Integrating the area under a curve

[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.

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.

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?

 

[Mark44]

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.

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.

 

[Luke77]

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]

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:
$$ \frac{1}{3}\int_1^4 1 - x^{1/2}~dx$$

 

[Luke77]

And from there, just regularly integrate?

 

[Luke77]

Thank you so much Mark

 

[Mark44]

And from there, just regularly integrate?

Yes.

 

[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.

 

[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.

Specifically, -1.667

 

[algebrat]

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]

The integral should be
$\frac{1}{3}\int_1^4 x^{1/2} - 1 ~dx$

The graph of y = (1/3)x^1/2 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.

 

[Luke77]

The integral should be
$\frac{1}{3}\int_1^4 x^{1/2} - 1 ~dx$

The graph of y = (1/3)x^1/2 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!

 

[Luke77]

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

 

[Mark44]

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

 

[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

?

 

[HallsofIvy]

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".

 

[Luke77]

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.

 

[Luke77]

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

 

[Mark44]

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

 

[Luke77]

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