another area under the curve qn

Hi I dont know where I went.wrong again! Really would appreciate the help! For y=rand, over the interval x=0 to pi/4, I am supposed to find the area between the x-axis,x=pi/4. I got my area to be -ln|sqrt(2)/(2)|by integrating tanx from 0 to pi/4, but it is wrong!

Re: another area under the curve qn

Sorry for the typo! It is y=tanx, that's the graph given. Thank you!

Re: another area under the curve qn

I was wondering! Who told you it was wrong? Another of those "computer" systems where you have to give the answer in exactly the right form? I hate those things! $\displaystyle -ln(\sqrt{2}/2)$ is perfectly correct but can be simplified. $\displaystyle -ln(\sqrt{2}/2)= ln(2/\sqrt{2})= ln(\sqrt{2})= ln(2)/2$. Try one of those forms.