [itex]\frac{x}{x^{.5}}[/itex] dx between 1 and 4
[tex]\frac{x^1}{x^{.5}} = x^{.5}[/tex]

So your integral should look like this:
$$ \frac{1}{3}\int_1^4 1 - x^{1/2}~dx$$
 
  • #5
42
0
And from there, just regularly integrate?
 
  • #6
42
0
Thank you so much Mark
 
  • #7
33,106
4,802
And from there, just regularly integrate?
Yes.
 
  • #8
42
0
Actually, one last question. Why do when I compute it, it is a negative number? Is there any rules I should be aware of... I just started learning about Calculus and the definite integral.
 
  • #9
42
0
Actually, one last question. Why do when I compute it, it is a negative number? Is there any rules I should be aware of... I just started learning about Calculus and the definite integral.
Specifically, -1.667
 
  • #10
428
1
1. Homework Statement [/b]
Integrate the area under [itex]\frac{x}{3x}[/itex] and above [itex]\frac{x}{3x^.5}[/itex] between x=1 and x=4.
Same as: [itex]\int[/itex] [itex]\frac{x}{3x}[/itex] - [itex]\frac{x}{3x^.5}[/itex] dx between x=1 and x=4.

could be a problem regarding the area below 1/3 and above √x/3, but as x varies from 1 to 4, √x/3 varies from 1/3 to 2/3, so the setup doesn't make much sense.
 
  • #11
33,106
4,802
The integral should be
## \frac{1}{3}\int_1^4 x^{1/2} - 1 ~dx##

The graph of y = (1/3)x1/2 lies above the graph of y = 1/3 on the interval [1, 4]. When you measure a distance, you need to subtract the smaller number from the larger number, otherwise you'll get a negative value for the length of the typical area element.
 
  • #12
42
0
The integral should be
## \frac{1}{3}\int_1^4 x^{1/2} - 1 ~dx##

The graph of y = (1/3)x1/2 lies above the graph of y = 1/3 on the interval [1, 4]. When you measure a distance, you need to subtract the smaller number from the larger number, otherwise you'll get a negative value for the length of the typical area element.
Thank you very much Mark!
 
  • #13
42
0
My final answer came out to be 1/3, does that sound right?
 
  • #14
33,106
4,802
No, you have made a mistake. Show us what you did to get 1/3 for your answer.
 
  • #15
42
0
I see what I did. Let me try it again:

[itex]\frac{1}{3} [/itex][itex]\int[/itex] (x^.5 - 1)dx

= [itex]\frac{1}{3}[/itex] (([itex]\frac{x}{1.5}[/itex] ^1.5) - x)) for 1 through 4

= [itex]\frac{1}{3}[/itex] ([itex]\frac{4}{1.5}[/itex] ^1.5) - 4) - ([itex]\frac{1}{1.5}[/itex] ^1.5 - 1))

=[itex]\frac{1}{3}[/itex] (((2.66^1.5) - 4) - ((.66^1.5) - 1))

=1/3((4.338-4)- (.536 - 1)) = 1/3(.338 + .464) = 1/3(.802) = .267

???
 
Last edited:
  • #16
HallsofIvy
Science Advisor
Homework Helper
41,777
911
I see what I did. Let me try it again:

[itex]\frac{1}{3} [/itex][itex]\int[/itex] (x^.5 - 1)dx

= [itex]\frac{1}{3}[/itex] (([itex]\frac{x}{1.5}[/itex] ^1.5) - x)) for 1 through 4


= [itex]\frac{1}{3}[/itex] ([itex]\frac{4}{1.5}[/itex] ^1.5) - 4) - ([itex]\frac{1}{1.5}[/itex] ^1.5 - 1))

=[itex]\frac{1}{3}[/itex] (((2.66^1.5) - 4) - ((.66^1.5) - 1))
Here is your error. The integral of [itex]x^{.5}[/itex] is [itex]x^{1.5}/1.5[/itex] NOT [itex](x/1.5)^{1.5}[/itex] so that, at x= 4, it is [itex]4^{1.5}/1.5= 8/1.5= 16/3[/itex], NOT [itex](4/1.5)^{1.5}= (8/3)^{1.5}[/itex].

=1/3((4.338-4)- (.536 - 1)) = 1/3(.338 + .464) = 1/3(.802) = .267

???
By the way, it is a lot easier to put entire formulas in LaTeX, not just "bits and pieces".
 
  • #17
42
0
Here is your error. The integral of [itex]x^{.5}[/itex] is [itex]x^{1.5}/1.5[/itex] NOT [itex](x/1.5)^{1.5}[/itex] so that, at x= 4, it is [itex]4^{1.5}/1.5= 8/1.5= 16/3[/itex], NOT [itex](4/1.5)^{1.5}= (8/3)^{1.5}[/itex].


By the way, it is a lot easier to put entire formulas in LaTeX, not just "bits and pieces".
Thanks once again.
 
  • #18
42
0
Now I get about .55. This doesn't seem right, again.
 
  • #19
33,106
4,802
The actual answer is 5/9, which is approximately .55.
 
  • #20
42
0
When I graph the original two lines it seems to have a much larger area, but an answer is an answer. Thanks.
 

Related Threads for: Integrating the area under a curve

  • Last Post
Replies
4
Views
1K
Replies
3
Views
2K
Replies
2
Views
866
Replies
1
Views
1K
Replies
0
Views
3K
  • Last Post
Replies
3
Views
4K
Replies
3
Views
2K
Replies
3
Views
1K
Top