# Factoring an Integral

1. Jun 24, 2012

### 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?

2. Jun 24, 2012

### SammyS

Staff Emeritus
You must have made an algebra mistake.

Show us your steps each way.

3. Jun 24, 2012

### 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.

4. Jun 24, 2012

### LCKurtz

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?

5. Jun 24, 2012

### Luke77

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.

Last edited: Jun 24, 2012
6. Jun 24, 2012

### SammyS

Staff Emeritus
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.

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 .

7. Jun 24, 2012

### Luke77

Thanks everyone, I guess it was just an error.

8. Jun 24, 2012

### Luke77

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

9. Jun 24, 2012

### Luke77

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