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!
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Rewrite as $\displaystyle 160{\pi}\left[\int_{0}^{300}dr-\int_{0}^{300}\frac{1}{r+1}dr\right]$ Now, it's rather straightforward.
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?
What makes you think it's wrong? $\displaystyle 160\pi \left[r\big|_{0}^{300} - ln(r + 1) \big|_{0}^{300}\right]$ $\displaystyle = 160 \pi \left[(300 - 0) - (\ln301 - \ln1)\right]$
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
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 $\displaystyle 160 \pi (300 - \ln 301)$ .....
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