Integrating the area under a curve

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.

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
See above

3. 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?

 

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

 

And from there, just regularly integrate?

 

Thank you so much Mark

 

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

 

Thank you very much Mark!

 

My final answer came out to be 1/3, does that sound right?

 

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

???

 

Thanks once again.

 

Now I get about .55. This doesn't seem right, again.

 

When I graph the original two lines it seems to have a much larger area, but an answer is an answer. Thanks.