# Silly Integration Question

1. Dec 20, 2008

### flying_young

Dear all,

Sorry for the silly integration question. I haven't done calculus in ages and my memory has got really rusty! ='(

The integral question is as follows:

$$\int\ (X_1-\frac{X_2}{2})^2(\frac{1}{X_2}) dX_1$$

Sorry, I could not figure out how to insert the limits on my intergral. It is from 0 to $$X_2$$.

= $$\int\ (\frac{X_1^2}{X_2}-X_1 + {\frac{X^2}{4}) dX_1$$

Which, after integrating, becomes

$$(\frac{X_2^2}{3}-1/2X_2^2 + X_2^2/8 ]$$

In the end, I get -X2^2/24, but the answer in the book is X2^2/12. What am I doing wrong? Thanks in advance for your help!

Last edited: Dec 20, 2008
2. Dec 20, 2008

### NoMoreExams

So let me rewrite what you have:

$$\int_{0}^{X_2} \left( X_1 - \frac{X_2}{2}\right)^2\frac{1}{X_2} dX_1$$

Note that you can take $$\frac{1}{X_2}$$ outside of the integral since it's a constant w.r.t. $$X_1$$ (as well as taking out the $$\frac{1}{4}$$)

So expanding out the square we get:

$$\frac{1}{4X_2} \int_{0}^{X_2} 4X_{1}^{2} - 4X_{1}X_{2} + X_{2}^{2} dX_{1} = \frac{1}{4X_2} \left( \frac{4X_{1}^{3}}{3} - 2X_{1}^{2}X_{2} + X_{1}X_{2}^{2} \right)_{0}^{X_2}$$

And finally plugging in the limits we get

$$\frac{1}{4X_2} \left(\frac{4}{3}X_{2}^{3} - 2X_{2}^{3} + X_{2}^{3} \right) = \frac{1}{12} X_{2}^{2}$$

3. Dec 20, 2008

### NoMoreExams

Just a side note the notation that you can with limits is

\int_{a}^{b} f(x) dx or \sum_{i=0}^{n} p(x)

4. Dec 20, 2008

### HallsofIvy

Staff Emeritus
NoMoreExams, that was nicely done but after making the point that X2 was a constant and that you can multiply the square out (or make the substitution v= X1- X2/2) it would have been better to let flying young finish the problem himself.

5. Dec 20, 2008

### NoMoreExams

Good point, I will do so in the future.

6. Dec 21, 2008

### flying_young

Thank you, NoMoreExams! I finally figured out what I did wrong - a silly integration mistake! Thanks again! You have been an immense help.

7. Dec 21, 2008

### flying_young

I know very much to treat X2 as a constant, but if NoMoreExams hasn't written out his steps, I wouldn't have figured out where I have gone wrong in the midst of calculations. I do understand from your stance that simply giving out the answer would do one no good. Perhaps in the future, I will be sure to write out my steps and have the pros here point out what I have done wrong. Thank you!