1/3πIntegrate[(((1/1-24/0.859216(cos(1.45p)+2cos(0.94576p)-3sin(0.94576p)/0.94576))-1)(5/6-Power[p,2]/1.9602+Power[(Power[p,2]-Power[0.99,2]),2]/3.881196Log[0.99+p/0.99-p])Power[(p),2]),{p,0,70}]
If you typed it right, this is what Mathematica says.
Could explain me how you calculated it please? I need to do it with other numbers and different limits as homework
Thank you very much
Click Make Your Own Copy. Open a free Wolfram account and change the values. When done do a shift+enter.
WL is case sensitive, Sin and Cos, not sin and cos, and function arguments are enclosed in [ ], not ( ). Those are the changes Marvin made. However, the expression does not look right
\frac{\int_0^{70} p^2 \left(-\frac{p^2}{1.9602}+\frac{\left(p^2-0.99^2\right)^2 \log \left(\frac{p}{0.99}-p+0.99\right)}{3.8812}+\frac{5}{6}\right) \left(\left(1 \frac{1}{1}-\frac{24 \left(-\frac{3 \sin (0.94576 p)}{0.94576}+2 \cos (0.94576 p)+\cos (1.45 p)\right)}{0.859216}\right)-1\right) \, dp}{3 \pi }
The 1/1 probably means the expression is not correctly parenthesized.
Now that you sent me the expression I see it’s not correct, I had put it online on Wolfram and it looked pretty different from that… I don’t know why it changed so much, I just copied and pasted what was on there.. It’s my first time using this app, is there a way I can write the expression how you sent me into the notebook and not write it as I did so I can check the parenthesis?
Hi Ana Clara,
Are you using WolframAlpha or the Wolfram Cloud? I suggest you use the latter. It is free, with some resource limitations. WolframAlpha uses a different syntax that I am not very familiar with.
A Wolfram Cloud notebook that you can make your own copy of and modify is here.
Thank you very much!