1. ## integration

I am having trouble solving this integration...I've tried about 10 times now and I'm getting the same answer but it's still wrong. help!

Integral from 0 to 300 of: (80/(1+r))*(2*pi*r)

thanks!

2. Rewrite as $\displaystyle 160{\pi}\left[\int_{0}^{300}dr-\int_{0}^{300}\frac{1}{r+1}dr\right]$

Now, it's rather straightforward.

3. ## same problem..help!

I did that...then I used substitution to solve and I get:

160pi(300-(ln(301)-ln(1)))

but that's still the wrong answer...am I doing something wrong?

4. What makes you think it's wrong?

$\displaystyle 160\pi \left[r\big|_{0}^{300} - ln(r + 1) \big|_{0}^{300}\right]$

$\displaystyle = 160 \pi \left[(300 - 0) - (\ln301 - \ln1)\right]$

5. ## same problem...still: integration

we turn in our answers into a website.. I turned in that answer and it says its wrong. I have more tries left but I don't know what I'm doing wrong

6. Originally Posted by amaya
I did that...then I used substitution to solve and I get:

160pi(300-(ln(301)-ln(1)))

but that's still the wrong answer...am I doing something wrong?
You do realise that ln(1) = 0 ....? So maybe the answer has to be submitted as

$\displaystyle 160 \pi (300 - \ln 301)$ .....