# exponential integral problem

• Jan 29th 2009, 04:37 PM
mech.engineer.major
exponential integral problem
What is the integral from -2 to 2 of (3u+1)^2 du?

Here is my work so far:

((3u+1)^3)/3

((3(2)+1)^3)/3 - ((3(-2)+1)^3)/3

= 343/3 + 125/3

= 468/3

= 156

I know that is wrong but the book says it is 52. I realized if I divide my final answer again I will get it so I was thinking that I multiplied by 3 one time too many or something. That or I made an error somewhere else.

Thanks for the help in advance.
• Jan 29th 2009, 04:52 PM
Chop Suey
You forgot something. If you were to sub $\displaystyle z = 3u+1 \implies dz = 3~du$, the integral becomes:

$\displaystyle \frac{1}{3}\int z^2~du = \frac{z^3}{9} + C$

Now evaluate the definite integral.

Of course, this can be tackled very easily without resorting to subs. But I had to use it here for clarity.
• Jan 29th 2009, 04:53 PM
Equals
Excuse me if you don't understand me, but this is my first time posting on these forums.

Your problem is finding your anti-derivative. To check yourself, try taking the derivative back. If you use the chain rule correctly, you will see that it is 3 times more than you wanted. So to fix this, just divide by another 3. this will leave you with

((3u+1)^3)/9. From here, just plug in your numbers(2 and -2)

When you plug them in you get 343/9 + 125/9. which equals 52.

Hope this helped,

-Equals
• Jan 29th 2009, 05:10 PM
mech.engineer.major
Hey thanks for the input Chop Suey and Equals. Yeah I forgot to balance it out with the 1/3 on the outside. One of those slap your forehead "oh duh!" things.