Exploring Integral Factoring: Understanding the Impact on Area

[Luke77]

Hey everybody I was wondering why when you factor an integral, the final answer, or area, is smaller than if you hadn't. Here's an example:
$\int$$\frac{x}{2x^2}$ - $\frac{x}{2x}$ between 1 and 2.

You would factor out $\frac{1}{2}$ and bring it in front of the integral, right? But, my final answer came out to be about .45 and when I graphed the original two lines, it seemed the actual area should have been around 1. Why is this? What is the correct answer?

 

[SammyS]

Hey everybody I was wondering why when you factor an integral, the final answer, or area, is smaller than if you hadn't. Here's an example:
$\int$$\frac{x}{2x^2}$ - $\frac{x}{2x}$ between 1 and 2.

You would factor out $\frac{1}{2}$ and bring it in front of the integral, right? But, my final answer came out to be about .45 and when I graphed the original two lines, it seemed the actual area should have been around 1. Why is this? What is the correct answer?

You must have made an algebra mistake.

Show us your steps each way.

 

[Luke77]

Integrate area below $\frac{x}{2x^2}$ and above $\frac{x}{2x}$ between x=1 and x=2.

$\int$ ($\frac{x}{2x^2}$ - $\frac{x}{2x}$
= .5 $\int$ ($\frac{x}{x^2}$ - 1)
=.5$\int$ (1 - $\frac{x}{x^2}$) because 1 is now above x/x^2
=.5$\int$ (1 - x^-1)
=.5 (x - ($\frac{x}{x^2}$ ^2)/2) between 1 and 2

=.5 ((2 - $\frac{2}{4}$^2)/2) - (1- $\frac{1}{4}$^2)/2)
=.5 (2-.125) - (1-.03125)
=.5 (1.875 - .96875)
=.5 (.90625) = .453125

The graph between the original two lines, however, seems to have a greater area than .45.

 

[LCKurtz]

Integrate area below $\frac{x}{2x^2}$ and above $\frac{x}{2x}$ between x=1 and x=2.

I'm curious where you got this problem. I can't imagine any book phrasing it that way.

$\int$ ($\frac{x}{2x^2}$ - $\frac{x}{2x}$
= .5 $\int$ ($\frac{x}{x^2}$ - 1)
=.5$\int$ (1 - $\frac{x}{x^2}$) because 1 is now above x/x^2
=.5$\int$ (1 - x^-1)
=.5 (x - ($\frac{x}{x^2}$ ^2)/2) between 1 and 2

Your antiderivative of $x^{-1}$ is incorrect. Have you studied the derivative on the natural logarithm function yet?

 

[Luke77]

I'm curious where you got this problem. I can't imagine any book phrasing it that way.
Your antiderivative of $x^{-1}$ is incorrect. Have you studied the derivative on the natural logarithm function yet?

Actually I have:
If f(x) = ln x, then f ' x=$\frac{1}{x}$

$\int$ $\frac{1}{x}$ = ln x
I didnt think to use it though.

 

[SammyS]

Integrate area below $\frac{x}{2x^2}$ and above $\frac{x}{2x}$ between x=1 and x=2.

$\int$ ($\frac{x}{2x^2}$ - $\frac{x}{2x}$
= .5 $\int$ ($\frac{x}{x^2}$ - 1)
=.5$\int$ (1 - $\frac{x}{x^2}$) because 1 is now above x/x^2
=.5$\int$ (1 - x^-1)
=.5 (x - ($\frac{x}{x^2}$ ^2)/2) between 1 and 2   I suppose you mean .5 (x - $\left(\frac{x}{x^2}\right)^2$ /2) 

=.5 ((2 - $\frac{2}{4}$^2)/2) - (1- $\frac{1}{4}$^2)/2)
=.5 (2-.125) - (1-.03125)   This should be .5 (1-.125) - (.5-.03125) in addition to the sign error.
=.5 (1.875 - .96875)
=.5 (.90625) = .453125

The graph between the original two lines, however, seems to have a greater area than .45.

... Also, I made this problem up entirely to purely show my problem with the factorization. Did I make an error, or does factoring an integral change the graph?

Yes, you did make several errors, one of which LCKurtz pointed out.

You have that the anti-derivative of x^-1 is $\displaystyle \frac{1}{2}\left(\frac{x}{x^2}\right)^2\,,$ as near as I can tell. This is equivalent to $\displaystyle \frac{1}{2}x^{-2}\,.$ But the derivative of $\displaystyle \frac{1}{2}x^{-2}\$ is $\displaystyle -x^{-1}\,,$ not $\displaystyle x^{-1}\ .$ So you have a sign error.

After that, you fail to distribute 1/2 of the anti-derivative, when you plug-in the limits of integration.

There may be additional errors in your working of the problem.

Also, in making up the problem, you failed to realize that $\displaystyle \frac{x}{2x^2}$ is below $\displaystyle \frac{x}{2x}$ on the interval, 1 < x < 2 .

 

[Luke77]

Thanks everyone, I guess it was just an error.

 

[Luke77]

Yes, you did make several errors, one of which LCKurtz pointed out.

You have that the anti-derivative of x^-1 is $\displaystyle \frac{1}{2}\left(\frac{x}{x^2}\right)^2\,,$ as near as I can tell. This is equivalent to $\displaystyle \frac{1}{2}x^{-2}\,.$ But the derivative of $\displaystyle \frac{1}{2}x^{-2}\$ is $\displaystyle -x^{-1}\,,$ not $\displaystyle x^{-1}\ .$ So you have a sign error.

After that, you fail to distribute 1/2 of the anti-derivative, when you plug-in the limits of integration.

There may be additional errors in your working of the problem.

Also, in making up the problem, you failed to realize that $\displaystyle \frac{x}{2x^2}$ is below $\displaystyle \frac{x}{2x}$ on the interval, 1 < x < 2 .

You're right about the graph! I entered it totally wrong! Thank you very much.

 

[Luke77]

Hey everybody I was wondering why when you factor an integral, the final answer, or area, is smaller than if you hadn't. Here's an example:
$\int$$\frac{x}{2x^2}$ - $\frac{x}{2x}$ between 1 and 2.

You would factor out $\frac{1}{2}$ and bring it in front of the integral, right? But, my final answer came out to be about .45 and when I graphed the original two lines, it seemed the actual area should have been around 1. Why is this? What is the correct answer?

I figured it all out thanks to Sammy and LCKurtz. Thanks everyone, I had just entered the graph wrong.