# Integrating the area under a curve

1. Jun 22, 2012

### Luke77

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 $\frac{x}{3x}$ and above $\frac{x}{3x^.5}$ between x=1 and x=4.
Same as: $\int$ $\frac{x}{3x}$ - $\frac{x}{3x^.5}$ 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?

2. Jun 22, 2012

### Staff: Mentor

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. Jun 22, 2012

### Luke77

Factoring's not my strong topic:

$\frac{1}{3}$$\int$ 1 - $\frac{x}{x^.5}$ dx between 1 and 4

4. Jun 22, 2012

### Staff: Mentor

$$\frac{x^1}{x^{.5}} = x^{.5}$$

So your integral should look like this:
$$\frac{1}{3}\int_1^4 1 - x^{1/2}~dx$$

5. Jun 22, 2012

### Luke77

And from there, just regularly integrate?

6. Jun 22, 2012

### Luke77

Thank you so much Mark

7. Jun 22, 2012

### Staff: Mentor

Yes.

8. Jun 22, 2012

### Luke77

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. Jun 22, 2012

### Luke77

Specifically, -1.667

10. Jun 22, 2012

### algebrat

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. Jun 23, 2012

### Staff: Mentor

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. Jun 23, 2012

### Luke77

Thank you very much Mark!

13. Jun 23, 2012

### Luke77

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

14. Jun 23, 2012

### Staff: Mentor

No, you have made a mistake. Show us what you did to get 1/3 for your answer.

15. Jun 23, 2012

### Luke77

I see what I did. Let me try it again:

$\frac{1}{3}$$\int$ (x^.5 - 1)dx

= $\frac{1}{3}$ (($\frac{x}{1.5}$ ^1.5) - x)) for 1 through 4

= $\frac{1}{3}$ ($\frac{4}{1.5}$ ^1.5) - 4) - ($\frac{1}{1.5}$ ^1.5 - 1))

=$\frac{1}{3}$ (((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: Jun 23, 2012
16. Jun 23, 2012

### HallsofIvy

Staff Emeritus
Here is your error. The integral of $x^{.5}$ is $x^{1.5}/1.5$ NOT $(x/1.5)^{1.5}$ so that, at x= 4, it is $4^{1.5}/1.5= 8/1.5= 16/3$, NOT $(4/1.5)^{1.5}= (8/3)^{1.5}$.

By the way, it is a lot easier to put entire formulas in LaTeX, not just "bits and pieces".

17. Jun 23, 2012

### Luke77

Thanks once again.

18. Jun 23, 2012

### Luke77

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

19. Jun 23, 2012

### Staff: Mentor

The actual answer is 5/9, which is approximately .55.

20. Jun 23, 2012

### Luke77

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