# 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.

## Homework Statement

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.

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?

Mark44
Mentor
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 $\frac{x}{3x}$ 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.

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.

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:

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

Mark44
Mentor
Factoring's not my strong topic:

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

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

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

And from there, just regularly integrate?

Thank you so much Mark

Mark44
Mentor
And from there, just regularly integrate?
Yes.

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.

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

1. Homework Statement [/b]
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.

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.

Mark44
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.

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!

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

Mark44
Mentor
No, you have made a mistake. Show us what you did to get 1/3 for your answer.

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:
HallsofIvy
Homework Helper
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))
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 $(4/1.5)^{1.5}= (8/3)^{1.5}$.

=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".

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".

Thanks once again.

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

Mark44
Mentor
The actual answer is 5/9, which is approximately .55.

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