# Thread: exponential integral problem

1. ## 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.

2. 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.

3. 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

4. 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.